Gauss Hypergeometric Representations of the Ferrers Function of the Second
Kind

Gauss Hypergeometric Representations
of the Ferrers Function of the Second Kind

Howard S. COHL a, Justin PARK b and Hans VOLKMER c

H.S. Cohl, J. Park and H. Volkmer

a) Applied and Computational Mathematics Division,
a) National Institute of Standards and Technology, Mission Viejo, CA 92694,
USA<EMAIL_ADDRESS>https://www.nist.gov/people/howard-cohl

b) Department of Mathematics, Massachusetts Institute of Technology,
b) Cambridge, MA 02139, USA<EMAIL_ADDRESS>
c) Department of Mathematical Sciences, University of Wisconsin-Milwaukee,
c) Milwaukee, WI 53211, USA<EMAIL_ADDRESS>
Received September 17, 2020, in final form May 04, 2021; Published online May
20, 2021

We derive all eighteen Gauss hypergeometric representations for the Ferrers
function of the second kind, each with a different argument. They are obtained
from the eighteen hypergeometric representations of the associated Legendre
function of the second kind by using a limit representation. For the 18
hypergeometric arguments which correspond to these representations, we give
geometrical descriptions of the corresponding convergence regions in the
complex plane. In addition, we consider a corresponding single sum Fourier
expansion for the Ferrers function of the second kind. In four of the eighteen
cases, the determination of the Ferrers function of the second kind requires
the evaluation of the hypergeometric function separately above and below the
branch cut at $[1,\infty)$. In order to complete these derivations, we use
well-known results to derive expressions for the hypergeometric function above
and below its branch cut. Finally we give a detailed review of the 1888 paper
by Richard Olbricht who was the first to study hypergeometric representations
of Legendre functions.

Ferrers functions; associated Legendre functions; Gauss hypergeometric
function

33C05; 33C55; 42B05

##  1 Introduction

In 1888, Richard Emil Olbricht, a student of Felix Klein, gave a list of 72
solutions of the associated Legendre differential equation [10, Section 3]
(see Appendix A for a detailed description of Olbricht’s analysis), [11,
equation (14.2.1)]

$\displaystyle\big{(}1-x^{2}\big{)}\frac{{\rm d}^{2}y}{{\rm
d}x^{2}}-2x\frac{{\rm d}y}{{\rm
d}x}+\left(\nu(\nu+1)-\frac{\mu^{2}}{1-x^{2}}\right)y=0,$ (1.1)

in terms of the Gauss hypergeometric series [11, Chapter 15]. For instance,
the first entry of the list is

$\displaystyle\left(\frac{1-x}{2}\right)^{\frac{1}{2}\mu}\left(\frac{1+x}{2}\right)^{\frac{1}{2}\mu}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\mu-\nu,\nu+\mu+1}{1+\mu};\frac{1-x}{2}\right)\\!.$

In fact, as Olbricht has pointed out, there are 18 possible arguments $w$ of
this type of hypergeometric series. Olbricht has arranged these arguments into
three separate groups which we now list.

Group I:

$\displaystyle w_{1}=\frac{1-x}{2},\quad\ w_{2}=\frac{1+x}{2},\quad\
w_{3}=\frac{x-1}{x+1},\quad\ w_{4}=\frac{x+1}{x-1},\quad\
w_{5}=\frac{2}{1+x},\quad\ w_{6}=\frac{2}{1\\!-x}.$

Group II:

$\displaystyle w_{7}=1-x^{2},\quad\ w_{8}=\frac{1}{1-x^{2}},\quad\
w_{9}=x^{2},\quad\ w_{10}=\frac{1}{x^{2}},\quad\
w_{11}=\frac{x^{2}-1}{x^{2}},\quad\ w_{12}=\frac{x^{2}}{x^{2}-1}.$

Group III:

$\displaystyle w_{13}=\frac{-x+\sqrt{x^{2}-1}}{2\sqrt{x^{2}-1}},\qquad
w_{14}=\frac{x-\sqrt{x^{2}-1}}{x+\sqrt{x^{2}-1}},\qquad
w_{15}=\frac{2\sqrt{x^{2}-1}}{x+\sqrt{x^{2}-1}},$ $\displaystyle
w_{16}=\frac{2\sqrt{x^{2}-1}}{-x+\sqrt{x^{2}-1}},\qquad
w_{17}=\frac{x+\sqrt{x^{2}-1}}{2\sqrt{x^{2}-1}},\qquad
w_{18}=\frac{x+\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}.$

