# Highest Cusped Waves for the Fractional KdV Equations

Joel Dahne

###### Abstract

In this paper we prove the existence of highest, cusped, traveling wave
solutions for the fractional KdV equations $f_{t}+ff_{x}=|D|^{\alpha}f_{x}$
for all $\alpha\in(-1,0)$ and give their exact leading asymptotic behavior at
zero. The proof combines careful asymptotic analysis and a computer-assisted
approach.

##  1 Introduction

This paper is concerned with the existence and regularity of highest, cusped,
periodic traveling-wave solutions to the fractional Korteweg-de Vries (KdV)
equations, in the periodic setting given by

$f_{t}+ff_{x}=|D|^{\alpha}f_{x},\quad\text{ for
}(x,t)\in\mathbb{T}\times\mathbb{R}.$ (1)

Here $|D|^{\alpha}$ is the Fourier multiplier operator given by

$\widehat{|D|^{\alpha}f}(\xi)=|\xi|^{\alpha}\widehat{f}(\xi)$

where the parameter $\alpha$ may in general take any real value. It can serve
as a model for investigating the balance between nonlinear and dispersive
effects [47]. For $\alpha=2$ and $\alpha=1$ it reduces to the classical KdV
and Benjamin-Ono equations, for $\alpha=-2$ one gets the reduced Ostrovsky
equation. For $\alpha=-1$ it reduces to the Burgers-Hilbert equation [31]

$f_{t}+ff_{x}=\mathbf{H}[f],\quad\text{ for
}(x,t)\in\mathbb{T}\times\mathbb{R}.$ (2)

Here $\mathbf{H}$ is the Hilbert transform which, for
$f:\mathbb{T}\to\mathbb{R}$, is defined by

$\mathbf{H}[f](x)=\frac{1}{2\pi}p.v.\int_{-\pi}^{\pi}\cot\left(\frac{x-y}{2}\right)f(y)\
dy,\quad\widehat{\mathbf{H}[f]}(k)=-i\operatorname{sgn}(k)\widehat{f}(k).$

The results in this paper, see Theorem 1.1, are for $\alpha\in(-1,0)$, though
the analysis of the Burgers-Hilbert case, $\alpha=-1$, is also important for
the proof.

For $\alpha\in(-1,0)$ and small initial data in $H^{N}(\mathbb{R})$ with
$N\geq 3$ estimates for the life span were proved by Ehrnström and Wang [21],
see also [32] and [33] for the Burgers-Hilbert equation. Well-posedness in
$H^{s}(\mathbb{R})$ with $s>3/2$ was established by Riaño [52].

For $\alpha\in(-1,0)$ the equation exhibits finite time blow [8, 34]. The
characterization as wave breaking (i.e. the functions stay bounded, but its
gradient blows up) was proved for $\alpha\in(-1,-1/3)$ by Hur and Tao [36, 35]
and for $\alpha\in(-1,0)$ by Oh and Pasqualotto [50]. See [45] for a numerical
study and [13] where they give a precise characterization of a chock forming
in finite time for $0<\alpha<\frac{1}{3}$.

The study of traveling waves is an important topic in fluid dynamics, see e.g.
[29] for a recent overview of traveling water waves. The traveling wave
assumption $f(x,t)=\varphi(x-ct)$, where $c>0$ denotes the wave speed, gives
us

$-c\varphi^{\prime}+\varphi\varphi^{\prime}=|D|^{\alpha}\varphi^{\prime}.$ (3)

For the fractional KdV equation there is a branch of even, zero-mean,
$2\pi$-periodic, smooth traveling wave solutions bifurcating from constant
solutions. For $\alpha<-1$ and $\alpha\in(-1,0)$ this branch has been studied
and proved to end in a highest cusped wave, for $\alpha<-1$ by Bruell and
Dhara [6] and for $\alpha\in(-1,0)$ by Hildrum and Xue [30]. For $\alpha=-1$
the branch has been studied by Castro, Córdoba and Zheng [9].

The notion of a highest traveling wave goes back to Stokes. For the free
boundary Euler equation Stokes argued that if there exists a singular solution
with a steady profile it must have an interior angle of $120^{\circ}$ at the
crest [53]. This is known as the Stokes conjecture and was proved in 1982 [1].
For the Whitham equation [56] the existence of a highest cusped traveling wave
was conjectured by Whitham in [55]. Its existence, together with its $C^{1/2}$
regularity, was proved by Ehrnström and Wahlén [20]. For the family of
fractional KdV equation and variants thereof there has recently been much
progress related to highest waves. Bruell and Dhara proved existence of
highest traveling waves which are Lipschitz at their cusp for $\alpha<-1$ [6].
Hildrum and Xue proved existence and the optimal $-\alpha$-Hölder regularity
for a family of equations including the fractional KdV equation with
$\alpha\in(-1,0)$ [30]. Ørke proved their existence for $\alpha\in(-1,0)$ for
the inhomogeneous fractional KdV equations as well as the fractional
Degasperis-Procesi equations, together with their optimal $-\alpha$-Hölder
regularity [51].

The results in [20, 6, 51, 30] are all based on global bifurcation arguments,
bifurcating from the constant solution and proving that the branch must end in
a highest wave which is not smooth at its crest. For the Burgers-Hilbert
equation Dahne and Gómez-Serrano proved the existence of a highest wave which
at its cusp behaves like $|x|\log|x|$ [15]. The proof uses a different method
where the problem is first reduced to a fixed point problem. The wave is
therefore not directly tied to a branch of solutions as in the other results.
This method of rewriting the problem into a fixed point problem was first used
by Enciso, Gómez-Serrano and Vergara for proving the convexity and the precise
asymptotic behavior of a highest wave solution to the Whitham equation [22],
answering a conjecture made by Ehrnström and Wahlén [20]. In this paper we use
a similar approach.

Similar to Hildrum and Xue we prove the existence of a highest wave for the
fractional KdV equation with $\alpha\in(-1,0)$. Our main contribution is a
more precise description of the asymptotic behavior of the wave at the cusp.
The description is in line with earlier results for the Whitham and Burgers-
Hilbert equation. Recently Ehrnström, Mæhlen and Varholm have obtained similar
results for the asymptotic behavior of two families of equations including the
uni- and bidirectional Whitham equations using different types of methods
[19].

We prove the following theorem:

######  Theorem 1.1.

There is a $2\pi$-periodic traveling wave $\varphi$ of (3) for every
$\alpha\in(-1,0)$, which behaves asymptotically at $x=0$ as

$\varphi(x)=c-\nu_{\alpha}|x|^{-\alpha}+\mathcal{O}(|x|^{p}).$

for $\nu_{\alpha}>0$ as given in Lemma 4.1 and some $-\alpha<p\leq 1$ to be
made explicit later on.

######  Remark 1.2.

The remainder term $\mathcal{O}(|x|^{p})$ in Theorem 1.1 is in general not
sharp. The estimate follows from the choice of our space.

######  Remark 1.3.

Both our results and Hildrum and Xue’s results assert the existence of a
highest wave. While, a priori, they don’t necessarily correspond to the same
wave, we do believe this is the case.

The ansatz $\varphi(x)=c-u(x)$ allows us to rewrite (3) as an equation that
does not explicitly depend on the wave speed $c$. Proving the existence of a
solution $u$ can be rewritten as a fixed point problem by considering the
ansatz

$u(x)=u_{\alpha}(x)+w_{\alpha}(x)v(x)$

where $u_{\alpha}(x)$ is an explicit, carefully chosen, approximate solution
and $w_{\alpha}(x)$ is an explicit weight factor. Proving the existence of a
fixed point can be reduced to checking an inequality involving three
constants, $D_{\alpha}$, $\delta_{\alpha}$ and $n_{\alpha}$, that only depend
on the choice of $u_{\alpha}$ and $w_{\alpha}$, see Proposition 2.2. This
inequality is checked by bounding $D_{\alpha}$, $\delta_{\alpha}$ and
$n_{\alpha}$ using a computer-assisted proof, see Section 12.

One of the key difficulties compared to the Whitham equation [22] and the
Burgers-Hilbert equation [15] is that we are treating a family of equations,
instead of one fixed equation. To handle the full family we need to understand
how the equation changes with respect to $\alpha$. In particular the endpoints
of the interval, $\alpha$ near $-1$ and $0$, require adapting the method to
those cases. To handle this we split the interval $(-1,0)$ into three parts,

$(-1,0)=(-1,-1+\delta_{1})\cup[-1+\delta_{1},-\delta_{2}]\cup(-\delta_{2},0)=I_{1}\cup
I_{2}\cup I_{3},$

and adapt the methods used for $I_{1}$ and $I_{3}$. For $\alpha\in I_{1}$ the
main complication comes from that as $\alpha\to-1$ the coefficient
$\nu_{\alpha}$ in Theorem 1.1 tends to infinity. For $\alpha\in I_{3}$ the
issue is that both sides of the inequality that needs to be verified for the
fixed point argument tend to zero as $\alpha\to 0$ and to assert that it holds
arbitrarily close to $\alpha=0$ we need an understanding of the rate at which
the two sides tend to zero.

An important part of the work is the construction of the approximate solution
$u_{\alpha}$. Due to the singularity at $x=0$ it is not possible to use a
trigonometric polynomial alone, it would converge very slowly and have the
wrong asymptotic behavior. Pure products of powers and logarithms,
$|x|^{a}\log^{b}|x|$, have the issue that they are not periodic and do not
interact well with the operator $|D|^{\alpha}$. Instead we take inspiration
from the construction in [22] and consider a combination of trigonometric
polynomials and Clausen functions of different orders, defined as

$C_{s}(x)=\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^{s}},\quad
S_{s}(x)=\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^{s}},$

for $s>1$ and by analytic continuation otherwise. We also make use of their
derivatives with respect to the order, for which we use the notation

$C_{s}^{(\beta)}(x):=\frac{d^{\beta}}{ds^{\beta}}C_{s}(x),\quad
S_{s}^{(\beta)}(x):=\frac{d^{\beta}}{ds^{\beta}}S_{s}(x).$

The Clausen functions are $2\pi$-periodic, non-analytic at $x=0$ and behave
well with respect to $|D|^{\alpha}$. In particular
$C_{s}(x)-C_{s}(0)\sim|x|^{s-1}$, which corresponds to the behavior we expect
in Theorem 1.1. See Section 3 and Appendix C for more details about the
Clausen functions. The main idea for the construction is the same as in [22],
to choose the coefficients for the Clausen functions according to the
asymptotic expansion of the solution at $x=0$. However, in our case we also
have to handle the limits $\alpha\to-1$ and $\alpha\to 0$, which requires a
good understanding of how the approximations depend on $\alpha$. In particular
understanding the limit $\alpha\to-1$ is important not only for this work, but
is also used to handle the Burgers-Hilbert case in [15].

An essential part of our work is the interplay between traditional
mathematical tools and rigorous computer calculations. Traditional numerical
methods typically only compute approximate results, to be able to use the
results in a proof we need the results to be rigorously verified. The basis
for rigorous calculations is interval arithmetic, pioneered by Moore in the
1970s [49]. Due to improvements in both computational power and great
improvements in software it is getting practical to use computer-assisted
tools in more and more complicated settings. The main idea with interval
arithmetic is to do arithmetic not directly on real numbers but on intervals
with computer representable endpoints. Given a function
$f:\mathbb{R}\to\mathbb{R}$, an interval extension of $f$ is an extension to
intervals satisfying that for an interval
$\textbf{{x}}=[\underline{x},\overline{x}]$, $f(\textbf{{x}})$ is an interval
satisfying $f(x)\in f(\textbf{{x}})$ for all $x\in\textbf{{x}}$. In particular
this allows us to prove inequalities for the function $f$, for example the
right endpoint of $f(\textbf{{x}})$ gives an upper bound of $f$ on the
interval x. For an introduction to interval arithmetic and rigorous numerics
we refer the reader to the books [49, 54] and to the survey [26] for a
specific treatment of computer-assisted proofs in PDE. For computer-assisted
proofs in fluid mechanics in particular some recent results are: [4, 2] for
the Navier-Stokes equation, [23, 24, 25] for the Kuramoto-Shivasinsky
equation, [12] for the Hou-Luo model, [7] for the compressible Euler and
Navier-Stokes equations and [10, 11] for blowup of the 2D Boussinesq and 3D
Euler equation.

For all the calculations in this paper we make use of the Arb library [41] for
ball (intervals represented as a midpoint and radius) arithmetic. It has good
support for many of the special functions we use [43, 37, 42], Taylor
arithmetic (see e.g. [39]) as well as rigorous integration [38].

The paper is organized as follows. In Section 2 we reduce the proof of Theorem
1.1 to a fixed point problem. In Section 3 we give a brief overview of
properties of the Clausen functions that are relevant for the construction of
$u_{\alpha}$, in Section 4 we give the construction of $u_{\alpha}$ and in
Section 5 we discuss the choice of the weight function $w_{\alpha}$ used.
Section 6 gives the general strategy for bounding $n_{\alpha}$,
$\delta_{\alpha}$ and $D_{\alpha}$, Section 7 and 8 discusses evaluation of
$u_{\alpha}$. Section 9 is devoted to the approach for bounding $n_{\alpha}$,
Section 10 to bounding $\delta_{\alpha}$ and Section 11 to studying the linear
operator that appears in the construction of the fixed point problem and
bounding $D_{\alpha}$. The computer-assisted proofs giving bounds for
$n_{\alpha}$, $\delta_{\alpha}$ and $D_{\alpha}$ are given in Section 12.
Finally we give the proof of Theorem 1.1 in Section 13.

Six appendices are given at the end of the paper. Appendix A gives some
technical details for how to compute enclosures of functions around removable
singularities. Appendix B gives a brief introduction to Taylor models which
are used in some parts of the proof. Appendix C is concerned with computing
enclosures of the Clausen functions and Appendix D with the rigorous numerical
integration needed for bounding $D_{0}$. Finally Appendix E, F and G contains
some of the details needed for $\alpha$ close to $-1$.

##  2 Reduction to a fixed point problem

In this section we reduce the problem of proving Theorem 1.1 to proving the
existence of a fixed point for a certain operator.

From [30, Theorem 12] we have the following characterization of even,
nondecreasing solutions of (1).

######  Lemma 2.1.

Let $\varphi\in C^{1}$ be a nonconstant, even solution of (1) which is
nondecreasing on $(-\pi,0)$, then

$\varphi^{\prime}>0\quad\text{ and }\quad\varphi<c$

on $(-\pi,0)$.

As a consequence, any continuous, nonconstant, even function which is
nondecreasing on $(-\pi,0)$ that satisfy (1) almost everywhere must satisfy
$\varphi\leq c$. The maximal possible height is thus given by $c$ and due to
the function being even and nondecreasing on $(-\pi,0)$ the maximal height has
to be attained at $x=0$.

Now, the ansatz $\varphi(x)=c-u(x)$ inserted in (3) gives an equation which
does not explicitly depend on the wave speed $c$. Indeed, inserting this gives
us

$uu^{\prime}=-|D|^{\alpha}u^{\prime}.$ (4)

Note that a solution of this equation gives a solution of (3) for any wave
speed $c$. This is to be expected due to the Galilean change of variables

$\varphi\mapsto\varphi+\gamma,\ c\mapsto c+\gamma$

which leaves (3) invariant. In particular, taking $c$ equal to the mean of $u$
gives a zero mean solution. For a highest wave we expect to have
$\varphi(0)=c$, giving us $u(0)=0$. Integrating (4) gives us

$\frac{1}{2}u^{2}=-\mathcal{H}^{\alpha}[u].$ (5)

Here $\mathcal{H}$ is the operator

$\mathcal{H}^{\alpha}[u](x)=|D|^{\alpha}u(x)-|D|^{\alpha}u(0).$ (6)

It is the integral of the right-hand side of (4) with the constant of
integration taken such that $\mathcal{H}^{\alpha}[f](0)=0$, this ensures that
any solution of (5) satisfies $u(0)=0$. Note that any solution of (5) is a
solution of (4) and hence gives a solution to (3).

To reduce the problem to a fixed point problem the idea is to write $u$ as
one, explicit, approximate solution of Equation (5) and one unknown term. More
precisely we make the ansatz

$u(x)=u_{\alpha}(x)+w_{\alpha}(x)v(x)$ (7)

where $u_{\alpha}(x)$ is an explicit, carefully chosen, approximate solution
of Equation (5), see Section 4, and $w_{\alpha}(x)$ is an explicit weight, see
Section 5, both of them depending on $\alpha$. By taking
$u_{\alpha}(x)\sim\nu_{\alpha}|x|^{-\alpha}$ and
$w_{\alpha}=\mathcal{O}(|x|^{p})$, proving Theorem 1.1 reduces to proving
existence of $v\in L^{\infty}(\mathbb{T})$ such that the given ansatz is a
solution of Equation (5).

Inserting the ansatz (7) into Equation (5) gives us

$\frac{1}{2}(u_{\alpha}+w_{\alpha}v)^{2}=-\mathcal{H}^{\alpha}[u_{\alpha}+w_{\alpha}v]\iff\frac{1}{2}u_{\alpha}^{2}+u_{\alpha}w_{\alpha}v+\frac{1}{2}w_{\alpha}^{2}v^{2}=-\mathcal{H}^{\alpha}[u_{\alpha}]-\mathcal{H}^{\alpha}[w_{\alpha}v]$

By collecting all the linear terms in $v$ we can write this as

$u_{\alpha}w_{\alpha}v+\mathcal{H}^{\alpha}[w_{\alpha}v]=-\mathcal{H}^{\alpha}u_{\alpha}-\frac{1}{2}u_{\alpha}^{2}-\frac{1}{2}w_{\alpha}^{2}v^{2}\iff
v+\frac{1}{w_{\alpha}u_{\alpha}}\mathcal{H}^{\alpha}[v]=-\frac{1}{w_{\alpha}u_{\alpha}}\left(\mathcal{H}^{\alpha}[u_{\alpha}]+u_{\alpha}^{2}\right)-\frac{w_{\alpha}}{2u_{\alpha}}v^{2}.$

Now let $T_{\alpha}$ denote the operator

$T_{\alpha}[v]=-\frac{1}{w_{\alpha}u_{\alpha}}\mathcal{H}^{\alpha}[w_{\alpha}v].$
(8)

Denote the weighted defect of the approximate solution $u_{\alpha}(x)$ by

$F_{\alpha}(x)=\frac{1}{w_{\alpha}(x)u_{\alpha}(x)}\left(\mathcal{H}^{\alpha}[u_{\alpha}](x)+\frac{1}{2}u_{\alpha}(x)^{2}\right),$
(9)

and let

$N_{\alpha}(x)=\frac{w_{\alpha}(x)}{2u_{\alpha}(x)}.$ (10)

Then we can write the above as

$(I-T_{\alpha})v=-F_{\alpha}-N_{\alpha}v^{2}.$

Assuming that $I-T_{\alpha}$ is invertible we rewrite this as

$v=(I-T_{\alpha})^{-1}\left(-F_{\alpha}-N_{\alpha}v^{2}\right)=:G_{\alpha}[v].$
(11)

Hence proving the existence of $v$ such that $u_{\alpha}+w_{\alpha}v$ is a
solution to Equation (5) reduces to proving existence of a fixed point of the
operator $G_{\alpha}$.

Next we reduce the problem of proving that $G_{\alpha}$ has a fixed point to
checking an inequality for three numbers (depending on $\alpha$) that depend
only on the choice of $u_{\alpha}$ and $w_{\alpha}$. We let $\|T\|$ denote the
$L^{\infty}(\mathbb{T})\to L^{\infty}(\mathbb{T})$ norm of a linear operator
$T$.

######  Proposition 2.2.

Let $D_{\alpha}=\|T_{\alpha}\|$,
$\delta_{\alpha}=\|F_{\alpha}\|_{L^{\infty}(\mathbb{T})}$ and
$n_{\alpha}=\|N_{\alpha}\|_{L^{\infty}(\mathbb{T})}$. If $D_{\alpha}<1$ and
they satisfy the inequality

$\delta_{\alpha}<\frac{(1-D_{\alpha})^{2}}{4n_{\alpha}}$

then for

$\epsilon_{\alpha}=\frac{1-D_{\alpha}-\sqrt{(1-D_{\alpha})^{2}-4\delta_{\alpha}n_{\alpha}}}{2n_{\alpha}}$

and

$X_{\epsilon}=\\{v\in
L^{\infty}(\mathbb{T}):v(x)=v(-x),\|v\|_{L^{\infty}(\mathbb{T})}\leq\epsilon\\}$

we have

  1. 1.

$G_{\alpha}(X_{\epsilon_{\alpha}})\subseteq X_{\epsilon_{\alpha}}$;

  2. 2.

$\|G_{\alpha}[v]-G_{\alpha}[w]\|_{L^{\infty}(\mathbb{T})}\leq
k_{\alpha}\|v-w\|_{L^{\infty}(\mathbb{T})}$ with $k_{\alpha}<1$ for all
$v,w\in X_{\epsilon_{\alpha}}$.

###### Proof.

Using that $N_{\alpha}$ and $F_{\alpha}$ are even it can be checked that

$G_{\alpha}(X_{\epsilon_{\alpha}})\subseteq(I-T)^{-1}X_{\delta_{\alpha}+n_{\alpha}\epsilon_{\alpha}^{2}}.$

Since $\|T_{\alpha}\|<1$ the operator $I-T_{\alpha}$ is invertible and an
upper bound of the norm of the inverse is given by $\frac{1}{1-D_{\alpha}}$,
moreover $T_{\alpha}$ takes even functions to even functions and hence so will
$(I-T_{\alpha})^{-1}$. This gives us

$G_{\alpha}(X_{\epsilon_{\alpha}})\subseteq(I-T)^{-1}X_{\delta_{\alpha}+n_{\alpha}\epsilon_{\alpha}^{2}}\subseteq
X_{\frac{\delta_{\alpha}+n_{\alpha}\epsilon_{\alpha}^{2}}{1-D_{\alpha}}}.$

The choice of $\epsilon_{\alpha}$ then gives

$\frac{\delta_{\alpha}+n_{\alpha}\epsilon_{\alpha}^{2}}{1-D_{\alpha}}=\epsilon_{\alpha}.$

Next we have
$G_{\alpha}[v]-G_{\alpha}[w]=(I-T_{\alpha})^{-1}(-N_{\alpha}(v^{2}-w^{2}))$
and hence

$\|G_{\alpha}[v]-G_{\alpha}[w]\|_{L^{\infty}(\mathbb{T})}\leq\frac{n_{\alpha}}{1-D_{\alpha}}\|v^{2}-w^{2}\|_{L^{\infty}(\mathbb{T})}\leq\frac{2n_{\alpha}\epsilon_{\alpha}}{1-D_{\alpha}}\|v-w\|_{L^{\infty}(\mathbb{T})}.$

Where $k_{\alpha}=\frac{2n_{\alpha}\epsilon_{\alpha}}{1-D_{\alpha}}<1$ since
$\epsilon_{\alpha}<\frac{1-D_{\alpha}}{2n_{\alpha}}$. ∎

##  3 Clausen functions

We here give definitions and properties of the Clausen functions that are used
in the construction of $u_{\alpha}$ and when bounding $n_{\alpha}$,
$\delta_{\alpha}$ and $D_{\alpha}$. For more details about the Clausen
functions see Appendix C.

For $s>1$ the Clausen functions can be defined through their Fourier
expansions as

$\displaystyle C_{s}(x)$
$\displaystyle=\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^{s}},$ $\displaystyle
S_{s}(x)$ $\displaystyle=\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^{s}}.$

For general $s$ they are more conveniently defined by their relation to the
polylogarithm through

$\displaystyle
C_{s}(x)=\frac{1}{2}\left(\operatorname{Li}_{s}(e^{ix})+\operatorname{Li}_{s}(e^{-ix})\right)=\operatorname{Re}\left(\operatorname{Li}_{s}(e^{ix})\right),$
$\displaystyle
S_{s}(x)=\frac{1}{2}\left(\operatorname{Li}_{s}(e^{ix})-\operatorname{Li}_{s}(e^{-ix})\right)=\operatorname{Im}\left(\operatorname{Li}_{s}(e^{ix})\right).$

They behave nicely with respect to the operator $|D|^{\alpha}$, for which we
have

$|D|^{\alpha}C_{s}=-C_{s-\alpha}(x),\quad|D|^{\alpha}S_{s}=-S_{s-\alpha}(x).$

In many cases we want to work with functions which are normalized to be zero
at $x=0$, for which we use the notation

$\tilde{C}_{s}(x)=C_{s}(x)-C_{s}(0),\quad\tilde{C}_{s}^{(\beta)}(x)=C_{s}^{(\beta)}(x)-C_{s}^{(\beta)}(0).$

Note that $C_{s}(0)$ is in general only finite for $s>1$, in which case we get
directly from the Fourier expansion that $C_{s}(0)=\zeta(s)$ and
$C_{s}^{(\beta)}(0)=\zeta^{(\beta)}(s)$. With this notation we get for the
operator $\mathcal{H}^{\alpha}$,

$\mathcal{H}^{\alpha}[\tilde{C}_{s}](x)=-\tilde{C}_{s-\alpha}(x),\quad\mathcal{H}^{\alpha}[S_{s}](x)=-S_{s-\alpha}(x).$

From [22] we have the following expansion for $C_{s}(x)$ and $S_{s}(x)$, valid
for $|x|<2\pi$,

$\displaystyle C_{s}(x)$
$\displaystyle=\Gamma(1-s)\sin\left(\frac{\pi}{2}s\right)|x|^{s-1}+\sum_{m=0}^{\infty}(-1)^{m}\zeta(s-2m)\frac{x^{2m}}{(2m)!};$
$\displaystyle S_{s}(x)$
$\displaystyle=\Gamma(1-s)\cos\left(\frac{\pi}{2}s\right)\operatorname{sgn}(x)|x|^{s-1}+\sum_{m=0}^{\infty}(-1)^{m}\zeta(s-2m-1)\frac{x^{2m+1}}{(2m+1)!}.$

For the functions $C_{s}^{(\beta)}$ and $S_{s}^{(\beta)}$ we will mainly make
use of $C_{2}^{(1)}(x)$ and $C_{3}^{(1)}(x)$, for which we have the following
expansions [3, Eq. 16], valid for $|x|<2\pi$,

$\displaystyle C_{2}^{(1)}(x)=$
$\displaystyle\zeta^{(1)}(2)-\frac{\pi}{2}|x|\log|x|-(\gamma-1)\frac{\pi}{2}|x|+\sum_{m=1}^{\infty}(-1)^{m}\zeta^{(1)}(2-2m)\frac{x^{2m}}{(2m)!};$
$\displaystyle C_{3}^{(1)}(x)=$
$\displaystyle\zeta^{(1)}(3)-\frac{1}{4}x^{2}\log^{2}|x|+\frac{3-2\gamma}{4}x^{2}\log|x|-\frac{36\gamma-12\gamma^{2}-24\gamma_{1}-42+\pi^{2}}{48}x^{2}$
$\displaystyle+\sum_{m=2}^{\infty}(-1)^{m}\zeta^{(1)}(3-2m)\frac{x^{2m}}{(2m)!}.$

Where $\gamma_{n}$ is the Stieltjes constant and $\gamma=\gamma_{0}$. Bounds
for the tails are given in Lemmas C.5 and C.6.

##  4 Construction of $u_{\alpha}$

In this section we give the construction of the approximate solution
$u_{\alpha}$ for $\alpha\in(-1,0)$. As a first step we determine the
coefficient for the leading term in the asymptotic expansion.

######  Lemma 4.1.

Let $\alpha\in(-1,0)$ and assume that $u$ is a bounded, even solution of
Equation (5) with the asymptotic behavior

$u(x)=\nu_{a}|x|^{-\alpha}+o(|x|^{-\alpha})$

close to zero, with $\nu_{\alpha}\not=0$. Then the coefficient is given by

$\nu_{\alpha}=\frac{2\Gamma(2\alpha)\cos\left(\pi\alpha\right)}{\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)}.$

###### Proof.

We directly get

$\frac{1}{2}u(x)^{2}=\frac{\nu_{\alpha}^{2}}{2}|x|^{-2\alpha}+o(|x|^{-2\alpha}).$

To get the asymptotic behavior of $\mathcal{H}^{\alpha}[u]$ hand side we go
through the Clausen function $\tilde{C}_{1-\alpha}(x)$. Based on the
asymptotic behavior of $\tilde{C}_{1-\alpha}(x)$ we can write $u$ as

$u(x)=\frac{\nu_{\alpha}}{\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)}\tilde{C}_{1-\alpha}(x)+o(|x|^{-\alpha}).$

This gives us

$\displaystyle\mathcal{H}^{\alpha}[u](x)$
$\displaystyle=-\frac{\nu_{\alpha}}{\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)}\tilde{C}_{1-2\alpha}(x)+\mathcal{H}^{\alpha}[o(|x|^{-\alpha})](x)$
$\displaystyle=-\nu_{\alpha}\frac{\Gamma(2\alpha)\cos\left(\pi\alpha\right)}{\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)}|x|^{-2\alpha}+o(|x|^{-2\alpha})+\mathcal{H}^{\alpha}[o(|x|^{-\alpha})](x).$

In a similar way as in [15, Lemma 4.1] it can be shown that
$\mathcal{H}^{\alpha}[o(|x|^{-\alpha})](x)=o(|x|^{-2\alpha})$, which implies
the lemma. ∎

In addition to having the correct asymptotic behavior we want $u_{\alpha}$ to
be a good approximate solution of (5), in the sense that we want the defect,

$F_{\alpha}(x)=\frac{1}{w_{\alpha}(x)u_{\alpha}(x)}\left(\mathcal{H}^{\alpha}[u_{\alpha}](x)+\frac{1}{2}u_{\alpha}(x)^{2}\right),$

to be small for $x\in[0,\pi]$. The hardest part is to make $F_{\alpha}(x)$
small locally near the singularity at $x=0$, this is done by studying the
asymptotic behavior of
$\mathcal{H}^{\alpha}[u_{\alpha}](x)+\frac{1}{2}u_{\alpha}(x)^{2}$. Once the
defect is sufficiently small near $x=0$ it can be made small globally by
adding a suitable trigonometric polynomial to $u_{\alpha}$.

The construction is similar to that in [22], but requires more work to handle
the limits $\alpha\to-1$ and $\alpha\to 0$. We take $u_{\alpha}$ to be a
combination of three parts:

  1. 1.

The first term is $a_{\alpha,0}\tilde{C}_{1-\alpha}$, where the coefficient is
chosen to give the right asymptotic behavior according to Lemma 4.1

  2. 2.

The second part is chosen to make the defect small near $x=0$, similarly to in
[22], it is given by a sum of Clausen functions.

  3. 3.

The third part is chosen to make the defect small globally and is given by a
trigonometric polynomial.

More precisely the approximation is given by

$u_{\alpha}(x)=\sum_{j=0}^{N_{\alpha,0}}a_{\alpha,j}\tilde{C}_{1-\alpha+jp_{\alpha}}(x)+\sum_{n=1}^{N_{\alpha,1}}b_{\alpha,n}(\cos(nx)-1),$
(12)

The values of $a_{\alpha,j}$ for $j\geq 1$ and $p_{\alpha}$ will be taken to
make the defect small near $x=0$ and $b_{\alpha,n}$ is taken to make the
defect small globally.

While the general form of the approximation is the same for all values of
$\alpha$ the limits $\alpha\to-1$ and $\alpha\to 0$ requires a slightly
different approach. For that reason we split it into three cases

  1. 1.

$\alpha\in(-1,-1+\delta_{1})=I_{1}$;

  2. 2.

$\alpha\in[-1+\delta_{1},-\delta_{2}]=I_{2}$;

  3. 3.

$\alpha\in(-\delta_{2},0)=I_{3}$,

where the precise values of $\delta_{1}$ and $\delta_{2}$ will be determined
later. We start with the approximation for the interval $I_{2}$ which has the
least amount of technical details. We then discuss the alterations required to
handle $I_{1}$ and $I_{3}$.

###  4.1 Construction of $u_{\alpha}$ for $I_{2}$

The asymptotic behavior of the approximation (12) is determined by
$a_{\alpha,0}$ and for it to agree with Lemma 4.1 we have to take

$a_{\alpha,0}=\frac{2\Gamma(2\alpha)\cos\left(\pi\alpha\right)}{\Gamma(\alpha)^{2}\cos\left(\frac{\pi}{2}\alpha\right)^{2}}.$
(13)

The behavior of $a_{\alpha,0}$ as a function in $\alpha$ is shown in Figure
1(a).

To choose the other parameters we have to study the defect of the
approximation. The defect is given by
$\mathcal{H}^{\alpha}[u_{\alpha}]+\frac{1}{2}u_{\alpha}^{2}$, where we have

$\mathcal{H}^{\alpha}[u_{\alpha}](x)=-\sum_{j=0}^{N_{\alpha,0}}a_{\alpha,j}\tilde{C}_{1-2\alpha+jp_{\alpha}}(x)-\sum_{n=1}^{N_{\alpha,1}}b_{\alpha,n}n^{\alpha}(\cos(nx)-1).$
(14)

For the defect close to $x=0$ we want to study the asymptotic expansion. We
get the asymptotic behavior of $u_{\alpha}$ and
$\mathcal{H}^{\alpha}[u_{\alpha}]$ in the following lemma, whose proof we omit
since it follows directly from the expansions of the Clausen and cosine
functions.

######  Lemma 4.2.

Let $u_{\alpha}$ be as in (12). Then the following asymptotic expansions hold
near $x=0$:

$u_{\alpha}(x)=\sum_{j=0}^{N_{\alpha,0}}a_{\alpha,j}^{0}|x|^{-\alpha+jp_{\alpha}}+\sum_{m=1}^{\infty}\frac{(-1)^{m}}{(2m)!}\left(\sum_{j=0}^{N_{\alpha,0}}a_{\alpha,j}\zeta(1-\alpha+jp_{\alpha}-2m)+\sum_{m=1}^{N_{\alpha,1}}b_{\alpha,j}n^{2m}\right)x^{2m}$
(15)
$\mathcal{H}^{\alpha}[u_{\alpha}](x)=-\sum_{j=0}^{N_{\alpha,0}}A_{\alpha,j}^{0}|x|^{-2\alpha+jp_{\alpha}}-\sum_{m=1}^{\infty}\frac{(-1)^{m}}{(2m)!}\left(\sum_{j=0}^{N_{\alpha,0}}a_{\alpha,j}\zeta(1-2\alpha+jp_{\alpha}-2m)+\sum_{m=1}^{N_{\alpha,1}}b_{\alpha,j}n^{2m+\alpha}\right)x^{2m}$
(16)

where

$\displaystyle a_{\alpha,j}^{0}$ $\displaystyle=\Gamma(\alpha-
jp_{\alpha})\cos\left(\frac{\pi}{2}(\alpha-jp_{\alpha})\right)a_{\alpha,j};$
$\displaystyle A_{\alpha,j}^{0}$ $\displaystyle=\Gamma(2\alpha-
jp_{\alpha})\cos\left(\frac{\pi}{2}(2\alpha-jp_{\alpha})\right)a_{\alpha,j}.$

From this we can compute the expansion of
$\mathcal{H}^{\alpha}[u_{\alpha}]+\frac{1}{2}u_{\alpha}^{2}$, given in the
following Lemma. Again we omit the proof which only involves standard
calculations.

######  Lemma 4.3.

Let $u_{\alpha}$ be as in (12). Then we have the following asymptotic
expansion near $x=0$

$\mathcal{H}^{\alpha}[u_{\alpha}]+\frac{1}{2}u_{\alpha}^{2}=\left(\frac{1}{2}(a_{\alpha,0}^{0})^{2}-A_{\alpha,0}^{0}\right)|x|^{-2\alpha}+\left(a_{\alpha,0}^{0}a_{\alpha,1}^{0}-A_{\alpha,1}^{0}\right)|x|^{-2\alpha+p_{\alpha}}\\\
+\sum_{k=2}^{N_{\alpha,0}}\left(\frac{1}{4}((-1)^{k}+1)(a_{\alpha,\lfloor\frac{k}{2}\rfloor}^{0})^{2}+\sum_{j=0}^{\lfloor\frac{k-1}{2}\rfloor}a_{\alpha,j}^{0}a_{\alpha,k-j}^{0}-A_{\alpha,k}^{0}\right)|x|^{-2\alpha+kp_{\alpha}}\\\
+\sum_{k=N_{\alpha,0}+1}^{2N_{\alpha,0}}\left(\frac{1}{4}((-1)^{k}+1)(a_{\alpha,\lfloor\frac{k}{2}\rfloor}^{0})^{2}+\sum_{j=0}^{max(0,\lfloor\frac{k-1}{2}\rfloor)}a_{\alpha,j}^{0}a_{\alpha,k-j}^{0}\right)|x|^{-2\alpha+kp_{\alpha}}\\\
+\left(\sum_{j=0}^{N_{\alpha,0}}a_{\alpha,j}^{0}|x|^{-\alpha+jp_{\alpha}}\right)S_{1}+\frac{1}{2}S_{1}^{2}-S_{2}$

where

$\displaystyle S_{1}$
$\displaystyle=\sum_{m=1}^{\infty}\frac{(-1)^{m}}{(2m)!}\left(\sum_{j=0}^{N_{\alpha
0}}a_{\alpha,j}\zeta(1-\alpha+jp_{\alpha}-2m)+\sum_{n=1}^{N_{\alpha,1}}b_{\alpha,n}n^{2m}\right)x^{2m},$
$\displaystyle S_{2}$
$\displaystyle=\sum_{m=1}^{\infty}\frac{(-1)^{m}}{(2m)!}\left(\sum_{j=0}^{N_{\alpha,0}}a_{\alpha,j}\zeta(1-2\alpha+jp_{\alpha}-2m)+\sum_{n=1}^{N_{\alpha,1}}b_{\alpha,n}n^{2m+\alpha}\right)x^{2m}.$

Taking $a_{\alpha,0}$ as in Equation (13) makes the leading coefficient
$\frac{1}{2}(a_{\alpha,0}^{0})^{2}-A_{\alpha,0}^{0}$ zero. If $p_{\alpha}$ is
not too large, $-\alpha+p_{\alpha}<2$, the next leading term is

$\left(a_{\alpha,0}^{0}a_{\alpha,1}^{0}-A_{\alpha,1}^{0}\right)|x|^{-2\alpha+p_{\alpha}}$

where the coefficient is given by

$a_{\alpha,0}^{0}a_{\alpha,1}^{0}-A_{\alpha,1}^{0}=\left(\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)\Gamma(\alpha-
p_{\alpha})\cos\left(\frac{\pi}{2}(\alpha-
p_{\alpha})\right)a_{\alpha,0}-\Gamma(2\alpha-
p_{\alpha})\cos\left(\frac{\pi}{2}(2\alpha-
p_{\alpha})\right)\right)a_{\alpha,1}.$

and the only way for this to be zero is to either take $a_{\alpha,1}=0$ or to
choose $p_{\alpha}$ such that

$\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)\Gamma(\alpha-
p_{\alpha})\cos\left(\frac{\pi}{2}(\alpha-
p_{\alpha})\right)a_{\alpha,0}-\Gamma(2\alpha-
p_{\alpha})\cos\left(\frac{\pi}{2}(2\alpha-p_{\alpha})\right)=0.$

Inserting the value of $a_{\alpha,0}$ and rearranging the terms, this can be
rewritten as

$\frac{\Gamma(2\alpha-p_{\alpha})\cos\left(\frac{\pi}{2}(2\alpha-
p_{\alpha})\right)}{\Gamma(\alpha-p_{\alpha})\cos\left(\frac{\pi}{2}(\alpha-
p_{\alpha})\right)}=\frac{2\Gamma(2\alpha)\cos(\pi\alpha)}{\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)}.$
(17)

This tells us how to pick the parameter $p_{\alpha}$, we take it to be the
smallest positive solution of (17). A plot of $p_{\alpha}$ as a function of
$\alpha$ is given in Figure 1(b).

(a)

(b)

Figure 1: Values of $a_{\alpha,0}$, defined by Equation (13), and
$p_{\alpha}$, defined by Equation (17), as functions of $\alpha$.

The expansion in Lemma 4.3 also tells us how to choose
$\\{a_{\alpha,j}\\}_{1\leq j\leq N_{\alpha,0}}$: identify the $N_{\alpha,0}$
leading coefficients in the expansion, ignoring $|x|^{-2\alpha}$ and
$|x|^{-2\alpha+p_{\alpha}}$ whose coefficients are zero from the choice of
$a_{\alpha,0}$ and $p_{\alpha}$, and take $\\{a_{\alpha,j}\\}_{1\leq j\leq
N_{\alpha,0}}$ to make them zero. There are three potential issues with this
approach for choosing $\\{a_{\alpha,j}\\}_{1\leq j\leq N_{\alpha,0}}$

  1. 1.

The coefficients in the expansion might also depend on $b_{\alpha,n}$ which we
have yet to determine;

  2. 2.

It is not clear if we indeed can pick $a_{\alpha,j}$ to make these
coefficients zero;

  3. 3.

Even if we can pick $a_{\alpha,j}$ to make the coefficients zero it is not
clear how we would find these values.

The first point is handled by simply ignoring the values of $b_{\alpha,n}$ at
this stage, taking them as zero. This means that some of the coefficients in
the expansion might not be zero in the end, once the values for $b_{\alpha,n}$
are taken into account. This however turns out to not be an issue, which is
related to how to handle the other two points. We summarize it in the
following remark

######  Remark 4.4.

Recall that $u_{\alpha}$ is only an approximate solution to the equation. In
the end the only important part is that the defect is sufficiently small. This
means that we don’t need to take exact values for $p_{\alpha}$,
$\\{a_{\alpha,j}\\}_{1\leq j\leq N_{\alpha,0}}$ or $\\{b_{\alpha,n}\\}_{1\leq
n\leq N_{\alpha,1}}$. We are free to use any means, in particular non-rigorous
numerical methods, for finding values which gives us a small defect. That the
defect indeed is small is then later rigorously verified.

Note that this does not apply to the parameter $a_{\alpha,0}$. For the defect
to be bounded near $x=0$ we need the leading coefficient in the expansion to
be exactly zero, it is therefore not enough to use a numerical approximation
of $a_{\alpha,0}$.

I light of this remark, points 2 and 3 above can be dealt with by setting up
the problem as a non-linear system of equations and finding a solution using
non-rigorous numerical methods. The numerical procedure might or might not
converge and the solution it gives might or might not correspond to a true
solution, but since we only need approximate value this doesn’t matter. That
the solutions we get are good enough will be verified when bounding the
defect. Similarly, we don’t need to take $p_{\alpha}$ to be an exact solution
of (17), a numerical approximation of the solution suffices.

Once $p_{\alpha}$ and $\\{a_{\alpha,j}\\}_{1\leq j\leq N_{\alpha,0}}$ have
been picked in the above way, to make the asymptotic defect small, we wish to
find $\\{b_{\alpha,n}\\}_{1\leq n\leq N_{\alpha,1}}$ to make the defect small
globally. This is done by taking $N_{\alpha,1}$ equally spaced points
$\\{x_{n}\\}_{1\leq n\leq N_{\alpha,1}}$ on the interval $(0,\pi)$ and
numerically solving the non-linear system

$\mathcal{H}^{\alpha}[u_{\alpha}](x_{n})+\frac{1}{2}u_{\alpha}(x_{n})^{2}=0\quad\text{
for }1\leq n\leq N_{\alpha,1}.$

So far we have not said anything about the values for $N_{\alpha,0}$ and
$N_{\alpha,1}$. These are tuning parameters, higher values will typically lead
to a smaller defect but also to a higher computational cost. For
$N_{\alpha,0}$ this is only true up to some point where the system for
determining the coefficients starts to become ill-conditioned. The values are
chosen to make sure that the inequality in Proposition 2.2 holds and depends
on $\alpha$. Some more details are given in Section 12.2.