The associated Legendre functions of the first kind $P_{\nu}^{\mu}(x)$ and of
the second kind $Q_{\nu}^{\mu}(x)$ are solutions of (1.1) with simple behavior
at the regular singularities $x=1$ and $x=\infty$, respectively. In Magnus,
Oberhettinger and Soni [9, pp. 155–163] (see also [2, pp. 124–139]) these
associated Legendre functions are expressed as linear combinations of one or
two of Olbricht’s 72 solutions for every of the 18 possible $w_{j}$. The
hypergeometric representations of the associated Legendre function
$P_{\nu}^{\mu}(x)$ involving $w_{j}$, $1\leq j\leq 18$, is labelled by the
number $j$, and the corresponding formulas for $Q_{\nu}^{\mu}(x)$ are labelled
by $j+18$.

The purpose of this paper is to generate a complete list of 18 such
hypergeometric representations for the Ferrers function of the second kind
$\mathsf{Q}_{\nu}^{\mu}(x)$ which is also a solution of (1.1). Some of these
18 representations can be found in the literature but we have not seen a
complete list. We will also discuss a related Fourier expansion and some
relevant geometric topics. We pay special attention to a precise formulation
of the domains of validity of the stated representations in the complex plane
including a discussion of branch cuts.

Of course, the theory of the associated Legendre differential equation is a
very classical topic. For instance, we refer to Heine [4, 5], Hobson [6],
Lebedev [8], Schäfke [12], Magnus, Oberhettinger and Soni [9], Andrews, Askey
and Roy [1], as well as Zhurina and Karmazina [13]. In such a well-
investigated area one cannot claim to have new results. This is a review paper
that, as we may hope, throws a new light on the theory of the Ferrers function
of the second kind. Since this review contains a large number of individual
formulas, we have summarized the main results in Table 1. We will use the
complex domains

$\displaystyle D_{1}:=\mathbb{C}\setminus((-\infty,-1]\cup[1,\infty)),$ (1.2)
$\displaystyle D_{1}^{+}:=D_{1}\cap\\{x\colon\mathop{\rm Re}x>0\\},$ (1.3)
$\displaystyle D_{2}:=\mathbb{C}\setminus(-\infty,1],$ (1.4) $\displaystyle
D_{2}^{+}:=D_{2}\cap\\{x\colon\mathop{\rm Re}x>0\\},$ (1.5) $\displaystyle
D_{3}:=\mathbb{C}\setminus(-1,\infty)=-D_{2}.$ (1.6)

######  Remark 1.1.

The DLMF tabulates hypergeometric representations for the Ferrers function of
the first kind [11, equations (14.3.1) and (14.3.11)] with arguments $(1-x)/2$
and $x^{2}$ respectively. One may also find representations with arguments
$(x-1)/(x+1)$, $1-x^{2}$, $1-x^{-2}$ in [9, pp. 166–167]. As far as we are
aware, an extensive list for all of the hypergeometric representations of the
Ferrers function of the first kind does not exist in the literature. However,
such a list could easily be generated by starting with the hypergeometric
representations of the associated Legendre function of the first kind
tabulated in [9] and using (2.7).