###  4.2 Construction of $u_{\alpha}$ for $I_{1}$

The approximation in the previous section encounters several issues as
$\alpha$ moves close to $-1$. To begin with $a_{\alpha,0}$ diverges towards
negative infinity, as can be seen in Figure 1(a). Furthermore $p_{\alpha}$
tends towards zero, as can be seen in Figure 1(b), which means that the
parameters for all the Clausen functions converge towards the same value.

We can get an idea for how to handle these issues by numerically studying the
behavior of

$\sum_{j=0}^{N_{\alpha,0}}a_{\alpha,j}\tilde{C}_{1-\alpha+jp_{\alpha}}(x),$
(18)

as $\alpha$ goes to $-1$. The first important observation is that as
$\alpha\to-1$ the parameter $a_{\alpha,1}$ goes to infinity in such a way that
$a_{\alpha,0}+a_{\alpha,1}$ remains bounded, while $\\{a_{\alpha,j}\\}_{2\leq
j\leq N_{\alpha,0}}$ remains individually bounded. The second important
observation is that as $\alpha\to-1$ the parameter $p_{\alpha}$ behaves like
$1+\alpha+(1+\alpha)^{2}/2+\mathcal{O}((1+\alpha)^{3})$. This hints at a
solution to the problem that $a_{\alpha,0}\tilde{C}_{1-\alpha}(x)$ doesn’t
converge: by taking $a_{\alpha,1}=-a_{\alpha,0}$ and
$p_{\alpha}=1+\alpha+(1+\alpha)^{2}/2$ the first two terms in the above sum
become

$a_{\alpha,0}(\tilde{C}_{1-\alpha}(x)-\tilde{C}_{2+(1+\alpha)^{2}/2}(x)).$

For which we have the following lemma.

######  Lemma 4.5.

The function

$a_{\alpha,0}(\tilde{C}_{1-\alpha}(x)-\tilde{C}_{2+(1+\alpha)^{2}/2}(x)),$

with $a_{\alpha,0}$ as in (13), converges to
$\frac{2}{\pi^{2}}\tilde{C}_{2}^{(1)}(x)$ pointwise in $x$ as $\alpha\to-1$.
Moreover, for $\alpha\in\left(-1,-\frac{3}{4}\right)$ it satisfies the
inequality

$a_{\alpha,0}(\tilde{C}_{1-\alpha}(x)-\tilde{C}_{2+(1+\alpha)^{2}/2}(x))\geq\frac{2}{\pi^{2}}\tilde{C}_{2}^{(1)}(x).$

###### Proof.

We have

$a_{\alpha,0}(\tilde{C}_{1-\alpha}(x)-\tilde{C}_{2+(1+\alpha)^{2}/2}(x))=-(1+\alpha)a_{\alpha,0}\cdot\frac{\tilde{C}_{2+(1+\alpha)^{2}/2}(x)-\tilde{C}_{1-\alpha}(x)}{1+\alpha}.$

Taking the limit as $\alpha\to-1$ for the first factor we get
$\lim_{\alpha\to-1}-(1+\alpha)a_{\alpha,0}=\frac{2}{\pi^{2}}.$ The second
factor converges to the derivative $\tilde{C}_{1-\alpha}^{(1)}(x)$. This gives
us the pointwise convergence.

To prove the inequality it is enough to prove that the left-hand side is
increasing in $\alpha$. For that it is enough to prove that both factors are
positive and increasing in $\alpha$.

The proof that $-(1+\alpha)a_{\alpha,0}$ is positive and increasing in
$\alpha$ is computer-assisted. It is done by rigorously computing a lower
bound of the value and the derivative for $-1<\alpha<-\frac{3}{4}$ and
asserting that these lower bounds are positive.

For the Clausen-factor we let $h=1+\alpha$ and expand the Clausen functions at
$s=2$, giving us

$\displaystyle\tilde{C}_{1-\alpha}(x)$
$\displaystyle=\sum_{n=0}^{\infty}\frac{\tilde{C}_{2}^{(n)}(x)}{n!}(-h)^{n},$
$\displaystyle\tilde{C}_{1+(1+\alpha)^{2}/2}(x)$
$\displaystyle=\sum_{n=0}^{\infty}\frac{\tilde{C}_{2}^{(n)}(x)}{n!}\left(\frac{h^{2}}{2}\right)^{n}.$

From which we get

$\frac{\tilde{C}_{2+(1+\alpha)^{2}/2}(x)-\tilde{C}_{1-\alpha}(x)}{1+\alpha}=\sum_{n=1}^{\infty}\frac{\tilde{C}_{2}^{(n)}(x)}{n!}h^{n-1}\left(\frac{h^{n}}{2^{n}}+(-1)^{n-1}\right).$
(19)

For $-1<\alpha<0$ we have $\left|\frac{h^{n}}{2^{n}}\right|<1$ and the sign of
the factor $\frac{h}{2^{n}}+(-1)^{n-1}$ is therefore the same as for
$(-1)^{n-1}$. Furthermore, we have

$\tilde{C}_{2}^{(n)}(x)=(-1)^{n}\sum_{k=1}^{\infty}\frac{\cos(kx)-1}{k^{2}}\log
k$

and since $\cos(kx)-1\leq 0$ this has the same sign as $(-1)^{n-1}$. It
follows that all the terms in (19) are positive, hence the sum is positive and
increasing in $\alpha$ for $-1<\alpha<0$. ∎

If we subtract
$a_{\alpha,0}(\tilde{C}_{1-\alpha}(x)-\tilde{C}_{2+(1+\alpha)^{2}/2}(x))$ from
(18) the remaining part of

$(a_{\alpha,0}+a_{\alpha,1})\tilde{C}_{1-\alpha+jp_{\alpha}}(x)+\sum_{j=2}^{N_{\alpha,0}}a_{\alpha,j}\tilde{C}_{1-\alpha+jp_{\alpha}}(x),$

Numerically we can see that this converges to a bounded function as $\alpha$
goes towards $-1$. Since we only need an approximation of the limit we can
simply fix some $\alpha>-1$ and use the values for this $\alpha$.

More precisely we construct the approximation for $I_{1}$ in the following
way. First fix $\hat{\alpha}$ close to $-1$ but such that
$\alpha<\hat{\alpha}$ for all $\alpha\in I_{1}$. Then compute
$\\{a_{\hat{\alpha},j}\\}_{0\leq j\leq N_{\hat{\alpha},1}}$ and
$p_{\hat{\alpha}}$ using the approach in Section 4.1. We then take

$u_{\alpha}(x)=a_{\alpha,0}(\tilde{C}_{1-\alpha}(x)-\tilde{C}_{2+(1+\alpha)^{2}/2}(x))+(a_{\hat{\alpha},0}+a_{\hat{\alpha},1})\tilde{C}_{1-\hat{\alpha}+p_{\hat{\alpha}}}(x)\\\
+\sum_{j=2}^{N_{\hat{\alpha},0}}a_{\hat{\alpha},j}\tilde{C}_{1-\hat{\alpha}+jp_{\hat{\alpha}}}(x)+\sum_{n=1}^{N_{-1,1}}b_{-1,n}(\cos(nx)-1)$
(20)

The parameters $\\{b_{-1,n}\\}_{1\leq n\leq N_{-1,1}}$ are determined by
considering the defect at the limit $\alpha\to-1$, where we from Lemma 4.5
have

$u_{-1}(x)=\frac{2}{\pi^{2}}\tilde{C}_{2}^{(1)}(x)+(a_{\hat{\alpha},0}+a_{\hat{\alpha},1})\tilde{C}_{1-\hat{\alpha}+p_{\hat{\alpha}}}(x)+\sum_{j=2}^{N_{\hat{\alpha},0}}a_{\hat{\alpha},j}\tilde{C}_{1-\hat{\alpha}+jp_{\hat{\alpha}}}(x)+\sum_{n=1}^{N_{-1,1}}b_{-1,n}(\cos(nx)-1).$
(21)

In the same way as in Section 4.1 we take $\\{b_{-1,n}\\}_{1\leq n\leq
N_{-1,1}}$ so that the defect is minimized on $N_{-1,1}$ equally spaced points
on the interval $(0,\pi)$. Note that by Lemma 4.5 $u_{-1}(x)$ gives a lower
bound of $u_{\alpha}(x)$.

With this approximation we get

$\mathcal{H}^{\alpha}[u_{\alpha}](x)=-a_{\alpha,0}(\tilde{C}_{1-2\alpha}(x)-\tilde{C}_{2-\alpha+(1+\alpha)^{2}/2}(x))+(a_{\hat{\alpha},0}+a_{\hat{\alpha},1})\tilde{C}_{1-\alpha-\hat{\alpha}+p_{\hat{\alpha}}}(x)\\\
-\sum_{j=2}^{N_{\hat{\alpha},0}}a_{\hat{\alpha},j}\tilde{C}_{1-\alpha-\hat{\alpha}+jp_{\hat{\alpha}}}(x)-\sum_{n=1}^{N_{-1,1}}b_{-1,n}n^{-\alpha}(\cos(nx)-1).$
(22)

Note that since $a_{\alpha,0}\to-\infty$ as $\alpha\to-1$ we cannot compute a
finite enclosure of $a_{\alpha,0}$ valid for an interval of the form
$(-1,-1+\delta)$. When computing enclosures we therefore have to treat

$a_{\alpha,0}(\tilde{C}_{1-\alpha}(x)-\tilde{C}_{2+(1+\alpha)^{2}/2}(x))\quad\text{and}\quad
a_{\alpha,0}(\tilde{C}_{1-2\alpha}(x)-\tilde{C}_{2-\alpha+(1+\alpha)^{2}/2}(x))$

as one part, and explicitly handle the removable singularity.

The asymptotic behavior are given in the following lemma, again we omit the
proof since it follows directly from the expansions of the Clausen functions.

######  Lemma 4.6.

The approximation for $\alpha\in I_{1}$, with $u_{\alpha}$ given by (20) and
$\mathcal{H}^{\alpha}[u_{\alpha}]$ given by (22) has the asymptotic expansions

$\displaystyle u_{\alpha}(x)=$ $\displaystyle
a_{\alpha,0}\left(\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)-\Gamma(-1-(1+\alpha)^{2}/2)\cos\left(\frac{\pi}{2}(1+(1+\alpha)^{2}/2)\right)|x|^{1+\alpha+(1+\alpha)^{2}/2}\right)|x|^{-\alpha}$
$\displaystyle+a_{\alpha,0}\sum_{m=1}^{\infty}\frac{(-1)^{m}}{(2m)!}\left(\zeta(1-\alpha-2m)-\zeta(2+(1+\alpha)^{2}/2-2m)\right)x^{2m}$
$\displaystyle+\sum_{j=1}^{N_{\hat{\alpha},0}}\hat{a}_{\alpha,j}^{0}|x|^{-\hat{\alpha}+jp_{\hat{\alpha}}}+\sum_{m=1}^{\infty}\frac{(-1)^{m}}{(2m)!}\left(\sum_{j=1}^{N_{\hat{\alpha},0}}a_{\hat{\alpha},j}\zeta(1-\hat{\alpha}+jp_{\hat{\alpha}}-2m)+\sum_{m=1}^{N_{-1,1}}b_{\hat{\alpha},j}n^{2m}\right)x^{2m}$

and

$\displaystyle\mathcal{H}^{\alpha}u_{\alpha}(x)=$ $\displaystyle-
a_{\alpha,0}\left(\Gamma(2\alpha)\cos\left(\pi\alpha\right)-\Gamma(-1+\alpha-(1+\alpha)^{2}/2)\cos\left(\frac{\pi}{2}(1-\alpha+(1+\alpha)^{2}/2)\right)|x|^{1+\alpha+(1+\alpha)^{2}/2}\right)|x|^{-2\alpha}$
$\displaystyle-
a_{\alpha,0}\sum_{m=1}^{\infty}\frac{(-1)^{m}}{(2m)!}\left(\zeta(1-2\alpha-2m)-\zeta(2-\alpha+(1+\alpha)^{2}/2-2m)\right)x^{2m}$
$\displaystyle-\sum_{j=1}^{N_{\hat{\alpha},0}}\hat{A}_{\alpha,j}^{0}|x|^{-\alpha-\hat{\alpha}+jp_{\hat{\alpha}}}-\sum_{m=1}^{\infty}\frac{(-1)^{m}}{(2m)!}\left(\sum_{j=1}^{N_{\hat{\alpha},0}}a_{\hat{\alpha},j}\zeta(1-\alpha-\hat{\alpha}+jp_{\hat{\alpha}}-2m)+\sum_{m=1}^{N_{-1,1}}b_{\hat{\alpha},j}n^{2m+\alpha}\right)x^{2m},$

where

$\displaystyle\hat{a}_{\alpha,j}^{0}$
$\displaystyle=\Gamma(\hat{\alpha}-jp_{\hat{\alpha}})\cos\left(\frac{\pi}{2}(\hat{\alpha}-jp_{\hat{\alpha}})\right)a_{\hat{\alpha},j};$
$\displaystyle\hat{A}_{j}^{0}$
$\displaystyle=\Gamma(\alpha+\hat{\alpha}-jp_{\hat{\alpha}})\cos\left(\frac{\pi}{2}(\alpha+\hat{\alpha}-jp_{\hat{\alpha}})\right)a_{\hat{\alpha},j}.$

There are a couple of things that make these expansions tricky to handle. To
begin with the terms
$a_{\alpha,0}(\zeta(1-\alpha-2m)-\zeta(2+(1+\alpha)^{2}/2-2m))$ in
$u_{\alpha}$ have removable singularities at $\alpha=-1$, this is handled
using the approach described in Appendix A. For
$\mathcal{H}^{\alpha}[u_{\alpha}]$ the exponent for the term

$-a_{\alpha,0}\left(\Gamma(2\alpha)\cos\left(\pi\alpha\right)-\Gamma(-1+\alpha-(1+\alpha)^{2}/2)\cos\left(\frac{\pi}{2}(1-\alpha+(1+\alpha)^{2}/2)\right)|x|^{1+\alpha+(1+\alpha)^{2}/2}\right)|x|^{-2\alpha}$

approaches $2$ and its interaction with the $x^{2}$ term

$-\frac{1}{2}a_{\alpha,0}\left(\zeta(1-2\alpha-2)-\zeta(2-\alpha+(1+\alpha)^{2}/2-2)\right)x^{2}$

needs special care. A similar thing happens for
$\hat{A}_{\alpha,j}^{0}|x|^{-\alpha-\hat{\alpha}+jp_{\hat{\alpha}}}$ and the
corresponding $x^{2}$ terms. How this is handled is discussed in detail in
Section 10.2 and Appendix E. Bounding the tails of the sum

$\sum_{m=1}^{\infty}\frac{(-1)^{m}}{(2m)!}a_{\alpha,0}\left(\zeta(1-\alpha-2m)-\zeta(2+(1+\alpha)^{2}/2-2m)\right)x^{2m}$

and the corresponding one for $\mathcal{H}^{\alpha}[u_{\alpha}]$ requires a
bit extra work, this is done in Lemma C.7 and C.8.

For the precise values of $\hat{\alpha}$, $N_{\hat{\alpha},0}$ and $N_{-1,1}$
see Section 12.1.

###  4.3 Construction of $u_{\alpha}$ for $I_{3}$

Using the approximation from Section 4.1 and letting $\alpha$ tend to $0$ it
turns out that we need fewer and fewer terms to get a small defect. For
$\alpha$ close to $0$ (around $-1/6$ or closer) it is enough to take
$N_{\alpha,0}=2$ and $N_{\alpha,1}=0$, giving us the approximation

$u_{\alpha}(x)=\sum_{j=0}^{2}a_{\alpha,j}\tilde{C}_{1-\alpha+jp_{\alpha}}(x)$
(23)

and

$\mathcal{H}^{\alpha}[u_{\alpha}](x)=-\sum_{j=0}^{2}a_{\alpha,j}\tilde{C}_{1-2\alpha+jp_{\alpha}}(x).$
(24)

With this approximation the defect tends to zero as $\alpha\to 0$, which is
good. However, in Section 11.2.3 we will see that the value of $D_{\alpha}$ in
Proposition 2.2 tends to $1$, meaning that $(1-D_{\alpha})^{2}/4n_{\alpha}$
also tends to zero. This means that we can never compute an enclosure for
$D_{\alpha}$ and $\delta_{\alpha}$ on any interval $(-\delta_{2},0)$ such that
the inequality in Proposition 2.2 holds. Instead, we compute expansions in
$\alpha$ at $\alpha=0$ of both $D_{\alpha}$ and $\delta_{\alpha}$, using these
expansions we can then prove that the inequality holds on the interval
$I_{3}$.

In this case it is not enough to use numerical approximations of
$a_{\alpha,1}$, $a_{\alpha,2}$ and $p_{\alpha}$, we need to take into account
their dependence on $\alpha$. For $p_{\alpha}$ we use the defining equation
(17). As $\alpha\to 0$ we have $p_{\alpha}\to 1.426\dots$, this gives us that
the four leading terms (in $x$) in the expansion from Lemma 4.3 are given by,
ordered by their exponents,

$\displaystyle\left(\frac{1}{2}(a_{\alpha,0}^{0})^{2}-A_{\alpha,0}^{0}\right)|x|^{-2\alpha},$
$\displaystyle\left(a_{\alpha,0}^{0}a_{\alpha,1}^{0}-A_{\alpha,1}^{0}\right)|x|^{-2\alpha+p_{\alpha}},$
$\displaystyle\frac{1}{2}\left(\sum_{j=0}^{2}a_{\alpha,j}\zeta(-1-2\alpha+jp_{\alpha})\right)x^{2},$
$\displaystyle-\frac{1}{2}a_{\alpha,0}^{0}\left(\sum_{j=0}^{2}a_{\alpha,j}\zeta(-1-\alpha+jp_{\alpha})\right)x^{2-\alpha}.$

The choice of $a_{\alpha,0}$ makes the first term zero and the choice of
$p_{\alpha}$ makes the second term zero. We then want to take $a_{\alpha,1}$
and $a_{\alpha,2}$ to make the third and fourth term zero. Solving for
$a_{\alpha,1}$ and $a_{\alpha,2}$ we get the linear system

$\begin{cases}a_{\alpha,1}\zeta(-1-2\alpha+p_{\alpha})+a_{\alpha,2}\zeta(-1-2\alpha+2p_{\alpha})&=-a_{\alpha,0}\zeta(-1-2\alpha)\\\
a_{\alpha,1}\zeta(-1-\alpha+p_{\alpha})+a_{\alpha,2}\zeta(-1-\alpha+2p_{\alpha})&=-a_{\alpha,0}\zeta(-1-\alpha)\end{cases},$

with the solution

$\begin{cases}a_{\alpha,1}&=a_{\alpha,0}\frac{\zeta(-1-\alpha)\zeta(-1-2\alpha+2p_{\alpha})-\zeta(-1-2\alpha)\zeta(-1-\alpha+2p_{\alpha})}{\zeta(-1-2\alpha+p_{\alpha})\zeta(-1-\alpha+2p_{\alpha})-\zeta(-1-2\alpha+2p_{\alpha})\zeta(-1-\alpha+p_{\alpha})}\\\
a_{\alpha,2}&=-a_{\alpha,0}\frac{\zeta(-1-\alpha)\zeta(-1-2\alpha+p_{\alpha})-\zeta(-1-2\alpha)\zeta(-1-\alpha+p_{\alpha})}{\zeta(-1-2\alpha+p_{\alpha})\zeta(-1-\alpha+2p_{\alpha})-\zeta(-1-2\alpha+2p_{\alpha})\zeta(-1-\alpha+p_{\alpha})}\end{cases}.$
(25)

The following lemma gives us information about the asymptotic behavior of
$a_{\alpha,j}$.

######  Lemma 4.7.

The expansion of $a_{\alpha,0}$ at $\alpha=0$ is given by

$a_{\alpha,0}=\alpha+\mathcal{O}(\alpha^{3}).$

For $a_{\alpha,1}$ and $a_{\alpha,2}$ it is given by

$a_{\alpha,j}=a_{\alpha,j,1}\alpha+\mathcal{O}(\alpha^{2})$

with

$a_{\alpha,1,1}=\lim_{\alpha\to
0}\frac{\zeta(-1-\alpha)\zeta(-1-2\alpha+2p_{\alpha})-\zeta(-1-2\alpha)\zeta(-1-\alpha+2p_{\alpha})}{\zeta(-1-2\alpha+p_{\alpha})\zeta(-1-\alpha+2p_{\alpha})-\zeta(-1-2\alpha+2p_{\alpha})\zeta(-1-\alpha+p_{\alpha})}$

and

$a_{\alpha,2,1}=\lim_{\alpha\to
0}\frac{\zeta(-1-2\alpha)\zeta(-1-\alpha+p_{\alpha})-\zeta(-1-\alpha)\zeta(-1-\alpha+2p_{\alpha})}{\zeta(-1-2\alpha+p_{\alpha})\zeta(-1-\alpha+2p_{\alpha})-\zeta(-1-2\alpha+2p_{\alpha})\zeta(-1-\alpha+p_{\alpha})}.$

###### Proof.

The expansion for $a_{\alpha,0}$ follows from Equation (13) by using that
$\Gamma(\alpha)=\alpha^{-1}+\gamma+\mathcal{O}(\alpha)$, giving us

$a_{\alpha,0}=\frac{2\left((2\alpha)^{-1}+\gamma+\mathcal{O}(\alpha)\right)(1+\mathcal{O}(\alpha^{2}))}{(\alpha^{-1}+\gamma+\mathcal{O}(\alpha))^{2}(1+\mathcal{O}(\alpha^{2}))}=\frac{\alpha^{-1}+2\gamma+\mathcal{O}(\alpha)}{\alpha^{-2}+2\gamma\alpha^{-1}+\mathcal{O}(1)}=\alpha\frac{1+2\gamma\alpha+\mathcal{O}(\alpha^{2})}{1+2\gamma\alpha+\mathcal{O}(\alpha^{2})}=\alpha+\mathcal{O}(\alpha^{3}).$

The expansions for $a_{\alpha,1}$ and $a_{\alpha,2}$ follows directly from
Equation (25). ∎

Using this we can get information about the asymptotic behavior of
$u_{\alpha}$ and $\mathcal{H}^{\alpha}[u_{\alpha}]$.