Table 1: This table lists all Gauss hypergeometric representations of the Ferrers function of the second kind presented in this paper. In regard to Theorem 6.1, $u=x+{\rm i}\sqrt{1-x^{2}}$, $v=x-{\rm i}\sqrt{1-x^{2}}$. | Olbricht
---
group
| Domain
---
of $x$
| Domain
---
of $\nu$
| Domain
---
of $\mu$
| Extra
---
restriction
Argument(s) | Theorem
I | $x\in D_{1}$ | $\nu\in{\mathbb{C}}$ | $\mu\in{\mathbb{C}}\setminus{\mathbb{Z}}$ | $\nu+\mu\not\in-{\mathbb{N}}$ |  $\frac{1-x}{2}$ | 3.1
I | $x\in D_{1}$ | $\nu\in{\mathbb{C}}$ | $\mu\in{\mathbb{C}}\setminus{\mathbb{Z}}$ | $\nu+\mu\not\in-{\mathbb{N}}$ |  $\frac{1+x}{2}$ | 3.2
I | $x\in D_{1}$ | $\nu\in{\mathbb{C}}$ | $\mu\in{\mathbb{C}}\setminus{\mathbb{Z}}$ | $\nu+\mu\not\in-{\mathbb{N}}$ |  $\frac{x-1}{x+1}$ | 3.3
I | $x\in D_{1}$ | $\nu\in{\mathbb{C}}$ | $\mu\in{\mathbb{C}}\setminus{\mathbb{Z}}$ | $\nu+\mu\not\in-{\mathbb{N}}$ |  $\frac{x+1}{x-1}$ | 3.4
I | $\pm\mathop{\rm Im}x\lessgtr 0$ | $\nu\in{\mathbb{C}},\ 2\nu\not\in{\mathbb{Z}}$ | $\mu\in{\mathbb{C}}$ | $\nu+\mu\not\in-{\mathbb{N}}_{0}$ |  $\frac{2}{1+x}$ | 3.5
I | $\pm\mathop{\rm Im}x\lessgtr 0$ | $\nu\in{\mathbb{C}},\ 2\nu\not\in{\mathbb{Z}}$ | $\mu\in{\mathbb{C}}$ | $\nu+\mu\not\in-{\mathbb{N}}_{0}$ |  $\frac{2}{1-x}$ | 3.6
I | $x\in D_{1}$ | $\nu\in{\mathbb{C}}$ | $\mu\in{\mathbb{C}}$ | $\nu+\mu\not\in-{\mathbb{Z}}$ |  $\frac{1-x}{2},\ \frac{1+x}{2}$ | 3.9
II |  $x\in D_{1}^{+}$ | $\nu\in{\mathbb{C}}$ | $\mu\in{\mathbb{C}}\setminus{\mathbb{Z}}$ | $\nu+\mu\not\in-{\mathbb{N}}$ |  $1-x^{2}$ | 4.1
II | $\pm\mathop{\rm Im}x\lessgtr 0$ | $\nu\in{\mathbb{C}},\ \nu+\frac{1}{2}\not\in{\mathbb{Z}}$ | $\mu\in{\mathbb{C}}$ | $\nu+\mu\not\in-{\mathbb{N}}$ |  $\frac{1}{1-x^{2}}$ | 4.2
II | $x\in D_{1}$ | $\nu\in{\mathbb{C}}$ | $\mu\in{\mathbb{C}}$ | $\nu+\mu\not\in-{\mathbb{N}}$ | $x^{2}$ | 4.3
II | $\pm\mathop{\rm Im}x\lessgtr 0$ | $\nu\in{\mathbb{C}},\ \nu+\frac{1}{2}\not\in{\mathbb{Z}}$ | $\mu\in{\mathbb{C}}$ | $\nu+\mu\not\in-{\mathbb{N}}$ |  $\frac{1}{x^{2}}$ | 4.4
II | $x\in D_{1}^{+}$ | $\nu\in{\mathbb{C}}$ | $\mu\in{\mathbb{C}}\setminus{\mathbb{Z}}$ | $\nu+\mu\not\in-{\mathbb{N}}$ |  $\frac{x^{2}-1}{x^{2}}$ | 4.5
II | $x\in D_{1}$ | $\nu\in{\mathbb{C}}$ | $\mu\in{\mathbb{C}}$ | $\nu+\mu\not\in{\mathbb{N}}$ |  $\frac{x^{2}}{x^{2}-1}$ | 4.6
III | $x\in D_{1}$ | $\nu\in{\mathbb{C}},\ \nu+\frac{1}{2}\not\in{\mathbb{Z}}$ | $\mu\in{\mathbb{C}}$ | $\nu+\mu\not\in-{\mathbb{N}}$ |  $\frac{\pm x+{\rm i}\sqrt{1-x^{2}}}{2{\rm i}\sqrt{1-x^{2}}}$ | 5.1
III | $x\in D_{1}$ | $\nu\in{\mathbb{C}},\ \nu+\frac{1}{2}\not\in{\mathbb{Z}}$ | $\mu\in{\mathbb{C}}$ | $\nu+\mu\not\in-{\mathbb{N}}$ |  $\frac{x\mp{\rm i}\sqrt{1-x^{2}}}{x\pm{\rm i}\sqrt{1-x^{2}}}$ | 5.2
III | $x\in D_{1}^{+}$ | $\nu\in{\mathbb{C}}$ | $\mu\in{\mathbb{C}},\ 2\mu\not\in{\mathbb{Z}}$ | $\nu+\mu\not\in-{\mathbb{N}}$ |  $\frac{2{\rm i}\sqrt{1-x^{2}}}{\pm x+{\rm i}\sqrt{1-x^{2}}}$ | 5.3
III | $x\in D_{1}$ | $\nu\in{\mathbb{C}}$ | $\mu\in{\mathbb{C}}$ | $\nu+\mu\not\in-{\mathbb{N}}$ |  $\frac{u}{v},\ \frac{v}{u}$ | 6.1