######  Lemma 4.8.

For $x\in(0,\pi]$ we have the following expansions at $\alpha=0$:

$\displaystyle u_{\alpha}(x)$
$\displaystyle=1+b(x)\alpha+\mathcal{O}(\alpha^{2}),$
$\displaystyle\mathcal{H}^{\alpha}[u_{\alpha}](x)$
$\displaystyle=-\frac{1}{2}-b(x)\alpha+\mathcal{O}(\alpha^{2}),$

with

$b(x)=C_{1}(x)-\gamma+a_{\alpha,1,1}\tilde{C}_{1+p_{0}}(x)+a_{\alpha,2,1}\tilde{C}_{1+2p_{0}}(x).$

###### Proof.

For $j\geq 1$ the Clausen functions are all finite at $\alpha=0$ and from
Lemma 4.7 we therefore get

$\displaystyle a_{\alpha,1}\tilde{C}_{1-\alpha+p_{\alpha}}(x)$
$\displaystyle=a_{\alpha,1,1}\tilde{C}_{1+p_{0}}(x)\alpha+\mathcal{O}(\alpha^{2}),$
$\displaystyle a_{\alpha,2}\tilde{C}_{1-\alpha+2p_{\alpha}}(x)$
$\displaystyle=a_{\alpha,2,1}\tilde{C}_{1+2p_{0}}(x)\alpha+\mathcal{O}(\alpha^{2}),$
$\displaystyle a_{\alpha,1}\tilde{C}_{1-2\alpha+p_{\alpha}}(x)$
$\displaystyle=a_{\alpha,1,1}\tilde{C}_{1+p_{0}}(x)\alpha+\mathcal{O}(\alpha^{2}),$
$\displaystyle a_{\alpha,2}\tilde{C}_{1-2\alpha+p_{\alpha}}(x)$
$\displaystyle=a_{\alpha,2,1}\tilde{C}_{1+2p_{0}}(x)\alpha+\mathcal{O}(\alpha^{2}).$

For $j=0$ the function $\tilde{C}_{1-\alpha}(x)$ has a singularity at
$\alpha=0$. More precisely

$\tilde{C}_{1-\alpha}(x)=C_{1-\alpha}(x)-\zeta(1-\alpha),$

where for $0<x<2\pi$ the factor $C_{1-\alpha}(x)$ is finite, but
$\zeta(1-\alpha)$ has a singularity at $\alpha=0$, similarly for
$\tilde{C}_{1-2\alpha}(x)$. After multiplying with $a_{\alpha,0}$ the
singularity is removable. To begin with we have

$a_{\alpha,0}C_{1-\alpha}(x)=C_{1}(x)\alpha+\mathcal{O}(\alpha^{2}).$

Using the Laurent series of the zeta function we have

$\zeta(1-\alpha)=-\alpha^{-1}+\sum_{n=0}^{\infty}\frac{1}{n!}\gamma_{n}\alpha^{n}=-\alpha^{-1}+\gamma+\mathcal{O}(\alpha),$

where $\gamma_{n}$ is the Stieltjes constant and $\gamma=\gamma_{0}$ is
Euler’s constant, from which we get

$a_{\alpha,0}\zeta(1-\alpha)=(\alpha+\mathcal{O}(\alpha^{3}))(-\alpha^{-1}+\gamma+\mathcal{O}(\alpha))=-1+\gamma\alpha+\mathcal{O}(\alpha^{2}).$

Giving us

$a_{\alpha,0}\tilde{C}_{1-\alpha}(x)=1+(C_{1}(x)-\gamma)\alpha+\mathcal{O}(\alpha^{2}).$

In the same way we get

$a_{\alpha,0}\tilde{C}_{1-2\alpha}(x)=\frac{1}{2}+(C_{1}(x)-\gamma)\alpha+\mathcal{O}(\alpha^{2}).$

Combining all the above gives us

$u_{\alpha}(x)=1+(C_{1}(x)-\gamma+\tilde{C}_{1+p_{0}}(x)+\tilde{C}_{1+2p_{0}}(x))\alpha+\mathcal{O}(\alpha^{2})$

and

$\mathcal{H}^{\alpha}[u_{\alpha}](x)=-\frac{1}{2}-(C_{1}(x)-\gamma+\tilde{C}_{1+p_{0}}(x)+\tilde{C}_{1+2p_{0}}(x))\alpha+\mathcal{O}(\alpha^{2}),$

from which the result follows. ∎

###  4.4 Hybrid cases

While it is technically possible to use the construction from Section 4.1 to
handle the entire interval $I_{2}$ it is not computationally feasible. As
$\alpha$ gets close to either $-1$ or $0$ the bounds that needs to be computed
becomes harder and harder to handle.

For $\alpha$ near $0$ it is mainly bounding $\delta_{\alpha}$ that becomes
hard to deal with. It goes to zero as $\alpha\to 0$, which means that absolute
errors in the bound that are negligible when $\delta_{\alpha}$ is larger
quickly grow to become very large relative errors.

Near $\alpha=-1$ it is mainly the behavior of $F_{\alpha}(x)$ and
$\mathcal{T}_{\alpha}(x)$ (from Equation (31)) around $x=0$ that become
problematic. This is related to the powers for the terms
$\nu_{\alpha}|x|^{-\alpha}$ and $\mathcal{O}(|x|^{p})$ in Theorem 1.1 getting
closer and closer to each other, giving us less space to work with. Another
issue is, as mentioned in Section 4.2, that the two Clausen functions
$a_{\alpha,0}\tilde{C}_{1-\alpha}$ and
$a_{\alpha,1}\tilde{C}_{1-\alpha+p_{\alpha}}$ grow quickly in opposite
directions as $\alpha$ gets closer to $-1$ and as a result there are large
cancellations to handle.

For $\alpha$ near $0$ we use the approximation from Equation (23), but instead
of using expansions centered at $\alpha=0$ we use expansions centered at
$\alpha$ (or the midpoint, when computing with intervals). Compared to the
case $\alpha=0$ there are no removable singularities to handle, which makes
the process simpler.

For $\alpha$ near $-1$ we use the ideas from Section 4.2 for handling
$\alpha\in I_{1}$. Instead of using

$u_{\alpha}(x)=\sum_{j=0}^{N_{\alpha,0}}a_{\alpha,j}\tilde{C}_{1-\alpha+jp_{\alpha}}(x)+\sum_{n=1}^{N_{\alpha,1}}b_{\alpha,n}(\cos(nx)-1),$

from Equation (12) we modify it in the same spirit as Equation (20). More
precisely we use

$u_{\alpha}(x)=a_{\alpha,0}(\tilde{C}_{1-\alpha}(x)-\tilde{C}_{1-\alpha+p_{\alpha}}(x))+\hat{a}_{\alpha,1}\tilde{C}_{1-\alpha+p_{\alpha}}(x)+\sum_{j=2}^{N_{\alpha,0}}a_{\alpha,j}\tilde{C}_{1-\alpha+jp_{\alpha}}(x)+\sum_{n=1}^{N_{\alpha,1}}b_{\alpha,n}(\cos(nx)-1).$

If we take $\hat{a}_{\alpha,1}=a_{\alpha,0}+a_{\alpha,1}$ this is equivalent
to Equation (20) The benefit with this approach is that we don’t have to use
an enclosure of $a_{\alpha,0}$ in $\hat{a}_{\alpha,1}$, but can take a
numerical approximation instead.

In both cases we make adjustments to how the bounds are computed, the details
are given in the relevant sections.

##  5 Choice of $w_{\alpha}$

In the above section we discussed how to construct the approximation
$u_{\alpha}$ for the ansatz (7). We here discuss the choice of the weight
$w_{\alpha}$ in the ansatz. To get the statement in Theorem 1.1 we need
$w_{\alpha}(x)=\mathcal{O}(|x|^{p})$ for some $p$ satisfying $-\alpha<p\leq
1$. The natural choice is $w_{\alpha}(x)=|x|$, however this turns out to not
work for all values of $\alpha$.

The main obstruction when choosing $w_{\alpha}$ is its effect on the value of
$D_{\alpha}$. For Proposition 2.2 to apply we require that $D_{\alpha}<1$, if
we take $w_{\alpha}(x)=|x|$ this is not always the case. Using the procedure
described in Section 11, in particular lemma 11.7, we can compute a lower
bound of $D_{\alpha}$ by computing $\mathcal{T}_{\alpha}(0)$, introduced in
(31) and (30). A plot of $\mathcal{T}_{\alpha}(0)$ as a function of $\alpha$,
using $w_{\alpha}(x)=|x|$, is given in Figure 2. From this plot we can see
that with the weight $w_{\alpha}(x)=|x|$ we have $D_{\alpha}>1$ for $\alpha$
between $-1$ and some value slightly smaller than $-0.5$, in which case
Proposition 2.2 won’t apply.

Figure 2: Value of $\mathcal{T}_{\alpha}(0)$, defined by (31), as a function
of $\alpha$ using the weight $w_{\alpha}(x)=|x|$. This gives a lower bound of
$D_{\alpha}$ and since we need $D_{\alpha}<1$ it means that this choice of
weight does not work for all values of $\alpha$.

By taking $w_{\alpha}(x)=|x|^{p}$ with $-\alpha<p<1$ the value of $D_{\alpha}$
can be made smaller than $1$, so that Proposition 2.2 applies. The precise
value of $p$ to use depends on $\alpha$ and is made more precise in Section
12.2.

The limits $\alpha\to-1$ and $\alpha\to 0$ both require extra attention. For
$\alpha\to 0$ with the weight $w_{\alpha}(x)=|x|$ the value of $D_{\alpha}$
tends to $1$. However, in this case $\delta_{\alpha}$ tends to $0$ and by
controlling the rate of convergence of $D_{\alpha}$ and $\delta_{\alpha}$ it
is still possible to apply Proposition 2.2, see Section 12.3 for more details.

For $\alpha\to-1$ the situation is more complicated. It is not possible to use
the weight $w_{\alpha}(x)=|x|$ because $D_{\alpha}$ would be strictly greater
than $1$ close to $\alpha=-1$. It is also not possible to use the weight
$w_{\alpha}(x)=|x|^{p}$ with $p$ satisfying $-\alpha<p<1$ since in the limit
we would have need to have $p\to 1$, and again we find that $D_{\alpha}$ is
greater than $1$. A natural choice would be to add a log-factor,
$w_{\alpha}(x)=|x|\log(1+1/|x|)$. Lengthy calculations however shows that this
would not work either. Instead, we have to take $w_{\alpha}(x)$ to behave like
$|x|^{p}\log(1+1/|x|)$ with $p<1$ and $p\to 1$ as $\alpha\to-1$. More
precisely we take $w_{\alpha}(x)=|x|^{(1-\alpha)/2}\log(2e+1/|x|)$, the
constant $2e$ does not change the asymptotic behavior close to $x=0$ but does
change the non-asymptotic behavior in a way that turns out to make it slightly
easier to satisfy the required inequality in Proposition 2.2. That we have to
take such a complicated weight, compared to just $w_{\alpha}(x)=|x|^{p}$,
makes the analysis of $D_{\alpha}$ in Section 11 significantly harder.

For the hybrid cases we mimic what happens at the endpoints. Summarizing we
have the following choice for $w_{\alpha}$:

  * •

For $\alpha\in I_{1}$ we take
$w_{\alpha}(x)=|x|^{(1-\alpha)/2}\log(2e+1/|x|)$.

  * •

For $\alpha\in I_{2}$ we have three cases

    * –

Near $\alpha=-1$ we take $w_{\alpha}(x)=|x|^{p}\log(2e+1/|x|)$ with
$-\alpha<p<1$.

    * –

Near $\alpha=0$ we take $w_{\alpha}(x)=|x|$

    * –

In between we take $w_{\alpha}(x)=|x|^{p}$ with $-\alpha<p<1$.

The precise choice of $p$ as a function of $\alpha$ is given in Section 12.2.

  * •

For $\alpha\in I_{3}$ we take $w_{\alpha}(x)=|x|$.

##  6 Proof strategy

The main part of the proof of Theorem 1.1 is the computation of the bounds of
$n_{\alpha}$, $\delta_{\alpha}$ and $D_{\alpha}$ required to apply Proposition
2.2. Explicit bounds of these values are given in Section 12. In this section
we describe the general procedure for computing these bounds. The details for
the different cases are given in Sections 7 to 11.

The values we want to bound are given by

$n_{\alpha}=\sup_{x\in[0,\pi]}|N_{\alpha}(x)|,\quad\delta_{\alpha}=\sup_{x\in[0,\pi]}|F_{\alpha}(x)|,\quad
D_{\alpha}=\sup_{x\in[0,\pi]}|\mathcal{T}_{\alpha}(x)|.$

With $N_{\alpha}$ and $F_{\alpha}$ as in (10) and (9) in Section 2 and
$\mathcal{T}_{\alpha}$ as in (31) in Section 11. At $x=0$ the functions all
have removable singularities, for small values of $x$ the functions therefore
need extra care when evaluating them. For that reason the computation of the
supremum is split into two parts, one on the interval $[0,\epsilon]$ and one
on the interval $[\epsilon,\pi]$, for some $\epsilon$ depending on $\alpha$
and the function. On the interval $[0,\epsilon]$ we use a method of evaluation
which takes into account the removable singularity.

To enclose the supremum on the intervals $[0,\epsilon]$ and $[\epsilon,\pi]$
we make use of methods from interval arithmetic and rigorous numerics. In this
section we give a brief description of these methods and also discuss what we
require of $N_{\alpha}$, $F_{\alpha}$ and $\mathcal{T}_{\alpha}$ to be able to
apply these methods.

###  6.1 Enclosing the supremum

Consider the general problem of enclosing the maximum of $f$ on some interval
$I$ to some predetermined tolerance. Assume that we can evaluate $f$ using
interval arithmetic, i.e. given any interval $\textbf{{x}}\subseteq I$ we can
compute an interval enclosing the set
$f(\textbf{{x}})=\\{f(x):x\in\textbf{{x}}\\}$. The main idea is to iteratively
bisect the interval $I$ into smaller and smaller subintervals. At every
iteration we compute an enclosure of $f$ on each subinterval. From these
enclosures a lower bound of the maximum can be computed. We then discard all
subintervals for which the enclosure is less than the lower bound of the
maximum, the maximum cannot be attained there. For the remaining subintervals
we check if their enclosure satisfies the required tolerance, in that case we
don’t bisect them further. If there are any subintervals left we bisect them
and continue with the next iteration. In the end, either when there are no
subintervals left to bisect or we have reached some maximum number of
iterations (to guarantee that the procedure terminates), we return the maximum
of all subintervals that were not discarded. This is guaranteed to give an
enclosure of the maximum of $f$ on the interval.

If we are able to compute Taylor expansions of the function $f$ we can improve
the performance of this procedure significantly (see e.g. [17, 16] where a
similar approach is used). Consider a subinterval $I_{i}$, instead of
computing an enclosure of $f(I_{i})$ we compute a Taylor polynomial $P$ at the
midpoint and an enclosure $R$ of the remainder term such that $f(x)\in P(x)+R$
for $x\in I_{i}$. We then have

$\sup_{x\in I_{i}}f(x)\in\sup_{x\in I_{i}}P(x)+R.$ (26)

To compute $\sup_{x\in I_{i}}P(x)$ we isolate the roots of $P^{\prime}$ on
$I_{i}$ and evaluate $P$ on the roots as well as the endpoints of the
interval. In practice the computation of $R$ involves computing an enclosure
of the Taylor expansion of $f$ on the full interval $I_{i}$. Since this
includes the derivative we can as an extra optimization check if the
derivative is non-zero, in which case $f$ is monotone, and it is enough to
evaluate $f$ on either the left or the right endpoint of $I_{i}$, depending on
the sign of the derivative.

The above procedures can easily be adapted to instead compute the minimum of
$f$ on the interval, joining them together we can thus compute the extrema on
the interval. In some cases we don’t care about computing an enclosure of the
maximum, but only to prove that it is bounded by some value. Instead of using
a tolerance we then discard any subintervals for which the enclosure of the
maximum is less than the bound.

###  6.2 Evaluation of $N_{\alpha}$, $F_{\alpha}$ and $\mathcal{T}_{\alpha}$
on intervals

To be able to use the above method we need to be able to evaluate
$N_{\alpha}$, $F_{\alpha}$ and $\mathcal{T}_{\alpha}$ using interval
arithmetic. That is, given some interval $\textbf{{x}}\subseteq[0,\pi]$ we
need to be able to compute enclosures of $N_{\alpha}(\textbf{{x}})$,
$F_{\alpha}(\textbf{{x}})$ and $\mathcal{T}_{\alpha}(\textbf{{x}})$
respectively.

In Sections 7 to 11 we discuss how to evaluate these functions for (non-
interval) $x\in[0,\pi]$. This then needs to be extended to also work for
intervals $\textbf{{x}}\subseteq(0,\pi)$. In general this is straight forward
and handled directly by just replacing the point $x\in[0,\pi]$ with the
interval $\textbf{{x}}\subseteq(0,\pi)$ in the computations, the operations on
the interval are then done using Arb [40]. In some cases we can make use of
monotonicity properties of the function. For example when computing the root
$r_{\alpha,x}$ introduced in Lemma 11.2 for an interval
$\textbf{{x}}=[\underline{x},\overline{x}]$ we can use that it is decreasing
in $x$, giving us
$r_{\alpha,\textbf{{x}}}=[r_{\alpha,\overline{x}},r_{\alpha,\underline{x}}].$.

To cover all values of $\alpha$ in $(-1,0)$ we need to use interval values
also for $\alpha$, and not only for $x$. Similar to $x$ this is in general
done by just replacing $\alpha\in(-1,0)$ with
$\boldsymbol{\alpha}\subseteq(-1,0)$ in the computations.

To be able to use the improved version of the method for enclosing the
supremum discussed in Section 6.1 we need to be able to compute Taylor
expansions of $N_{\alpha}$, $F_{\alpha}$ and $\mathcal{T}_{\alpha}$. This is
possible in many, but not all cases. For $x\in[\epsilon,\pi]$ this is in
general straight forward for $N_{\alpha}$ and $F_{\alpha}$, for example the
methods in Appendix C.2 can be used for computing Taylor expansions of the
involved Clausen functions, but not for $\mathcal{T}_{\alpha}$.

For $x\in[0,\epsilon]$ the functions are in general expanded at $x=0$ and you
have to be slightly careful when computing Taylor expansions from this
expansion. For example, for $\frac{\tilde{C}_{1-\alpha}(x)}{x^{\alpha}}$ we
have the expansion

$\frac{\tilde{C}_{1-\alpha}(x)}{x^{\alpha}}=\Gamma(\alpha)\sin\left(\frac{\pi}{2}(1-\alpha)\right)+\sum_{m=1}^{\infty}(-1)^{m}\zeta(1-\alpha-2m)\frac{x^{2m+\alpha}}{(2m)!}.$

When computing it we need to bound the tail. If we let $R$ be an interval
bounding the remainder term given in Lemma C.5 on the interval $[0,\epsilon]$
then for $x\in[0,\epsilon]$ we have

$\frac{\tilde{C}_{1-\alpha}(x)}{x^{\alpha}}\in\Gamma(\alpha)\sin\left(\frac{\pi}{2}(1-\alpha)\right)+\sum_{m=1}^{M-1}(-1)^{m}\zeta(1-\alpha-2m)\frac{x^{2m+\alpha}}{(2m)!}+Rx^{2M+\alpha}.$

If we consider $R$ to be fixed then it is straight forward to compute Taylor
expansions in $x$ from this representation. This will in general not give us a
Taylor expansion of $\frac{\tilde{C}_{1-\alpha}(x)}{x^{\alpha}}$ since we have
lost the remainder terms dependence on $x$. When enclosing the supremum this
is however not an issue, we can apply the method on

$\Gamma(\alpha)\sin\left(\frac{\pi}{2}(1-\alpha)\right)+\sum_{m=1}^{M-1}(-1)^{m}\zeta(1-\alpha-2m)\frac{x^{2m+\alpha}}{(2m)!}+Rx^{2M+\alpha}$

directly.

##  7 Evaluation of $u_{\alpha}$ and $\mathcal{H}^{\alpha}[u_{\alpha}]$

To compute bounds for $D_{\alpha}$, $\delta_{\alpha}$ and $n_{\alpha}$ we need
to be able to compute enclosures of $u_{\alpha}(x)$ and
$\mathcal{H}^{\alpha}[u_{\alpha}](x)$. In this section we describe how this is
done for the approximations used on the intervals $I_{1}$, $I_{2}$ and $I_{3}$
respectively.

###  7.1 Evaluation of $u_{\alpha}$ and $\mathcal{H}^{\alpha}[u_{\alpha}]$ for
$I_{2}$

In this case the expressions for $u_{\alpha}$ and
$\mathcal{H}^{\alpha}[u_{\alpha}]$ are given by (12) and (14). The parameter
$a_{\alpha,0}$ is given by (13), which is easily enclosed for a given
$\alpha$, and $p_{\alpha}$, $a_{\alpha,j}$ for $j\geq 1$ and $b_{\alpha,n}$
are fixed, numerically determined, numbers. Evaluation is straight forward,
using the approach in Appendix C to enclose the Clausen functions.

###  7.2 Evaluation of $u_{\alpha}$ and $\mathcal{H}^{\alpha}[u_{\alpha}]$ for
$I_{1}$

In this case the expressions for $u_{\alpha}$ and
$\mathcal{H}^{\alpha}[u_{\alpha}]$ are given by (20) and (22). Most of the
parameters are fixed, numerically determined, numbers, these parts of the
functions are straight forward to enclose. The only problematic part are the
leading terms,

$a_{\alpha,0}(\tilde{C}_{1-\alpha}(x)-\tilde{C}_{2+(1+\alpha)^{2}/2}(x))\text{
and
}-a_{\alpha,0}(\tilde{C}_{1-2\alpha}(x)-\tilde{C}_{2-\alpha+(1+\alpha)^{2}/2}(x)),$

which both have removable singularities at $\alpha=-1$. They are handled by
rewriting them as

$a_{\alpha,0}(\tilde{C}_{1-\alpha}(x)-\tilde{C}_{2+(1+\alpha)^{2}/2}(x))=((\alpha+1)a_{\alpha,0})\cdot\frac{\tilde{C}_{1-\alpha}(x)-\tilde{C}_{2+(1+\alpha)^{2}/2}(x)}{\alpha+1},$

as well as

$-a_{\alpha,0}(\tilde{C}_{1-2\alpha}(x)-\tilde{C}_{2-\alpha+(1+\alpha)^{2}/2}(x))=-((\alpha+1)a_{\alpha,0})\cdot\frac{\tilde{C}_{1-2\alpha}(x)-\tilde{C}_{2-\alpha+(1+\alpha)^{2}/2}(x)}{\alpha+1}$

and using the approach in Appendix A to handle the removable singularities.

###  7.3 Evaluation of $u_{\alpha}$ and $\mathcal{H}^{\alpha}[u_{\alpha}]$ for
$I_{3}$

In this case the expressions for $u_{\alpha}$ and
$\mathcal{H}^{\alpha}[u_{\alpha}]$ are, as for $I_{2}$, given by (23) and
(24). The parameters $p_{\alpha}$ and $a_{\alpha,j}$ are not fixed numbers,
but all depend on $\alpha$. Moreover, it is not enough to just compute
enclosures of $u_{\alpha}(x)$ and $\mathcal{H}^{\alpha}[u_{\alpha}]$, we need
their expansions in $\alpha$. For this we make use of Taylor models [44],
which we give a brief introduction of in Appendix B, centered at $\alpha=0$.
For $a_{\alpha,0}$, $a_{\alpha,1}$ and $a_{\alpha,2}$ we have explicit
formulas in (13) and (25) and Taylor models can be computed directly from
them, there are several removable singularities which are handled as described
in Appendix A. The value of $p_{\alpha}$ is given implicitly as a function of
$\alpha$ by (17), from this we can compute a Taylor model using the approach
described in Appendix B.2. Once these Taylor models are computed it is
straight forward to compute Taylor models of (23) and (24).

###  7.4 Evaluation of $u_{\alpha}$ and $\mathcal{H}^{\alpha}[u_{\alpha}]$ in
hybrid cases

For $\alpha$ near $-1$ the only part that needs special care is the evaluation
of $\tilde{C}_{1-\alpha}(x)-\tilde{C}_{1-\alpha+p_{\alpha}}(x)$ for
$u_{\alpha}$ and
$\tilde{C}_{1-2\alpha}(x)-\tilde{C}_{1-2\alpha+p_{\alpha}}(x)$ for
$\mathcal{H}^{\alpha}[u_{\alpha}]$, in both cases due to the large
cancellations between the two terms. For interval arguments x and s with
midpoints $x_{0}$ and $s_{0}$ respectively we compute
$\tilde{C}_{\textbf{{s}}}(\textbf{{x}})-\tilde{C}_{\textbf{{s}}+p_{\alpha}}(\textbf{{x}})$
using a midpoint approximation as

$\tilde{C}_{\textbf{{s}}}(\textbf{{x}})-\tilde{C}_{\textbf{{s}}+p_{\alpha}}(\textbf{{x}})=(\tilde{C}_{s_{0}}(x_{0})-\tilde{C}_{s_{0}+p_{\alpha}}(x_{0}))+(\textbf{{s}}-s_{0})(\tilde{C}_{\textbf{{s}}}^{(1)}(x_{0})-\tilde{C}_{\textbf{{s}}+p_{\alpha}}^{(1)}(x_{0}))-(\textbf{{x}}-x_{0})(\tilde{S}_{\textbf{{s}}-1}(\textbf{{x}})-\tilde{S}_{\textbf{{s}}+p_{\alpha}-1}(\textbf{{x}})).$

For $\alpha$ near $0$ we follow exactly the same approach as for $\alpha\in
I_{3}$, the only difference being that the Taylor models are centered at
$\alpha$ (or the midpoint of $\boldsymbol{\alpha}$ in the case of intervals)
instead of at $0$.

##  8 Division by $u_{\alpha}$

The computations of the values $n_{\alpha}$, $\delta_{\alpha}$ and
$D_{\alpha}$ all involve functions given by fractions where the denominator
contains $u_{\alpha}$. For $x>0$ this is not an issue since the denominators
are non-zero, we can enclose the numerator and denominator separately and then
perform the division. For $x=0$ we get a removable singularity and to handle
that we need to understand the asymptotic behavior of $u_{\alpha}(x)$ at
$x=0$. For this we start of with the following, straight forward, lemma

######  Lemma 8.1.

For $-1<\alpha<0$ the function $\frac{|x|^{-\alpha}}{u_{\alpha}(x)}$ is non-
zero and bounded at $x=0$.

###### Proof.

From the asymptotic expansion of $u_{\alpha}(x)$ at $x=0$ we have

$u_{\alpha}(x)=a_{\alpha,0}\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)|x|^{-\alpha}+o(|x|^{-\alpha}).$

This gives us

$\frac{|x|^{-\alpha}}{u_{\alpha}(x)}=\frac{1}{a_{\alpha,0}\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)+o(1)}.$

This is non-zero and bounded as long as

$a_{\alpha,0}\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)$

is non-zero and bounded. We get

$a_{\alpha,0}\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)=\frac{2\Gamma(2\alpha)\cos\left(\pi\alpha\right)}{\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)}$

The denominator is easily seen to be non-zero and bounded for
$\alpha\in(-1,0)$. For the numerator we can rewrite it as

$2\Gamma(2\alpha)\cos\left(\pi\alpha\right)=\frac{2\Gamma(2\alpha+2)\cos\left(\pi\alpha\right)}{2\alpha(2\alpha+1)}=\frac{\Gamma(2\alpha+2)}{\alpha}\frac{\sin\left(\frac{\pi}{2}(2\alpha+1)\right)}{(2\alpha+1)},$

which is bounded and non-zero. ∎

Computing $\frac{|x|^{-\alpha}}{u_{\alpha}(x)}$ near $x=0$ is straight
forward. Compute the expansion of $u_{\alpha}(x)$ at $x=0$ and subtract
$-\alpha$ from the exponents of all terms in the expansion, this gives an
expansion where all exponents are non-negative. Evaluate the expansion at $x$
and then compute the inverse.

While Lemma 8.1 is valid for the full interval $(-1,0)$ the bound is not
uniform in $\alpha$. For $\alpha\to 0$ it converges to $1$, and we are fine,
but for $\alpha\to-1$ the value of $\frac{x^{-\alpha}}{u_{\alpha}(x)}$ goes to
zero, and we need an understanding of the rate in which it does so. For this
we use the following, modified version of the lemma

######  Lemma 8.2.

For $\alpha\in I_{1}$ the function

$\frac{\Gamma(1+\alpha)|x|^{-\alpha}(1-|x|^{1+\alpha+(1+\alpha)^{2}/2})}{u_{\alpha}(x)}$

is non-zero and uniformly bounded in $\alpha$ at $x=0$.

###### Proof.

Since the function is even with respect to $x$ we can take $x\geq 0$.

From Lemma 4.6 we have that the leading term in the asymptotic expansion at
$x=0$ for $u_{\alpha}$ is given by

$a_{\alpha,0}\left(\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)-\Gamma(-1-(1+\alpha)^{2}/2)\cos\left(\frac{\pi}{2}(1+(1+\alpha)^{2}/2)\right)x^{1+\alpha+(1+\alpha)^{2}/2}\right)x^{-\alpha}.$

We therefore study the asymptotic behavior of

$\frac{\Gamma(1+\alpha)x^{-\alpha}(1-x^{1+\alpha+(1+\alpha)^{2}/2})}{a_{\alpha,0}\left(\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)-\Gamma(-1-(1+\alpha)^{2}/2)\cos\left(\frac{\pi}{2}(1+(1+\alpha)^{2}/2)\right)x^{1+\alpha+(1+\alpha)^{2}/2}\right)x^{-\alpha}}.$
(27)

We can cancel the $x^{-\alpha}$ factor and split it as

$\frac{\Gamma(1+\alpha)}{a_{\alpha,0}}\frac{1-x^{1+\alpha+(1+\alpha)^{2}/2}}{\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)-\Gamma(-1-(1+\alpha)^{2}/2)\cos\left(\frac{\pi}{2}(1+(1+\alpha)^{2}/2)\right)x^{1+\alpha+(1+\alpha)^{2}/2}}.$

From the definition of $a_{\alpha,0}$ as given in Equation (13) we get that
the first factor is given by

$\frac{\Gamma(1+\alpha)\Gamma(\alpha)^{2}\cos\left(\frac{\pi}{2}\alpha\right)^{2}}{2\Gamma(2\alpha)\cos\left(\pi\alpha\right)}=\frac{\alpha\Gamma(\alpha)^{3}\cos\left(\frac{\pi}{2}\alpha\right)^{2}}{2\Gamma(2\alpha)\cos\left(\pi\alpha\right)}.$

This has a removable singularity at $\alpha=-1$. It is then easily checked to
be non-zero.

For the second factor we let
$c(\alpha)=\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)$ and
$\tilde{p}_{\alpha}=1+\alpha+(1+\alpha)^{2}/2$, allowing us to write it as

$\frac{1-x^{\tilde{p}_{\alpha}}}{c(\alpha)-c(\alpha-\tilde{p}_{\alpha})x^{\tilde{p}_{\alpha}}}=\frac{1}{c(\alpha)}\frac{1-x^{\tilde{p}_{\alpha}}}{1-\frac{c(\alpha-\tilde{p}_{\alpha})}{c(\alpha)}x^{\tilde{p}_{\alpha}}}$

Here $c(\alpha)$ has a removable singularity, which can be handled in the same
way as the one for $\frac{\Gamma(1+\alpha)}{a_{0}}$. Again it can easily be
checked to be non-zero, so that $\frac{1}{c(\alpha)}$ is bounded. What remains
to handle is thus

$\frac{1-x^{\tilde{p}_{\alpha}}}{1-\frac{c(\alpha-\tilde{p}_{\alpha})}{c(\alpha)}x^{\tilde{p}_{\alpha}}}.$

Computing an enclosure of the derivative of
$\frac{c(\alpha-\tilde{p}_{\alpha})}{c(\alpha)}$ allows us to check that it is
negative. This together with the fact that
$\lim_{\alpha\to-1}\frac{c(\alpha-\tilde{p}_{\alpha})}{c(\alpha)}=1$ means
that an upper bound is given by

$\frac{1-x^{\tilde{p}_{\alpha}}}{1-x^{\tilde{p}_{\alpha}}}=1.$

For a lower bound we note that since
$\frac{c(\alpha-\tilde{p}_{\alpha})}{c(\alpha)}<1$ it is decreasing in $x$ as
long as $x<1$. It is therefore enough to check the value at some $0<x_{0}<1$
to get a lower bound. For this fixed $x_{0}>0$ we can handle the removable
singularity of

$\frac{1-x_{0}^{\tilde{p}_{\alpha}}}{1-\frac{c(\alpha-\tilde{p}_{\alpha})}{c(\alpha)}x_{0}^{\tilde{p}_{\alpha}}}$

at $\alpha=-1$ to compute an enclosure of the lower bound. ∎

The proof of this lemma also tells us how to compute an upper bound of

$\frac{\Gamma(1+\alpha)|x|^{-\alpha}(1-|x|^{1+\alpha+(1+\alpha)^{2}/2})}{u_{\alpha}(x)}.$

It is enough to check that the terms in the asymptotic expansion of
$u_{\alpha}$ that we skipped are positive, in that case Equation (27) gives an
upper bound. For $x$ not overlapping zero we instead split it as

$\left(\Gamma(1+\alpha)(1-|x|^{1+\alpha+(1+\alpha)^{2}/2})\right)\cdot\frac{|x|^{-\alpha}}{u_{\alpha}(x)}$

and handle the removable singularity for the first factor and enclose the
second factor directly using the asymptotic expansion.

##  9 Evaluation of $N_{\alpha}$

Recall that

$n_{\alpha}:=\|N_{\alpha}\|_{L^{\infty}(\mathbb{T})}=\sup_{x\in[0,\pi]}\left|N_{\alpha}(x)\right|$

where

$N_{\alpha}(x)=\frac{w_{\alpha}(x)}{2u_{\alpha}(x)}.$

In this section we discuss how to evaluate $N_{\alpha}$. The interval
$[0,\pi]$ is split into two subintervals, $[0,\epsilon]$ and $[\epsilon,\pi]$.
On the interval $[0,\epsilon]$ we make use of asymptotic expansions, on
$[\epsilon,\pi]$ the function is evaluated directly. The value of $\epsilon$
depends on $\alpha$ and is discussed more in Section 12.

We start by describe the procedure for evaluation for $\alpha\in I_{2}$ and
then discuss the alterations required to handle $I_{1}$ and $I_{3}$.

###  9.1 Evaluation of $N_{\alpha}$ for $I_{2}$

For $x\in[\epsilon,\pi]$ both $w_{\alpha}(x)$, $u_{\alpha}(x)$ and the
resulting quotient are straight forward to evaluate.

For $x\in[0,\epsilon]$ we use that the weight is given by
$w_{\alpha}(x)=x^{p}$ and write the function as

$N_{\alpha}(x)=\frac{1}{2}x^{p+\alpha}\cdot\frac{x^{-\alpha}}{u_{\alpha}(x)}.$

The first factor is easily enclosed and we use Lemma 8.1 for enclosing the
second factor.

###  9.2 Evaluation of $N_{\alpha}$ for $I_{1}$

For $x\in[\epsilon,\pi]$ we make one optimization compared to $I_{2}$, instead
of computing an enclosure we only compute an upper bound. For this we use that
$u_{\alpha}(x)\geq u_{-1}(x)$ for $\alpha\in I_{1}$ and hence

$N_{\alpha}(x)\leq\frac{w_{\alpha}(x)}{2u_{-1}(x)},$

as long as $u_{-1}(x)$ is positive.

For $x\in[0,\epsilon]$ we use that the weight is given by
$w_{\alpha}(x)=x^{(1-\alpha)/2}\log(2e+1/x)$ and write the function as

$\displaystyle N_{\alpha}(x)$
$\displaystyle=\frac{1}{2}\frac{w_{\alpha}(x)}{\Gamma(1+\alpha)x^{-\alpha}(1-x^{1+\alpha+(1+\alpha)^{2}/2})}\cdot\frac{\Gamma(1+\alpha)x^{-\alpha}(1-x^{1+\alpha+(1+\alpha)^{2}/2})}{u_{\alpha}(x)}$
$\displaystyle=\frac{1}{\Gamma(2+\alpha)}\cdot\frac{\log(2e+1/x)}{2\log(1/x)}\cdot\frac{x^{(1+\alpha)/2}\log(1/x)(1+\alpha)}{1-x^{1+\alpha+(1+\alpha)^{2}/2}}\cdot\frac{\Gamma(1+\alpha)x^{-\alpha}(1-x^{1+\alpha+(1+\alpha)^{2}/2})}{u_{\alpha}(x)}.$

The first factor is well-behaved, the second factor can be enclosed using that
it is increasing in $x$ and the limit as $x\to 0$ is $\frac{1}{2}$. The fourth
factor can be enclosed using Lemma 8.2. For the third factor we get a bound
from the following lemma

######  Lemma 9.1.

Let $-1<\alpha<0$ and $0<x<1$, then

$0<\frac{x^{(1+\alpha)/2}\log(1/x)(1+\alpha)}{1-x^{1+\alpha+(1+\alpha)^{2}/2}}<1.$

###### Proof.

The lower bound follows directly from that all the factors are positive. What
remains is proving the upper bound.

If we let $s=(1+\alpha)\log x$ we can write the quotient as

$\frac{-se^{s/2}}{1-e^{s(1+(1+\alpha)/2)}}.$

From $-1<\alpha<0$ and $0<x<1$ we get that $s<0$. We want to prove that

$-se^{s/2}<1-e^{s(1+(1+\alpha)/2)},$

which is equivalent to proving that

$f(s)=1-e^{s(1+(1+\alpha)/2)}+se^{s/2}$

is positive for $s<0$. We have $\lim_{s\to 0}f(s)=0$, and it is hence enough
to prove that $f$ is decreasing for $s<0$. The derivative is given by

$f^{\prime}(s)=-(1+(1+\alpha)/2)e^{s(1+(1+\alpha)/2)}+\left(1+\frac{s}{2}\right)e^{s/2}=e^{s/2}\left(1+\frac{s}{2}-(1+(1+\alpha)/2)e^{s(1+(1+\alpha))/2}\right)$

so it is enough to prove that

$g(s)=1+\frac{s}{2}-(1+(1+\alpha)/2)e^{s(1+(1+\alpha))/2}$

is negative for $s<0$. We have $g(0)=-\frac{1+\alpha}{2}<0$ and
$\lim_{s\to-\infty}g(s)=-\infty$. The unique critical point with respect to
$s$ is

$-2\frac{\log\left((1+(1+\alpha)/2)(1+(1+\alpha))\right)}{1+(1+\alpha)},$

which is negative. Inserting this into $g$ gives us

$1-\frac{\log\left((1+(1+\alpha)/2)(1+(1+\alpha))\right)}{1+(1+\alpha)}-(1+(1+\alpha)/2)e^{-\log\left((1+(1+\alpha)/2)(1+(1+\alpha))\right)}\\\
=1-\frac{1+\log\left((1+(1+\alpha)/2)(1+(1+\alpha))\right)}{1+(1+\alpha)}\\\
=1-\frac{1+\log(1+(1+\alpha)/2)+\log(1+(1+\alpha))}{1+(1+\alpha)}$

For this to be negative we need

$\log(1+(1+\alpha)/2)+\log(1+(1+\alpha))>1+\alpha,$

which holds for $-1<\alpha<0$. It follows that $g(s)$ is negative for $s<0$
and hence $f(s)$ is decreasing in $s$ and from $\lim_{s\to 0}f(s)=0$ we then
get that $f(s)$ is positive for $s<0$ and the result follows. ∎

###  9.3 Evaluation of $N_{\alpha}$ for $I_{3}$

Compared to $\delta_{\alpha}$ and $D_{\alpha}$ we don’t need to compute an
expansion in $\alpha$ of $n_{\alpha}$ for $I_{3}$, it is therefore enough to
just compute enclosures of $N_{\alpha}(x)$. The asymptotic expansions perform
very well in this case and it is possible to take $\epsilon=\pi$, we hence
only have to consider the interval $[0,\epsilon]$. The evaluation is done in
the same way as for $I_{2}$ (with $p=1$), by rewriting it as

$N_{\alpha}(x)=\frac{1}{2}x^{1+\alpha}\cdot\frac{x^{-\alpha}}{u_{\alpha}(x)}.$

and using Lemma 8.1.

###  9.4 Evaluation of $N_{\alpha}$ in hybrid cases

For $\alpha$ near $-1$ we don’t have to do anything special and follow the
same approach as in Section 9.1. For $\alpha$ near $0$ the approach is the
same as for $\alpha\in I_{3}$.

##  10 Evaluation of $F_{\alpha}$

Recall that

$\delta_{\alpha}=\|F_{\alpha}\|_{L^{\infty}(\mathbb{T})}=\sup_{x\in[0,\pi]}|F_{\alpha}(x)|,$

where

$F_{\alpha}(x)=\frac{\mathcal{H}^{\alpha}[u_{\alpha}](x)+\frac{1}{2}u_{\alpha}(x)^{2}}{w_{\alpha}(x)u_{\alpha}(x)}.$

In this section we discuss how to evaluate $F_{\alpha}$. The interval
$[0,\pi]$ is split into two subintervals, $[0,\epsilon]$ and $[\epsilon,\pi]$.
On the interval $[0,\epsilon]$ we make use of asymptotic expansions, on
$[\epsilon,\pi]$ the function is evaluated directly. The value of $\epsilon$
depends on $\alpha$ and is discussed more in Section 12.

We start by describe the procedure for evaluation for $\alpha\in I_{2}$ and
then discuss the alterations required to handle $I_{1}$ and $I_{3}$.

###  10.1 Evaluation of $F_{\alpha}$ for $I_{2}$

For $x\in[\epsilon,\pi]$ we evaluate each part separately. That is, we compute
$w_{\alpha}(x)$, $u_{\alpha}(x)$ and $\mathcal{H}^{\alpha}[u_{\alpha}](x)$, it
is then straight forward to compute $F_{\alpha}(x)$ from these values.