This paper is organized as follows. In Section 2 we provide definitions and
properties of the special functions used in this paper. In Sections 3 (Group
I), 4 (Group II), and 5 (Group III), we give the list of 18 Gauss
hypergeometric representations of $\mathsf{Q}_{\nu}^{\mu}(x)$ in terms of
Olbricht’s solutions and include some discussion of these results. In Section
6 we look at the Fourier expansion of the function
$(\sin\theta)^{-\mu}\mathsf{Q}_{\nu}^{\mu}(\cos\theta)$, and in Section 7 we
list the regions in the complex $x$-plane, where the hypergeometric arguments
$|w_{j}|<1$ for each $j=1,2,\dots,18$.

##  2 Special functions

We will use the following notations. Let $\mathbb{N}:=\\{1,2,3,\ldots\\}$,
$\mathbb{N}_{0}:=\mathbb{N}\cup\\{0\\}$, $\mathbb{Z}:=\\{0,\pm 1,\pm
2,\ldots\\}$, $\mathbb{R}$ real numbers, $\mathbb{C}$ complex numbers.

###  2.1 The gamma and Gauss hypergeometric functions

The gamma function
$\Gamma\colon\mathbb{C}\setminus-\mathbb{N}_{0}\to{\mathbb{C}}$ generalizes
the factorial $n!=\Gamma(n+1)$, $n\in\mathbb{N}_{0}$ (see [11, Chapter 5]). We
will use Euler’s reflection formula [11, equation (5.5.3)]

$\displaystyle\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}{,}$ (2.1)

and the duplication formula [11, equation (5.5.5)]

$\displaystyle\Gamma(2z)=\frac{2^{2z-1}}{\sqrt{\pi}}\Gamma(z)\Gamma\big{(}z+\tfrac{1}{2}\big{)}.$

The hypergeometric function ${}_{2}F_{1}(a,b;c;w)$, where
$(a,b,c,w)\in\mathbb{C}^{2}\times(\mathbb{C}\setminus-\mathbb{N}_{0})\times(\mathbb{C}\setminus[1,\infty))$
is a complex valued analytic function which can be defined in terms of the
infinite series for $|w|<1$,

$\displaystyle{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,b}{c};w\right):=\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}}\frac{w^{n}}{n!},$
(2.2)

where $(a)_{n}=a(a+1)\cdots(a+n-1)$ is the Pochhammer symbol [11, equation
(5.2.4)], and elsewhere by analytic continuation. Frequently, we will use the
following linear transformations of the Gauss hypergeometric function, valid
for $w\in\mathbb{C}\setminus[1,\infty)$, namely Euler’s and and Pfaff’s
transformations [11, equation (15.8.1)]

$\displaystyle{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,b}{c};w\right)$
$\displaystyle=(1-w)^{c-a-b}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{c-a,c-b}{c};w\right)$
(2.3)
$\displaystyle=(1-w)^{-a}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,c-b}{c};\frac{w}{w-1}\right)$
(2.4)
$\displaystyle=(1-w)^{-b}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{b,c-a}{c};\frac{w}{w-1}\right)\\!.$
(2.5)

Please see Appendix B for a presentation of the values of the hypergeometric
function from above and below its branch cut $[1,\infty)$. Unless stated
otherwise the complex power $x^{y}$ denotes its principal value defined for
$x\in\mathbb{C}\setminus(-\infty,0]$.