For $x\in[0,\epsilon]$ we use that the weight is given by
$w_{\alpha}(x)=x^{p}$ and write the function as

$F_{\alpha}(x)=\frac{x^{-\alpha}}{u_{\alpha}(x)}\cdot\frac{\mathcal{H}^{\alpha}[u_{\alpha}](x)+\frac{1}{2}u_{\alpha}(x)^{2}}{x^{p-\alpha}}.$

The first factor is enclosed using Lemma 8.1. For the second factor we use the
expansion from Lemma 4.3 and explicitly cancel the $x^{p-\alpha}$ from the
denominator. Note that since $-2\alpha<p-\alpha$ we need the leading term in
the expansion to be identically zero, hence our choice of $a_{\alpha,0}$. For
the second term in the expansion we have $p-\alpha<-2\alpha+p_{\alpha}$, so it
is bounded at $x=0$, as required.

###  10.2 Evaluation of $F_{\alpha}$ for $I_{1}$

For $x\in[\epsilon,\pi]$ we make one optimization compared to $I_{2}$, instead
of computing an enclosure we only compute an upper bound. For this we use that
$u_{\alpha}(x)\geq u_{-1}(x)$ for $\alpha\in I_{1}$ and hence

$|F_{\alpha}(x)|\leq\left|\frac{\mathcal{H}^{\alpha}[u_{\alpha}](x)+\frac{1}{2}u_{\alpha}(x)^{2}}{w_{\alpha}(x)u_{-1}(x)}\right|,$

as long as $u_{-1}(x)$ is positive.

For $x\in[0,\epsilon]$ we use that the weight is given by
$w_{\alpha}(x)=x^{(1-\alpha)/2}\log(2e+1/x)$ and write the function as

$F_{\alpha}(x)=\frac{\log(1/x)}{\log(2e+1/x)}\cdot\frac{\Gamma(1+\alpha)x^{-\alpha}(1-x^{1+\alpha+(1+\alpha)^{2}/2})}{u_{\alpha}(x)}\cdot\frac{\mathcal{H}^{\alpha}[u_{\alpha}](x)+\frac{1}{2}u_{\alpha}(x)^{2}}{\Gamma(1+\alpha)\log(1/x)(1-x^{1+\alpha+(1+\alpha)^{2}/2})x^{(1-\alpha)/2-\alpha}}.$

The first two factors are handled in the same way as when handling
$N_{\alpha}$ in Section 9.2. For the third factor we use that $(1-\alpha)/2<1$
and hence, if we require $\epsilon\leq 1$, an upper bound, in absolute value,
is given by the absolute value of

$\frac{\mathcal{H}^{\alpha}[u_{\alpha}](x)+\frac{1}{2}u_{\alpha}(x)^{2}}{\Gamma(1+\alpha)\log(1/x)(1-x^{1+\alpha+(1+\alpha)^{2}/2})x^{1-\alpha}}.$
(28)

A bound for this is given in Appendix E.

###  10.3 Evaluation of $F_{\alpha}$ for $I_{3}$

In this case we need to not only compute an enclosure of $F_{\alpha}(x)$, but
understand it behavior in $\alpha$. We therefore compute Taylor models of
degree $1$ centered at $\alpha=0$.

We start with the following lemma that gives us information about the first
two terms in the Taylor model.

######  Lemma 10.1.

For $x\in[0,\pi]$ the constant and linear terms in the expansion at $\alpha=0$
of $F_{\alpha}(x)$ are both zero.

###### Proof.

From lemma 4.8 we get

$F_{\alpha}(x)=\frac{\left(-\frac{1}{2}-b(x)\alpha+\mathcal{O}(\alpha^{2})\right)+\frac{1}{2}\left(1+b(x)\alpha+\mathcal{O}(\alpha^{2})\right)^{2}}{x(1+b(x)\alpha+\mathcal{O}(\alpha^{2}))}=\frac{\mathcal{O}(\alpha^{2})}{x(1+b(x)\alpha+\mathcal{O}(\alpha^{2}))}=\mathcal{O}(\alpha^{2}),$

which gives us the result. ∎

For $x\in[\epsilon,\pi]$ we compute the Taylor models of $u_{\alpha}(x)$ and
$\mathcal{H}^{\alpha}[u_{\alpha}](x)$ using the approach described in Section
7.3. For this case we have $w_{\alpha}(x)=|x|$ and hence $w_{\alpha}$ does not
depend on $\alpha$ and its corresponding Taylor model is just a constant.

For $x\in[0,\epsilon]$ we, similarly to for $I_{2}$, write the function as

$F_{\alpha}(x)=\frac{x^{-\alpha}}{u_{\alpha}(x)}\cdot\frac{\mathcal{H}^{\alpha}[u_{\alpha}](x)+\frac{1}{2}u_{\alpha}(x)^{2}}{x^{1-\alpha}}$

and compute Taylor models of the two factors independently, which are then
multiplied. See Appendix B.1 for how to compute Taylor models of expansions in
$x$.

###  10.4 Evaluation of $F_{\alpha}$ in hybrid cases

Near $\alpha=-1$ we don’t have to do anything special for
$x\in[\epsilon,\pi]$, the approach is thus the same as described in Section
10.1. For $x\in[0,\epsilon]$ it is in principle the same as well, but to get
good enclosures it requires slightly more care. The reasons this needs to be
done is that for $\alpha$ near $-1$ the factor

$\frac{\mathcal{H}^{\alpha}[u_{\alpha}](x)+\frac{1}{2}u_{\alpha}(x)^{2}}{x^{p-\alpha}}$

tends to zero very slowly and there are several terms in the asymptotic
expansions between which there are large cancellations. The main adjustment is
to more carefully handle the cancellations between the leading terms of
$\mathcal{H}^{\alpha}[u_{\alpha}](x)$ and $\frac{1}{2}u_{\alpha}(x)^{2}$. We
also make heavy use of Lemma C.3 when evaluating the expansion of
$\mathcal{H}^{\alpha}[u_{\alpha}](x)$.

Near $\alpha=0$ the approach is similar to that used for $\alpha\in I_{3}$.
Instead of the Taylor models being centered at zero they are centered at (the
midpoint) of $\alpha$. Since we in the end don’t need an expansion in $\alpha$
we enclose the computed Taylor models over the interval, giving us an
enclosure of $F_{\alpha}(x)$.

##  11 Analysis of $T_{\alpha}$

In this section we give more details about the operator $T_{\alpha}$ defined
by

$T_{\alpha}v=-\frac{1}{w_{\alpha}u_{\alpha}}\mathcal{H}^{\alpha}[w_{\alpha}v]$

and discuss how to bound $D_{\alpha}:=\|T_{\alpha}\|$.

The operator $\mathcal{H}^{\alpha}$ is defined by

$\mathcal{H}^{\alpha}[v](x)=|D|^{\alpha}[v](x)-|D|^{\alpha}[v](0).$

For even functions $v(x)$ and $0<x<\pi$ it has the integral representation

$\mathcal{H}^{\alpha}[v](x)=-\frac{1}{\pi}\int_{0}^{\pi}(C_{-\alpha}(x-y)+C_{-\alpha}(x+y)-2C_{-\alpha}(y))v(y)\
dy.$

This gives us

$T_{\alpha}[v](x)=\frac{1}{\pi
w_{\alpha}(x)u_{\alpha}(x)}\int_{0}^{\pi}(C_{-\alpha}(x-y)+C_{-\alpha}(x+y)-2C_{-\alpha}(y))w_{\alpha}(y)v(y)\
dy.$

Using the above expressions it is standard that the norm of $T_{\alpha}$ is
given by

$D_{\alpha}=\|T_{\alpha}\|=\sup_{0<x<\pi}\frac{1}{\pi|w_{\alpha}(x)||u_{\alpha}(x)|}\int_{0}^{\pi}|C_{-\alpha}(x-y)+C_{-\alpha}(x+y)-2C_{-\alpha}(y)||w_{\alpha}(y)|\
dy.$ (29)

Let

$I_{\alpha}(x,y)=C_{-\alpha}(x-y)+C_{-\alpha}(x+y)-2C_{-\alpha}(y)$

and

$U_{\alpha}(x)=\int_{0}^{\pi}|I_{\alpha}(x,y)|w_{\alpha}(y)\ dt,$

where we have removed the absolute value around $w_{\alpha}$ since it is
positive. We are then interested in computing

$D_{\alpha}=\|T_{\alpha}\|=\sup_{0<x<\pi}\frac{U_{\alpha}(x)}{\pi|w_{\alpha}(x)||u_{\alpha}(x)|}.$
(30)

We use the notation

$\mathcal{T}_{\alpha}(x)=\frac{U_{\alpha}(x)}{\pi|w_{\alpha}(x)||u_{\alpha}(x)|}.$
(31)

###  11.1 Properties of $U_{\alpha}$

Before discussing how to evaluate $\mathcal{T}_{\alpha}$ we give some general
properties about $U_{\alpha}$ that will be useful.

The integrand of $U_{\alpha}(x)$ has a singularity at $y=x$. It is therefore
natural to split the integral into two parts

$U_{\alpha,1}(x)=\int_{0}^{x}|I_{\alpha}(x,y)|w_{\alpha}(y)\ dy,\quad
U_{\alpha,2}(x)=\int_{x}^{\pi}|I_{\alpha}(x,y)|w_{\alpha}(y)\ dy.$

In some cases it will be beneficial to make the change of variables $y=tx$,
giving us

$U_{\alpha,1}(x)=x\int_{0}^{1}|\hat{I}_{\alpha}(x,t)|w_{\alpha}(tx)\ dt,\quad
U_{\alpha,2}(x)=x\int_{1}^{\pi/x}|\hat{I}_{\alpha}(x,t)|w_{\alpha}(tx)\
dt,\quad$

where

$\hat{I}_{\alpha}(x,t)=C_{-\alpha}(x(1-t))+C_{-\alpha}(x(1+t))-2C_{-\alpha}(xt).$

The following lemmas give information about the sign of $I(x,y)$ and
$\hat{I}(x,t)$, allowing us to remove the absolute value.

######  Lemma 11.1.

For all $\alpha\in(-1,0)$ the function

$(1-t)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}$

is increasing and continuous in $t$ for $t\in(0,1)$ and has a unique root
$r_{\alpha,0}$. For $x\in(0,\pi)$ and $t\in(0,r_{\alpha,0})$ the function
$\hat{I}_{\alpha}(x,t)$ is increasing in $x$.

###### Proof.

Differentiating the function with respect to $t$ gives us

$(-\alpha-1)(-(1-t)^{-\alpha-2}+(1+t)^{-\alpha-2}-2t^{-\alpha-2}),$

we want to prove that this is positive. Since $-\alpha-1<0$ and
$t^{-\alpha-2}>0$ it is enough to prove that

$-(1-t)^{-\alpha-2}+(1+t)^{-\alpha-2}<0,$

which follows immediately from that $-\alpha-2<0$ and $1-t<1+t$. This proves
that the function is increasing. Noticing that the limit $t\to 0$ is negative
and $t\to 1$ is positive gives us the existence of a unique root
$r_{\alpha,0}$.

To get the monotonicity for $\hat{I}_{\alpha}$ we differentiate with respect
to $x$,

$\frac{d}{dx}\hat{I}_{\alpha}(x,t)=-(1-t)S_{-\alpha-1}(x(1-t))-(1+t)S_{-\alpha-1}(x(1+t))+2tS_{\alpha-1}(xt),$

and expand,

$\frac{d}{dx}\hat{I}_{\alpha}(x,t)=-\Gamma(2+\alpha)\cos\left(-\frac{\pi}{2}(\alpha+1)\right)x^{-\alpha-2}((1-t)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{\alpha-1})\\\
-\sum_{m=0}^{\infty}(-1)^{m}\zeta(-\alpha-1-2m)\frac{x^{2m}}{(2m)!}((1-t)^{2m+1}+(1+t)^{2m+1}-2t^{2m+1}).$

We have
$-\Gamma(2+\alpha)\cos\left(-\frac{\pi}{2}(\alpha+1)\right)x^{-\alpha-2}<0$
and since $0<t<r_{\alpha,0}$ we have from the above that
$(1-t)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{\alpha-1}<0$, the first term is hence
positive. Due to the location of the zeros of the zeta function on the
negative real axis the factor $(-1)^{m}\zeta(-\alpha-1-2m)$ is negative. Hence
all terms in the sum are negative and the sum is decreasing in $x$. Taking
into account the minus sign in front of the sum gives us something increasing.

All terms in the sum are, due to the location of the zeta function on the
negative real axis, adding the minus sign in front means that the whole
expression is positive. It follows that $\hat{I}_{\alpha}(x,t)$ is increasing
in $x$. ∎

######  Lemma 11.2.

For all $\alpha\in(-1,0)$ and $x\in(0,\pi)$ the function
$\hat{I}_{\alpha}(x,t)$ is increasing and continuous in $t$ for $t\in(0,1)$
and has the limits

$\displaystyle\lim_{t\to 0^{+}}\hat{I}_{\alpha}(x,y)$ $\displaystyle=-\infty,$
$\displaystyle\lim_{t\to 1^{-}}\hat{I}_{\alpha}(x,y)$ $\displaystyle=\infty.$

Moreover, the unique root, $r_{\alpha,x}$, in $t$ is decreasing in $x$ and
satisfies the inequality

$\frac{1}{2}<r_{\alpha,x}<r_{\alpha,0},$

with $r_{\alpha,0}$ as in lemma 11.1.

###### Proof.

The continuity in $t$ follows from that $C_{\alpha}$ is continuous on the
interval $(0,2\pi)$ and that the arguments all stay in this range. For the
left limit we note that $C_{-\alpha}(x(1-t))$ and $C_{-\alpha}(x(1+t))$ both
remain finite, while $C_{-\alpha}(xt)$ diverges towards $\infty$. For the
right limit $C_{-\alpha}(x(1+t))$ and $C_{-\alpha}(xt)$ are finite while
$C_{-\alpha}(x(1-t))$ diverges towards $\infty$.

To show that it is increasing in $t$ we differentiate, giving us

$\frac{d}{dt}\hat{I}_{\alpha}(x,t)=x(S_{-\alpha-1}(x(1-t))-S_{-\alpha-1}(x(1+t))+2S_{-\alpha-1}(xt)).$

We want to prove that this is positive. Note that $0<xt<\pi$ and from Lemma
C.1 we have that $S_{-\alpha-1}$ is positive on the interval $(0,\pi)$. Since
$x$ is also positive it thus satisfies to check that

$S_{-\alpha-1}(x(1-t))-S_{-\alpha-1}(x(1+t))>0,$

for which it is enough to assert that $S_{-\alpha-1}$ is decreasing on
$(0,2\pi)$, which is the result of Lemma C.2. This proves the existence of a
unique root $r_{\alpha,x}$ on the interval $(0,1)$.

To prove that $r_{\alpha,x}$ is decreasing in $x$ we first prove that it is
upper bounded by $r_{\alpha,0}$. Expanding the Clausen functions gives us

$\hat{I}_{\alpha}(x,t)=\Gamma(1+\alpha)\sin\left(-\frac{\pi}{2}\alpha\right)x^{-\alpha-1}((1-t)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{\alpha-1})\\\
+\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{x^{2m}}{(2m)!}((1-t)^{2m}+(1+t)^{2m}-2t^{2m}).$

All terms in the sum are positive, due to the location of the roots of the
zeta function on the negative real axis, and to have a root we must hence have

$\Gamma(1+\alpha)\sin\left(-\frac{\pi}{2}\alpha\right)x^{-\alpha-1}((1-t)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{\alpha-1})<0.$

Since $\Gamma(1+\alpha)\sin\left(-\frac{\pi}{2}\alpha\right)x^{-\alpha-1}>0$
this means we must have

$(1-t)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{\alpha-1}<0,$

but from the previous lemma we know that this only holds for $t<r_{\alpha,0}$,
it follows that $r_{\alpha,x}<r_{\alpha,0}$. The monotonicity in $x$ now
follows directly from that $\hat{I}_{\alpha}(x,t)$ is increasing in $x$ for
$t\in(0,r_{\alpha,0})$. To get the lower bound $\frac{1}{2}<r_{\alpha,x}$ it
is enough to notice that

$\hat{I}_{\alpha}(\pi,1/2)=C_{-\alpha}(\pi/2)+C_{-\alpha}(\pi/2)-2C_{-\alpha}(\pi/2)=0.$

∎

In practice the root $r_{\alpha,x}$ is decreasing also in $\alpha$. However,
instead of proving this in the general case the following lemma can be used as
an a posteriori check.

######  Lemma 11.3.

Let
$\boldsymbol{\alpha}=[\underline{\alpha},\overline{\alpha}]\subseteq(-1,0)$
and $x\in(0,\pi)$. If

$\frac{d}{d\alpha}\hat{I}_{\alpha}(x,r_{\underline{\alpha},x})>0\quad\text{for
all}\quad\alpha\in\boldsymbol{\alpha},$

then $r_{\alpha,x}\leq r_{\underline{\alpha},x}$ for all
$\alpha\in\boldsymbol{\alpha}$. Similarly, if
$\frac{d}{d\alpha}\hat{I}_{\alpha}(x,r_{\overline{\alpha},x})>0$ for all
$\alpha\in\boldsymbol{\alpha}$, then $r_{\alpha,x}\geq
r_{\overline{\alpha},x}$ for all $\alpha\in\boldsymbol{\alpha}$.

###### Proof.

We only prove the first statement, the second one is similar. By definition of
$r_{x,\underline{\alpha}}$ we have
${\hat{I}_{\underline{\alpha}}(x,r_{x,\underline{\alpha}})=0}$ and hence if
$\frac{d}{d\alpha}\hat{I}_{\alpha}(x,r_{x,\underline{\alpha}})>0$ we have
$\hat{I}_{\alpha}(x,r_{x,\underline{\alpha}})\geq 0$. Since
$\hat{I}_{\alpha}(x,t)$ is increasing in $t$ this means that $r_{x,\alpha}\leq
r_{x,\underline{\alpha}}$. ∎

######  Lemma 11.4.

For all $\alpha\in(-1,0)$ and $x\in(0,\pi)$ we have $I_{\alpha}(x,y)>0$ for
$y\in(x,\pi)$ and $\hat{I}_{\alpha}(x,t)>0$ for $t\in(1,\pi/x)$.

###### Proof.

The function $C_{-\alpha}(y)$ is strictly convex for $y\in(0,2\pi)$, this can
be seen from the integral representation of
$\frac{d^{2}}{dy^{2}}C_{-\alpha}(y)=-C_{-\alpha-2}(y)$ given in the proof of
Lemma C.2. It immediately follows that

$I_{\alpha}(x,y)=C_{-\alpha}(y-x)+C_{-\alpha}(y+x)-2C_{-\alpha}(y)<0.$

A simple change of variables gives the result for $\hat{I}_{\alpha}(x,t)$. ∎

With the help of Lemma 11.2 and 11.4 we can slightly simplify $U_{1}$ and
$U_{2}$, we get

$U_{\alpha,1}(x)=-x\int_{0}^{r_{\alpha,x}}\hat{I}(x,t)w_{\alpha}(tx)\
dt+x\int_{r_{\alpha,x}}^{1}\hat{I}(x,t)w_{\alpha}(tx)\
dt=U_{\alpha,1,1}(x)+U_{\alpha,1,2}(x)$

and

$U_{\alpha,2}(x)=\int_{x}^{\pi}I(x,y)w_{\alpha}(y)\ dy.$

In general the integrals cannot be explicitly computed, the exception is when
$w_{\alpha}(x)=|x|$ in which case we have the following lemma.

######  Lemma 11.5.

For $\alpha\in(-1,0)$, $x\in(0,\pi)$ and $w_{\alpha}(x)=|x|$ we have

$U_{\alpha}(x)=2\tilde{C}_{2-\alpha}(x)+2(C_{2-\alpha}(x+\pi)-C_{2-\alpha}(\pi))\\\
-2\left(C_{2-\alpha}(x(1-r_{\alpha,x}))+C_{2-\alpha}(x(1+r_{\alpha,x}))-2C_{2-\alpha}(xr_{\alpha,x})\right)\\\
-2xr_{\alpha,x}\left(-S_{1-\alpha}(x(1-r_{\alpha,x}))+S_{1-\alpha}(x(1+r_{\alpha,x}))-2S_{1-\alpha}(xr_{\alpha,x})\right).$

###### Proof.

For $w_{\alpha}(x)=|x|$ we have

$U_{\alpha}(x)=-U_{\alpha,1,1}(x)+U_{\alpha,1,2}(x)+U_{\alpha,2}(x)$

with

$\displaystyle U_{\alpha,1,1}(x)$
$\displaystyle=x^{2}\int_{0}^{r_{\alpha,x}}(C_{-\alpha}(x(1-t))+C_{-\alpha}(x(1+t))-2C_{-\alpha}(xt))t\
dt,$ $\displaystyle U_{\alpha,1,2}(x)$
$\displaystyle=x^{2}\int_{r_{\alpha,x}}^{1}(C_{-\alpha}(x(1-t))+C_{-\alpha}(x(1+t))-2C_{-\alpha}(xt))t\
dt,$ $\displaystyle U_{\alpha,2}(x)=$ $\displaystyle
x^{2}\int_{1}^{\pi/x}(C_{-\alpha}(x(1-t))+C_{-\alpha}(x(1+t))-2C_{-\alpha}(xt))t\
dt.$

If we let

$\displaystyle J_{\alpha}(x,t)$
$\displaystyle=\int(C_{-\alpha}(x(1-t))+C_{-\alpha}(x(1+t))-2C_{-\alpha}(xt))t\
dt$
$\displaystyle=\frac{1}{x^{2}}\left(C_{2-\alpha}(x(1-t))+C_{2-\alpha}(x(1+t))-2C_{2-\alpha}(xt)\right)+\frac{t}{x}\left(-S_{1-\alpha}(x(1-t))+S_{1-\alpha}(x(1+t))-2S_{1-\alpha}(xt)\right)$

we get

$\displaystyle U_{\alpha}(x)$
$\displaystyle=-x^{2}(J_{\alpha}(x,r_{\alpha,x})-J_{\alpha}(x,0))+x^{2}(J_{\alpha}(x,1)-J_{\alpha,x}(x,r_{\alpha,x}))+x^{2}(J_{\alpha}(x,\pi/x)-J_{\alpha}(x,1))$
$\displaystyle=x^{2}J_{\alpha}(x,0)-2x^{2}J_{\alpha}(x,r_{x,\alpha})+x^{2}J_{\alpha}(x,\pi/x),$

where we have used that $J_{\alpha}(x,1)$ is finite. Both $J_{\alpha}(x,0)$
and $J_{\alpha}(x,\pi/x)$ can be simplified further. We have

$J_{\alpha}(x,0)=\frac{2}{x^{2}}\left(C_{2-\alpha}(x)-C_{2-\alpha}(0)\right)=\frac{2}{x^{2}}\tilde{C}_{2-\alpha}(x).$

and

$\displaystyle J_{\alpha}(x,\pi/x)$
$\displaystyle=\frac{1}{x^{2}}\left(C_{2-\alpha}(x-\pi)+C_{2-\alpha}(x+\pi)-2C_{2-\alpha}(\pi)\right)+\frac{\pi}{x}\left(-S_{1-\alpha}(x-\pi)+S_{1-\alpha}(x+\pi)-2S_{1-\alpha}(\pi)\right)$
$\displaystyle=\frac{2}{x^{2}}(C_{2-\alpha}(x+\pi)-C_{2-\alpha}(\pi)),$

where we have used that $C_{s}(x-\pi)=C_{s}(x+\pi)$,
$S_{s}(x-\pi)=S_{s}(x+\pi)$ and $S_{s}(\pi)=0$. Putting all of this together
gives us the result. ∎

###  11.2 Evaluation of $\mathcal{T}_{\alpha}$

Similarly to in the previous sections we divide the interval $[0,\pi]$, on
which we take the supremum, into two parts, $[0,\epsilon]$ and
$[\epsilon,\pi]$.

####  11.2.1 Evaluation of $\mathcal{T}_{\alpha}$ for $I_{2}$

For the weight $w_{\alpha}(x)=|x|$ evaluation is straight forward for
$x\in[\epsilon,\pi]$, using Lemma 11.5. For $x\in[0,\epsilon]$ we write

$\mathcal{T}_{\alpha}(x)=\frac{1}{\pi}\frac{x^{-\alpha}}{u_{\alpha}(x)}\cdot\frac{U_{\alpha}(x)}{x^{1-\alpha}}.$
(32)

The first factor is enclosed using Lemma 8.1. For the second factor we use
Lemma 11.5 and expand the Clausen functions at zero to be able to cancel the
removable singularity. For both $x\in[0,\epsilon]$ and $x\in[\epsilon,\pi]$ we
need to compute enclosures of the root of the integrand $r_{\alpha,x}$, this
is done using standard interval methods for root enclosures, the monotonicity
in $x$ from Lemma 11.2 and Lemma 11.3.

When the weight is $w_{\alpha}(x)=|x|^{p}$ with $p<1$ more work is required
when evaluating $U_{\alpha}$. For $x\in[\epsilon,\pi]$ it uses a combination
of asymptotic analysis near the singularities of the integrand and a rigorous
integrator, as described in Appendix D. For $x\in[0,\epsilon]$ we split it as
in (32) and use the following lemma to handle the second factor.

######  Lemma 11.6.

Let $0<\epsilon<\frac{\pi}{2}$, for $\alpha\in(-1,0)$, $x\in[0,\epsilon]$ and
$w_{\alpha}(x)=|x|^{p}$ with $-\alpha<p<1$ and $1+\alpha\not=p$ we have

$\frac{U_{\alpha,1}(x)}{x^{-\alpha+p}}\leq
c_{\alpha,1}+d_{\alpha,1}x^{3+\alpha}$

and

$\frac{U_{\alpha,2}(x)}{x^{-\alpha+p}}\leq
c_{\alpha,2}+d_{\alpha,2}x^{2+\alpha-p},$

where

$\displaystyle c_{\alpha,1}=$
$\displaystyle\Gamma(1+\alpha)\sin\left(-\frac{\pi}{2}\alpha\right)\Bigg{(}\frac{2}{\alpha-p}+\frac{\Gamma(-\alpha)\Gamma(1+p)}{\Gamma(1-\alpha+p)}+\frac{{}_{2}F_{1}(1+\alpha,1+p;2+p;-1)}{1+p}$
$\displaystyle-2r_{\alpha,0}^{p}\left(\frac{2r_{\alpha,0}^{-\alpha}}{\alpha-p}+\frac{r_{\alpha,0}\
{}_{2}F_{1}(1+\alpha,1+p,2+p,-r_{\alpha,0})}{1+p}+\frac{r_{\alpha,0}\
{}_{2}F_{1}(1+\alpha,1+p,2+p,r_{\alpha,0})}{1+p}\right)\Bigg{)}$
$\displaystyle c_{\alpha,2}=$
$\displaystyle\Gamma(1+\alpha)\sin\left(-\frac{\pi}{2}\alpha\right)\left(\frac{\Gamma(-\alpha)\Gamma(\alpha-p)}{\Gamma(-p)}+\frac{{}_{2}F_{1}(1+\alpha,\alpha-p;1+\alpha-p;-1)-2}{\alpha-p}\right);$
$\displaystyle d_{\alpha,1}=$ $\displaystyle
2\sum_{m=1}^{M-1}(-1)^{m}\zeta(-\alpha-2m)\frac{\epsilon^{2m-2}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{2k+1+p}+\frac{1}{\epsilon^{2}}\sum_{m=M}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{(2\epsilon)^{2m}}{(2m)!};$
$\displaystyle d_{\alpha,2}=$
$\displaystyle-\Gamma(1+\alpha)\sin\left(-\frac{\pi}{2}\alpha\right)\frac{(1+\alpha)(2+\alpha)}{(2+\alpha-p)\pi^{2+\alpha-p}}$
$\displaystyle+2\pi^{p-1}\sum_{m=1}^{M-1}(-1)^{m}\zeta(-\alpha-2m)\frac{\pi^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{\left(\frac{\epsilon}{\pi}\right)^{2(m-1-k)}}{2k+1+p}+6\pi^{p-1}\sum_{m=M}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{(\frac{3\pi}{2})^{2m}}{(2m)!}$

The tails for $d_{\alpha,1}$ and $d_{\alpha,2}$ are of the same form as in
those for the Clausen functions and can be bounded using Lemma C.5.

###### Proof.

Recall that

$\displaystyle U_{\alpha,1}(x)$
$\displaystyle=x^{1+p}\int_{0}^{1}\left|C_{-\alpha}(x(1-t))+C_{-\alpha}(x(1+t))-2C_{-\alpha}(xt)\right|t^{p}\
dt,$ $\displaystyle U_{\alpha,2}(x)=$ $\displaystyle
x^{1+p}\int_{1}^{\pi/x}(C_{-\alpha}(x(1-t))+C_{-\alpha}(x(1+t))-2C_{-\alpha}(xt))t^{p}\
dt.$

The idea is to expand the integrals in $x$ and integrate the expansions
termwise.

From the asymptotic expansion of $C_{s}(x)$ we get, with $x,t>0$,

$C_{-\alpha}(x(1-t))+C_{-\alpha}(x(1+t))-2C_{-\alpha}(xt)=\Gamma(1+\alpha)\sin\left(-\frac{\pi}{2}\alpha\right)(|1-t|^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1})x^{-\alpha-1}\\\
+\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)((1-t)^{2m}+(1+t)^{2m}-2t^{2m})\frac{x^{2m}}{(2m)!}.$
(33)

Using that

$(1-t)^{2m}+(1+t)^{2m}-2t^{2m}=\sum_{k=0}^{2m}\binom{2m}{k}(1+(-1)^{k})t^{k}-2t^{2m}=2\sum_{k=0}^{m-1}\binom{2m}{2k}t^{2k}$

the sum can be rewritten as

$\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)((1-t)^{2m}+(1+t)^{2m}-2t^{2m})\frac{x^{2m}}{(2m)!}=2\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{x^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}t^{2k}.$

We can further note that $(-1)^{m}\zeta(-\alpha-2m)>0$ for $-1<\alpha<0$ and
$m=1,2,\dots$, so all terms in the sum are positive.

For $U_{\alpha,1}$ we get

$U_{\alpha,1}(x)\leq\Gamma(1+\alpha)\sin\left(-\frac{\pi}{2}\alpha\right)x^{-\alpha+p}\int_{0}^{1}\left|(1-t)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}\right|t^{p}\
dt\\\
+2x^{1+p}\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{x^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\int_{0}^{1}t^{2k+p}\
dt.$

Where we have use that $1-t>0$ and the positivity of the terms in the sum to
remove some absolute values. For the first term we get from Lemma 11.2 that
the integrand has the unique root $r_{\alpha,0}$ on the interval $[0,1]$. This
gives us

$\int_{0}^{1}\left|(1-t)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}\right|t^{p}\
dt\\\
=-\int_{0}^{r_{\alpha,0}}((1-t)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1})t^{p}\
dt\\\
+\int_{r_{\alpha,0}}^{1}((1-t)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1})t^{p}\
dt.$

Integrating this gives us

$\int_{0}^{1}\left|(1-t)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}\right|t^{p}\
dt=\frac{2}{\alpha-p}+\frac{\Gamma(-\alpha)\Gamma(1+p)}{\Gamma(1-\alpha+p)}+\frac{{}_{2}F_{1}(1+\alpha,1+p;2+p;-1)}{1+p}\\\
-2r_{\alpha,0}^{p}\left(\frac{2r_{\alpha,0}^{-\alpha}}{\alpha-p}+\frac{r_{\alpha,0}\
{}_{2}F_{1}(1+\alpha,1+p,2+p,-r_{\alpha,0})}{1+p}+\frac{r_{\alpha,0}\
{}_{2}F_{1}(1+\alpha,1+p,2+p,r_{\alpha,0})}{1+p}\right)$

This, together with the factor
$\Gamma(1+\alpha)\sin\left(-\frac{\pi}{2}\alpha\right)$ gives us
$c_{\alpha,1}$.

For the sum we have $\int_{0}^{1}t^{2k+p}\ dt=\frac{1}{2k+1+p}$, giving us

$2x^{1+p}\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{x^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{2k+1+p}.$

Factoring out $x^{2}$ and using that the sum is increasing in $x$ we get the
bound

$2x^{3+p}\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{\epsilon^{2m-2}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{2k+1+p}.$

To bound the tail we use that

$\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{2k+1+p}\leq\sum_{k=0}^{m}\binom{2m}{2k}\frac{1}{2k+1}=\frac{2^{2m}}{1+2m}\leq
2^{2m}$

and hence

$2\sum_{m=M}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{\epsilon^{2m-2}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{2k+1+p}\leq\frac{1}{\epsilon^{2}}\sum_{m=M}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{(2\epsilon)^{2m}}{(2m)!}$

This together with the first $M-1$ terms in the sum gives us $d_{\alpha,1}$.

For $U_{\alpha,2}$ there is no absolute value and integrating termwise gives
us

$U_{\alpha,2}(x)=\Gamma(1+\alpha)\sin\left(-\frac{\pi}{2}\alpha\right)x^{-\alpha+p}\int_{1}^{\pi/x}((t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1})t^{p}\
dt\\\
+2x^{1+p}\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{x^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\int_{1}^{\pi/x}t^{2k+p}\
dt$

For the first term we get after long calculations that

$\int_{1}^{\pi/x}((t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1})t^{p}\
dt=\frac{\Gamma(-\alpha)\Gamma(\alpha-p)}{\Gamma(-p)}+\frac{{}_{2}F_{1}(1+\alpha,\alpha-p;1+\alpha-p;-1)-2}{\alpha-p}\\\
-\frac{1}{\alpha-p}\left(\frac{x}{\pi}\right)^{\alpha-p}\left({}_{2}F_{1}\left(1+\alpha,\alpha-p;1+\alpha-p;\frac{x}{\pi}\right)+{}_{2}F_{1}\left(1+\alpha,\alpha-p;1+\alpha-p;-\frac{x}{\pi}\right)-2\right).$

To handle the hypergeometric functions we use the series representation

${}_{2}F_{1}\left(1+\alpha,\alpha-p;1+\alpha-p;\frac{x}{\pi}\right)=\sum_{k=0}^{\infty}\frac{(1+\alpha)_{k}(\alpha-p)_{k}}{(1+\alpha-p)_{k}}\frac{1}{k!}\left(\frac{x}{\pi}\right)^{k},$

and similarly for $-\frac{x}{\pi}$, which holds since $\frac{x}{\pi}<1$ and
$1+\alpha-p$ is not equal to a non-positive integer. This gives us

${}_{2}F_{1}\left(1+\alpha,\alpha-p;1+\alpha-p;\frac{x}{\pi}\right)+{}_{2}F_{1}\left(1+\alpha,\alpha-p;1+\alpha-p;-\frac{x}{\pi}\right)-2\\\
=\sum_{k=0}^{\infty}\frac{(1+\alpha)_{k}(\alpha-p)_{k}}{(1+\alpha-p)_{k}}\frac{1}{k!}\left(\left(\frac{x}{\pi}\right)^{k}+\left(-\frac{x}{\pi}\right)^{k}\right)-2=2\sum_{k=1}^{\infty}\frac{(1+\alpha)_{2k}(\alpha-p)_{2k}}{(1+\alpha-p)_{2k}}\frac{1}{(2k)!}\left(\frac{x}{\pi}\right)^{2k}$

Putting this together we have

$\int_{1}^{\pi/x}((t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1})t^{p}\
dt=\frac{\Gamma(-\alpha)\Gamma(\alpha-p)}{\Gamma(-p)}+\frac{{}_{2}F_{1}(1+\alpha,\alpha-p;1+\alpha-p;-1)-2}{\alpha-p}\\\
-\frac{2}{\alpha-p}\left(\frac{x}{\pi}\right)^{\alpha-p}\sum_{k=1}^{\infty}\frac{(1+\alpha)_{2k}(\alpha-p)_{2k}}{(1+\alpha-p)_{2k}}\frac{1}{(2k)!}\left(\frac{x}{\pi}\right)^{2k}$

Noticing that

$\frac{1}{\alpha-p}\frac{(1+\alpha)_{2k}(\alpha-p)_{2k}}{(1+\alpha-p)_{2k}}=\frac{(1+\alpha)_{2k}}{2k+\alpha-p}>0$

for $k\geq 1$ we see that all terms in the sum are positive. Since we are
subtracting the sum we get an upper bound even if we truncate the sum to any
finite number of terms. In particular, keeping only the first term we get

$\int_{1}^{\pi/x}((t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1})t^{p}\
dt\leq\frac{\Gamma(-\alpha)\Gamma(\alpha-p)}{\Gamma(-p)}+\frac{{}_{2}F_{1}(1+\alpha,\alpha-p;1+\alpha-p;-1)-2}{\alpha-p}\\\
-\frac{(1+\alpha)(2+\alpha)}{2+\alpha-p}\left(\frac{x}{\pi}\right)^{2+\alpha-p}$

Multiplying this with $\Gamma(1+\alpha)\sin\left(-\frac{\pi}{2}\alpha\right)$
(which is positive) we get $c_{\alpha,2}$ from the constant term and the first
part of $d_{\alpha,2}$ from the non-constant term.

For the sum we get

$2x^{1+p}\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{x^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{\left(\frac{\pi}{x}\right)^{2k+1+p}-1}{2k+1+p}$

Factoring out $(\pi/x)^{2m-1+p}$ gives us

$2x^{1+p}\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{x^{2m}}{(2m)!}\left(\frac{\pi}{x}\right)^{2m-1+p}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{\left(\frac{x}{\pi}\right)^{2(m-1-k)}-\left(\frac{x}{\pi}\right)^{2m-1+p}}{2k+1+p}\\\
=2x^{2}\pi^{p-1}\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{\pi^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{\left(\frac{x}{\pi}\right)^{2(m-1-k)}-\left(\frac{x}{\pi}\right)^{2m-1+p}}{2k+1+p}.$

Looking at the inner sum an upper bound is given by

$\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{\left(\frac{x}{\pi}\right)^{2(m-1-k)}}{2k+1+p}.$

This is increasing in $x$ and an upper bound is hence given by

$2x^{2}\pi^{p-1}\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{\pi^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{\left(\frac{\epsilon}{\pi}\right)^{2(m-1-k)}}{2k+1+p}.$

To bound the tail we use that $\epsilon\leq\pi/2$ and hence

$\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{\left(\frac{\epsilon}{\pi}\right)^{2(m-1-k)}}{2k+1+p}\leq\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{\left(\frac{1}{2}\right)^{2(m-1-k)}}{2k+1}=\frac{2^{-2m}(1+3^{1+2m})-4}{1+2m}<3\left(\frac{3}{2}\right)^{2m}$

Inserting this we have

$\sum_{m=M}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{\pi^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{\left(\frac{\epsilon}{\pi}\right)^{2(m-1-k)}}{2k+1}\leq
3\sum_{m=M}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{(\frac{3\pi}{2})^{2m}}{(2m)!}.$

This together with the first $M-1$ terms in the sum gives us the second part
of $d_{\alpha,2}$. ∎

####  11.2.2 Evaluation of $\mathcal{T}_{\alpha}$ for $I_{1}$

For $x\in[\epsilon,\pi]$ we make one optimization compared to $I_{2}$, instead
of computing an enclosure we only compute an upper bound. For this we use that
$u_{\alpha}(x)\geq u_{-1}(x)$ for $\alpha\in I_{1}$ and hence

$\mathcal{T}_{\alpha}(x)\leq\frac{U_{\alpha}(x)}{\pi|w_{\alpha}(x)||u_{-1}(x)|}.$

The different weight also means that the asymptotic analysis near the
singularities of the integrand has to be adjusted, this is discussed in
Appendix D.

For $x\in[0,\epsilon]$ we write

$\mathcal{T}_{\alpha}(x)=\frac{\log(1/x)}{\pi\log(2e+1/x)}\cdot\frac{\Gamma(1+\alpha)x^{-\alpha}(1-x^{1+\alpha+(1+\alpha)^{2}/2})}{u_{\alpha}(x)}\cdot\frac{U_{\alpha}(x)}{x^{(1-\alpha)/2-\alpha}\log(1/x)(1-x^{1+\alpha+(1+\alpha)^{2}/2})\Gamma(1+\alpha)}.$

The first two factors are handled in the same way as in Section 10.2. For the
third factor we start with the following lemma

######  Lemma 11.7.

For $\alpha\in(-1,0)$ and $w_{\alpha}(x)=|x|^{(1-\alpha)/2}\log(2e+1/|x|)$ we
have

$\displaystyle\frac{U_{\alpha}(x)}{x^{(1-\alpha)/2-\alpha}\log(1/x)(1-x^{1+\alpha+(1+\alpha)^{2}/2})\Gamma(1+\alpha)}\leq$
$\displaystyle\sin\left(-\frac{\pi}{2}\alpha\right)\left(G_{\alpha,1}(x)+G_{\alpha,2}(x)\right)$
$\displaystyle+\frac{x^{1+\alpha}}{\Gamma(1+\alpha)\log(1/x)(1-x^{1+\alpha+(1+\alpha)^{2}/2})}R_{\alpha}(x)$

with

$\displaystyle G_{\alpha,1}(x)$
$\displaystyle=\frac{\int_{0}^{1}\left|(1-t)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}\right|t^{(1-\alpha)/2}\log(2e+1/(xt))\
dt}{\log(1/x)(1-x^{1+\alpha+(1+\alpha)^{2}/2})};$ (34) $\displaystyle
G_{\alpha,2}(x)$
$\displaystyle=\frac{\int_{1}^{\pi/x}\left((t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}\right)t^{(1-\alpha)/2}\log(2e+1/(xt))\
dt}{\log(1/x)(1-x^{1+\alpha+(1+\alpha)^{2}/2})};$ (35) $\displaystyle
R_{\alpha}(x)$
$\displaystyle=2\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{x^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\int_{0}^{\pi/x}t^{2k+(1-\alpha)/2}\log(2e+1/(xt))\
dt.$ (36)

###### Proof.

With the given weight we have

$U_{\alpha}(x)=x\int_{0}^{\pi/x}\left|C_{-\alpha}(x(1-t))+C_{-\alpha}(x(1+t))-2C_{-\alpha}(xt)\right|(xt)^{(1-\alpha)/2}\log(2e+1/(xt))\
dt.$

This means that we want to bound

$\frac{x^{1+\alpha}\int_{0}^{\pi/x}\left|C_{-\alpha}(x(1-t))+C_{-\alpha}(x(1+t))-2C_{-\alpha}(xt)\right|t^{(1-\alpha)/2}\log(2e+1/(xt))\
dt}{\log(1/x)(1-x^{1+\alpha+(1+\alpha)^{2}/2})\Gamma(1+\alpha)}.$

Using the asymptotic expansion of the Clausen terms in the integrand from (33)
we can split the integral as

$\int_{0}^{\pi/x}\left|C_{-\alpha}(x(1-t))+C_{-\alpha}(x(1+t))-2C_{-\alpha}(xt)\right|t^{(1-\alpha)/2}\log(2e+1/(xt))dt\\\
\leq\Gamma(1+\alpha)\sin(-\pi\alpha/2)x^{-\alpha-1}\int_{0}^{\pi/x}\left||1-t|^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}\right|t^{(1-\alpha)/2}\log(2e+1/(xt))\
dt\\\
+2\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{x^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\int_{0}^{\pi/x}t^{2k+(1-\alpha)/2}\log(2e+1/(xt))\
dt$

From this we get

$\frac{\sin(-\pi\alpha/2)\int_{0}^{\pi/x}\left||1-t|^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}\right|t^{(1-\alpha)/2}\log(2e+1/(xt))\
dt}{\log(1/x)(1-x^{1+\alpha+(1+\alpha)^{2}/2})}\\\
+\frac{x^{1+\alpha}2\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{x^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\int_{0}^{\pi/x}t^{2k+(1-\alpha)/2}\log(2e+1/(xt))\
dt}{\log(1/x)(1-x^{1+\alpha+(1+\alpha)^{2}/2})\Gamma(1+\alpha)}$

as an upper bound. For the integral in the first term we can split the
interval of integration at $t=1$ to get

$\int_{0}^{\pi/x}\left||1-t|^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}\right|t^{(1-\alpha)/2}\log(2e+1/(xt))\
dt\\\
=\int_{0}^{1}\left|(1-t)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}\right|t^{(1-\alpha)/2}\log(2e+1/(xt))\
dt\\\
+\int_{1}^{\pi/x}\left((t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}\right)t^{(1-\alpha)/2}\log(2e+1/(xt))\
dt.$

Here we have removed the absolute values around $1-t$ according to its sign
and also used that

$(t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}$

is positive for $t>1$, which can be shown in the same way as in Lemma 11.4. ∎

Bounds of $G_{\alpha,1}$, $G_{\alpha,2}$ and $R_{\alpha}$ are given in
Appendix F. The factor

$\frac{x^{1+\alpha}}{\Gamma(1+\alpha)\log(1/x)(1-x^{1+\alpha+(1+\alpha)^{2}/2})}$

has a removable singularity at $\alpha=-1$ that needs to be treated but is
otherwise straightforward to enclose.

####  11.2.3 Evaluation of $\mathcal{T}_{\alpha}$ for $I_{3}$

In this case we need to not only compute an enclosure of
$\mathcal{T}_{\alpha}(x)$, but understand it behavior in $\alpha$. We
therefore compute Taylor models of degree $0$ centered at $\alpha=0$. The
weight is given by $w_{\alpha}(x)=|x|$, and we can therefore use the explicit
expression for the integral given in Lemma 11.5.

We start with the following lemma that gives us information about the first
term in the Taylor model.

######  Lemma 11.8.

For $x\in[0,\pi]$ the constant term in the expansion at $\alpha=0$ of
$\mathcal{T}_{\alpha}(x)$ is $1$.

###### Proof.

Recall that

$\mathcal{T}_{\alpha}(x)=\frac{U_{\alpha}(x)}{\pi|w_{\alpha}(x)||u_{\alpha}(x)|}.$

In this case $w_{\alpha}(x)=|x|$, giving us

$\mathcal{T}_{\alpha}(x)=\frac{U_{\alpha}(x)}{\pi x|u_{\alpha}(x)|}.$

By Lemma 4.8 the constant term in the expansion at $\alpha=0$ of
$u_{\alpha}(x)$ is $1$, it is therefore enough to show that the constant term
for the numerator is $\pi x$.

From Lemma 11.5 we have that

$U_{\alpha}(x)=2\tilde{C}_{2-\alpha}(x)+2(C_{2-\alpha}(x+\pi)-C_{2-\alpha}(\pi))\\\
-2\left(C_{2-\alpha}(x(1-r_{\alpha,x}))+C_{2-\alpha}(x(1+r_{\alpha,x}))-2C_{2-\alpha}(xr_{\alpha,x})\right)\\\
-2xr_{\alpha,x}\left(-S_{1-\alpha}(x(1-r_{\alpha,x}))+S_{1-\alpha}(x(1+r_{\alpha,x}))-2S_{1-\alpha}(xr_{\alpha,x})\right).$

To get the constant term in the expansion we want to compute the limit of this
as $\alpha\to 0$, which we denote by $U_{0}(x)$.

To do that we first need to compute $\lim_{\alpha\to 0^{-}}r_{\alpha,x}$. The
function $\hat{I}_{\alpha}(x,t)$ for which $r_{\alpha,x}$ is a zero converges
to $0$ everywhere as $\alpha\to 0$. If we divide by $\alpha$ and compute the
zero of $\frac{\hat{I}_{\alpha}(x,t)}{\alpha}$ instead, we get a well-defined
limit, and we get that $r_{0,x}$ is the zero of

$\frac{d}{d\alpha}\hat{I}_{\alpha}(x,t)=C_{-\alpha}^{(1)}(x(1-t))+C_{-\alpha}^{(1)}(x(1+t))-2C_{-\alpha}^{(1)}(xt).$

This function satisfies the same properties as in Lemma 11.2 with respect to
the root $r_{0,x}$.

Taking the limit in $\alpha$ now gives us

$U_{0}(x)=2\tilde{C}_{2}(x)+2(C_{2}(x+\pi)-C_{2}(\pi))\\\
-2\left(C_{2}(x(1-r_{0,x}))+C_{2}(x(1+r_{0,x}))-2C_{2}(xr_{0,x})\right)\\\
-2xr_{0,x}\left(-S_{1}(x(1-r_{0,x}))+S_{1}(x(1+r_{0,x}))-2S_{1}(xr_{0,x})\right).$

For these parameters to the Clausen functions we have the explicit expressions

$C_{2}(x)=\frac{\pi^{2}}{6}-\frac{\pi}{2}x+\frac{1}{4}x^{2},\quad
S_{1}(x)=\frac{\pi}{2}-\frac{1}{2}x,$

valid when $x\in[0,2\pi]$. For the different parts of $U_{\alpha}(x)$ we get

$2\tilde{C}_{2}(x)+2(C_{2}(x+\pi)-C_{2}(\pi))\\\
=2\left(\left(-\frac{\pi}{2}x+\frac{1}{4}x^{2}\right)+\left(\frac{\pi^{2}}{6}-\frac{\pi}{2}(x+\pi)+\frac{1}{4}(x+\pi)^{2}\right)-\left(\frac{\pi^{2}}{6}-\frac{\pi}{2}\pi+\frac{1}{4}\pi^{2}\right)\right)\\\
=2\left(-\frac{\pi}{2}x+\frac{1}{2}x^{2}\right)=-\pi x+x^{2},$
$2\left(C_{2}(x(1-r_{0,x}))+C_{2}(x(1+r_{0,x}))-2C_{2}(xr_{0,x})\right)\\\
=2\bigg{(}\left(\frac{\pi^{2}}{6}-\frac{\pi}{2}x(1-r_{0,x})+\frac{1}{4}x^{2}(1-r_{0,x})^{2}\right)+\left(\frac{\pi^{2}}{6}-\frac{\pi}{2}x(1+r_{0,x})+\frac{1}{4}x^{2}(1+r_{0,x})^{2}\right)\\\
-2\left(\frac{\pi^{2}}{6}-\frac{\pi}{2}xr_{0,x}+\frac{1}{4}x^{2}r_{0,x}^{2}\right)\bigg{)}=2\left(-\pi
x(1-r_{0,x})+\frac{1}{2}x^{2}\right)=-2\pi x(1-r_{0,x})+x^{2},$
$2xr_{0,x}\left(-S_{1}(x(1-r_{0,x}))+S_{1}(x(1+r_{0,x}))-2S_{1}(xr_{0,x})\right)\\\
=2xr_{0,x}\left(-\left(\frac{\pi}{2}-\frac{1}{2}x(1-r_{0,x})\right)+\left(\frac{\pi}{2}-\frac{1}{2}x(1+r_{0,x})\right)-2\left(\frac{\pi}{2}-\frac{1}{2}xr_{0,x}\right)\right)=-2\pi
xr_{0,x}.$

Putting this together we get

$U_{0}(x)=(-\pi x+x^{2})-(-2\pi x(1-r_{0,x})+x^{2})+2\pi xr_{0,x}=\pi x,$

which is exactly what we needed to show. ∎

For $x\in[\epsilon,\pi]$ we compute the Taylor model of $u_{\alpha}(x)$ using
the approach described in Section 7.3. For $U_{\alpha}$ we use the expression

$U_{\alpha}(x)=2\tilde{C}_{2-\alpha}(x)+2(C_{2-\alpha}(x+\pi)-C_{2-\alpha}(\pi))\\\
-2\left(C_{2-\alpha}(x(1-r_{\alpha,x}))+C_{2-\alpha}(x(1+r_{\alpha,x}))-2C_{2-\alpha}(xr_{\alpha,x})\right)\\\
-2xr_{\alpha,x}\left(-S_{1-\alpha}(x(1-r_{\alpha,x}))+S_{1-\alpha}(x(1+r_{\alpha,x}))-2S_{1-\alpha}(xr_{\alpha,x})\right)$

from Lemma 11.5. For the Clausen functions not depending on $r_{\alpha,x}$ we
can compute Taylor models directly. If we let

$K_{\alpha}(x)=\left(C_{2-\alpha}(x(1-r_{\alpha,x}))+C_{2-\alpha}(x(1+r_{\alpha,x}))-2C_{2-\alpha}(xr_{\alpha,x})\right)\\\
+xr_{\alpha,x}\left(-S_{1-\alpha}(x(1-r_{\alpha,x}))+S_{1-\alpha}(x(1+r_{\alpha,x}))-2S_{1-\alpha}(xr_{\alpha,x})\right)$

then the part of $U_{\alpha}(x)$ depending on $r_{\alpha,x}$ is given by
$-2K_{\alpha}(x)$. For the Taylor model of $K_{\alpha}(x)$ we need to take
into account that $r_{\alpha,x}$ also depends on $\alpha$. The constant term
in the Taylor model is given by $K_{0}(x)$, which we can compute. For the
remainder term we want to enclose $\frac{d}{d\alpha}K_{\alpha}(x)$ for
$\alpha\in I_{3}$. Differentiation gives us

$\frac{d}{d\alpha}K_{\alpha}(x)=-\left(C_{2-\alpha}^{(1)}(x(1-r_{\alpha,x}))+C_{2-\alpha}^{(1)}(x(1+r_{\alpha,x}))-2C_{2-\alpha}^{(1)}(xr_{\alpha,x})\right)\\\
-x\left(\frac{d}{d\alpha}r_{\alpha,x}\right)\left(-S_{1-\alpha}(x(1-r_{\alpha,x}))+S_{1-\alpha}^{(1)}(x(1+r_{\alpha,x}))-2S_{1-\alpha}^{(1)}(xr_{\alpha,x})\right)\\\
+x\left(\frac{d}{d\alpha}r_{\alpha,x}\right)\left(-S_{1-\alpha}(x(1-r_{\alpha,x}))+S_{1-\alpha}(x(1+r_{\alpha,x}))-2S_{1-\alpha}(xr_{\alpha,x})\right)\\\
-xr_{\alpha,x}\left(-S_{1-\alpha}^{(1)}(x(1-r_{\alpha,x}))+S_{1-\alpha}^{(1)}(x(1+r_{\alpha,x}))-2S_{1-\alpha}^{(1)}(xr_{\alpha,x})\right)\\\
+x^{2}r_{\alpha,x}\left(\frac{d}{d\alpha}r_{\alpha,x}\right)\left(-C_{-\alpha}(x(1-r_{\alpha,x}))+C_{-\alpha}(x(1+r_{\alpha,x}))-2C_{-\alpha}(xr_{\alpha,x})\right).$

We see that two of the terms cancel out, and the last term is zero since
$r_{\alpha,x}$ is a root of exactly this function, this gives us

$\frac{d}{d\alpha}K_{\alpha}(x)=-\left(C_{2-\alpha}^{(1)}(x(1-r_{\alpha,x}))+C_{2-\alpha}^{(1)}(x(1+r_{\alpha,x}))-2C_{2-\alpha}^{(1)}(xr_{\alpha,x})\right)\\\
-xr_{\alpha,x}\left(-S_{1-\alpha}^{(1)}(x(1-r_{\alpha,x}))+S_{1-\alpha}^{(1)}(x(1+r_{\alpha,x}))-2S_{1-\alpha}^{(1)}(xr_{\alpha,x})\right).$

In particular we see that the derivative only depends on $r_{\alpha,x}$, for
which we can easily compute an enclosure, and not on
$\frac{d}{d\alpha}r_{\alpha,x}$.

For $x\in[0,\epsilon]$ we write the function as in 32 and the only problematic
part is the evaluation of $U_{\alpha}(x)x^{\alpha-1}$. The terms not depending
on $r_{\alpha,x}$ are handled by expanding the Clausen functions, following
Appendix B.1 to get a Taylor model of the expansion. What remains is to
compute a Taylor model of $K_{\alpha}(x)x^{\alpha-1}$. For that we compute the
constant term and an enclosure of the derivative separately. The constant term
is done by just expanding the Clausen functions. For the derivative we get

$\frac{d}{d\alpha}K_{\alpha}(x)x^{\alpha-1}=\left(\frac{d}{d\alpha}K_{\alpha}(x)\right)x^{\alpha-1}+K_{\alpha}(x)\log(x)x^{\alpha-1}.$

When expanding the Clausen functions for this we have to handle the
cancellations between the two terms since they individually blow up at $x=0$.

######  Lemma 11.9.

We have the following expansion for $0<x<\pi$

$\displaystyle\frac{d}{d\alpha}K_{\alpha}(x)x^{\alpha-1}=$
$\displaystyle-\frac{d}{ds}\left(\Gamma(1-s)\sin\left(\frac{\pi}{2}s\right)\left((1-r_{\alpha,x})^{s-1}+(1+r_{\alpha,x})^{s-1}-2r_{\alpha,x}^{s-1}\right)\right)\Bigg{|}_{s=2-\alpha}$
$\displaystyle-
r_{\alpha,x}\frac{d}{ds}\left(\Gamma(1-s)\cos\left(\frac{\pi}{2}s\right)\left(-(1-r_{\alpha,x})^{s-1}+(1+r_{\alpha,x})^{s-1}-2r_{\alpha,x}^{s-1}\right)\right)\Bigg{|}_{s=1-\alpha}$
$\displaystyle+\log(x)\sum_{m=1}^{\infty}\frac{(-1)^{m}}{(2m)!}\zeta(2-\alpha-2m)\left((1-r_{\alpha,x})^{2m}+(1+r_{\alpha,x})^{2m}-2r_{\alpha,x}^{2m}\right)x^{2m-1+\alpha}$
$\displaystyle+r_{\alpha,x}\log(x)\sum_{m=0}^{\infty}\frac{(-1)^{m}}{(2m+1)!}\zeta(-\alpha-2m)\left(-(1-r_{\alpha,x})^{2m+1}+(1+r_{\alpha,x})^{2m+1}-2r_{\alpha,x}^{2m+1}\right)x^{2m+1+\alpha}$
$\displaystyle-\sum_{m=1}^{\infty}\frac{(-1)^{m}}{(2m)!}\zeta^{\prime}(2-\alpha-2m)\left((1-r_{\alpha,x})^{2m}+(1+r_{\alpha,x})^{2m}-2r_{\alpha,x}^{2m}\right)x^{2m-1+\alpha}$
$\displaystyle-
r_{\alpha,x}\sum_{m=0}^{\infty}\frac{(-1)^{m}}{(2m+1)!}\zeta^{\prime}(-\alpha-2m)\left(-(1-r_{\alpha,x})^{2m+1}+(1+r_{\alpha,x})^{2m+1}-2r_{\alpha,x}^{2m+1}\right)x^{2m+1+\alpha}.$

###### Proof.

We get directly from expanding the Clausen functions that

$K_{\alpha}(x)\log(x)x^{\alpha-1}=\Gamma(\alpha-1)\sin\left(\frac{\pi}{2}(2-\alpha)\right)\left((1-r_{\alpha,x})^{1-\alpha}+(1+r_{\alpha,x})^{1-\alpha}-2r_{\alpha,x}^{1-\alpha}\right)\log(x)\\\
+r_{\alpha,x}\Gamma(\alpha)\cos\left(\frac{\pi}{2}(1-\alpha)\right)\left((1-r_{\alpha,x})^{-\alpha}+(1+r_{\alpha,x})^{-\alpha}-2r_{\alpha,x}^{-\alpha}\right)\log(x)\\\
+\log(x)\sum_{m=1}^{\infty}\frac{(-1)^{m}}{(2m)!}\zeta(2-\alpha-2m)\left((1-r_{\alpha,x})^{2m}+(1+r_{\alpha,x})^{2m}-2r_{\alpha,x}^{2m}\right)x^{2m-1+\alpha}\\\
+r_{\alpha,x}\log(x)\sum_{m=0}^{\infty}\frac{(-1)^{m}}{(2m+1)!}\zeta(-\alpha-2m)\left(-(1-r_{\alpha,x})^{2m+1}+(1+r_{\alpha,x})^{2m+1}-2r_{\alpha,x}^{2m+1}\right)x^{2m+1+\alpha}$
(37)

and

$\left(\frac{d}{d\alpha}K_{\alpha}(x)\right)x^{\alpha-1}=-\frac{d}{ds}\left(\Gamma(1-s)\sin\left(\frac{\pi}{2}s\right)\left((1-r_{\alpha,x})^{s-1}+(1+r_{\alpha,x})^{s-1}-2r_{\alpha,x}^{s-1}\right)x^{s-1}\right)\Bigg{|}_{s=2-\alpha}x^{\alpha-1}\\\
-r_{\alpha,x}\frac{d}{ds}\left(\Gamma(1-s)\cos\left(\frac{\pi}{2}s\right)\left((1-r_{\alpha,x})^{s-1}+(1+r_{\alpha,x})^{s-1}-2r_{\alpha,x}^{s-1}\right)x^{s-1}\right)\Bigg{|}_{s=1-\alpha}x^{\alpha}\\\
-\sum_{m=1}^{\infty}\frac{(-1)^{m}}{(2m)!}\zeta^{\prime}(2-\alpha-2m)\left((1-r_{\alpha,x})^{2m}+(1+r_{\alpha,x})^{2m}-2r_{\alpha,x}^{2m}\right)x^{2m-1+\alpha}\\\
-r_{\alpha,x}\sum_{m=0}^{\infty}\frac{(-1)^{m}}{(2m+1)!}\zeta^{\prime}(-\alpha-2m)\left(-(1-r_{\alpha,x})^{2m+1}+(1+r_{\alpha,x})^{2m+1}-2r_{\alpha,x}^{2m+1}\right)x^{2m+1+\alpha}.$
(38)

For the two first terms in the expansion for
$\left(\frac{d}{d\alpha}K_{\alpha}(x)\right)x^{\alpha-1}$ we have

$\frac{d}{ds}\left(\Gamma(1-s)\sin\left(\frac{\pi}{2}s\right)\left((1-r_{\alpha,x})^{s-1}+(1+r_{\alpha,x})^{s-1}-2r_{\alpha,x}^{s-1}\right)x^{s-1}\right)\Bigg{|}_{s=2-\alpha}x^{\alpha-1}\\\
=\frac{d}{ds}\left(\Gamma(1-s)\sin\left(\frac{\pi}{2}s\right)\left((1-r_{\alpha,x})^{s-1}+(1+r_{\alpha,x})^{s-1}-2r_{\alpha,x}^{s-1}\right)\right)\Bigg{|}_{s=2-\alpha}\\\
+\Gamma(1-s)\sin\left(\frac{\pi}{2}s\right)\left((1-r_{\alpha,x})^{s-1}+(1+r_{\alpha,x})^{s-1}-2r_{\alpha,x}^{s-1}\right)\log(x)$

and

$r_{\alpha,x}\frac{d}{ds}\left(\Gamma(1-s)\cos\left(\frac{\pi}{2}s\right)\left((1-r_{\alpha,x})^{s-1}+(1+r_{\alpha,x})^{s-1}-2r_{\alpha,x}^{s-1}\right)x^{s-1}\right)\Bigg{|}_{s=1-\alpha}x^{\alpha}\\\
=r_{\alpha,x}\frac{d}{ds}\left(\Gamma(1-s)\cos\left(\frac{\pi}{2}s\right)\left((1-r_{\alpha,x})^{s-1}+(1+r_{\alpha,x})^{s-1}-2r_{\alpha,x}^{s-1}\right)\right)\Bigg{|}_{s=1-\alpha}\\\
+r_{\alpha,x}\Gamma(1-s)\cos\left(\frac{\pi}{2}s\right)\left((1-r_{\alpha,x})^{s-1}+(1+r_{\alpha,x})^{s-1}-2r_{\alpha,x}^{s-1}\right)\log(x).$

For both of these the second term exactly cancels the corresponding one in
(37), this gives us the result. ∎

####  11.2.4 Evaluation of $\mathcal{T}_{\alpha}$ in hybrid cases

For $\alpha$ near $0$ this is straight forward, the weight is in this case
given by $w_{\alpha}(x)=|x|$, and we can use the same approach as described in
Section 11.2.1.

For $\alpha$ near $-1$ we can use the same approach for $x\in[\epsilon,\pi]$.
For $x\in[0,\epsilon]$ this doesn’t work since the weight is not on the form
$w_{\alpha}(x)=|x|^{p}$ so Lemma 11.6 doesn’t apply. Instead, we have to use
an approach more similar to that used for $\alpha\in I_{1}$. We write
$\mathcal{T}_{\alpha}$ as

$\mathcal{T}_{\alpha}(x)=\frac{\log(1/x)}{\pi\log(2e+1/x)}\cdot\frac{x^{-\alpha}}{u_{\alpha}(x)}\cdot\frac{U_{\alpha}(x)}{\log(1/x)x^{-\alpha+p}}.$

The first two factors are bounded in the same way as in the above sections.
For the third one we give a bound in Appendix G.

##  12 Bounds for $D_{\alpha}$, $\delta_{\alpha}$ and $n_{\alpha}$

We are now ready to give bounds for $D_{\alpha}$, $\delta_{\alpha}$ and
$n_{\alpha}$. Recall that they are given by

$n_{\alpha}=\sup_{x\in[0,\pi]}|N_{\alpha}(x)|,\quad\delta_{\alpha}=\sup_{x\in[0,\pi]}|F_{\alpha}(x)|,\quad
D_{\alpha}=\sup_{x\in[0,\pi]}|\mathcal{T}_{\alpha}(x)|.$