###  2.2 The associated Legendre and Ferrers functions

We will follow the notation for Legendre functions as used in [11, Chapter
14]. The associated Legendre functions of the first kind $P_{\nu}^{\mu}(z)$ is
given by [11, equation (14.3.6)]

$\displaystyle
P_{\nu}^{\mu}(z):=\frac{1}{\Gamma(1-\mu)}\left(\frac{z+1}{z-1}\right)^{\frac{1}{2}\mu}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1-\mu};\frac{1-z}{2}\right)\\!.$

It is defined for all $\nu,\mu\in\mathbb{C}$ and all $z\in D_{2}$. The
associated Legendre function of the second kind $Q_{\nu}^{\mu}({z})$ is given
by [11, equation (14.3.7)]

$\displaystyle Q_{\nu}^{\mu}(z):=\frac{\sqrt{\pi}\,{\rm e}^{{\rm
i}\pi\mu}\Gamma(\nu+\mu+1)\big{(}z^{2}-1\big{)}^{\frac{1}{2}\mu}}{2^{\nu+1}\Gamma\big{(}\nu+\frac{3}{2}\big{)}z^{\nu+\mu+1}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{\nu+\mu+2}{2},\frac{\nu+\mu+1}{2}}{\nu+\frac{3}{2}};\frac{1}{z^{2}}\right)\\!,$
(2.6)

where $z\in D_{2}$ and $\nu+\mu\in\mathbb{C}\setminus-\mathbb{N}$. Note that
for any expression of the form $\big{(}z^{2}-1\big{)}^{\alpha}$, read this as

$\displaystyle\big{(}z^{2}-1\big{)}^{\alpha}:=(z+1)^{\alpha}(z-1)^{\alpha},$

for any fixed $\alpha\in\mathbb{C}$ and $z\in D_{2}$.

The Ferrers functions of the first and second kind respectively ${\sf
P}_{\nu}^{\mu}(x)$, $\mathsf{Q}_{\nu}^{\mu}(x)$, can be defined by [11, cf.
equations (14.23.4) and (14.23.5)]

$\displaystyle{\sf P}_{\nu}^{\mu}(x)={\rm e}^{\frac{1}{2}{\rm
i}\pi\mu}P_{\nu}^{\mu}(x+{\rm i}0),$ (2.7)
$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=\tfrac{1}{2}\big{[}{\rm
e}^{-\frac{1}{2}{\rm i}\pi\mu}\big{(}{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(x+{\rm i}0)\big{)}+{\rm e}^{\frac{1}{2}{\rm
i}\pi\mu}\big{(}{\rm e}^{-{\rm i}\pi\mu}Q_{\nu}^{\mu}(x-{\rm
i}0)\big{)}\big{]},$ (2.8)

for $x\in(-1,1)$, although the Ferrers functions can be extended analytically
to the domain $D_{1}$. Note that we have written (2.8) in this particular form
since the table of associated Legendre functions of the second kind found in
Magnus, Oberhettinger and Soni (1966) [9, pp. 155–163] (see also [2, pp.
124–139]), are listed in terms of ${\rm e}^{-{\rm i}\pi\mu}Q_{\nu}^{\mu}(z)$,
so this form of the limit representation is particularly useful.

In regard to the name Ferrers functions (also known as Legendre functions on
the cut), this is the name adopted for these functions in the NIST DLMF [11,
Chapter 14]. This is because the mathematician Norman Macleod Ferrers
(1829–1903), was the first to extensively describe their properties [3]. Of
course, since they are intimately connected to associated Legendre functions,
another name as suggested by a referee could be Legendre–Ferrers
functions.111One may compute the Ferrers function of the first and second kind
respectively with the computer algebra system Mathematica by using the
commands LegendreP[$\nu$,$\mu$,p,$x$] and LegendreQ[$\nu$,$\mu$,p,$x$], where
$p=1,2$. In the computer algebra system MAPLE if you define _ EnvLegendreCut
:= -1..1; (default), then you get the standard associated Legendre functions.
Alternatively, if you define _ EnvLegendreCut :=1..infinity; then you get the
Ferrers functions. 222 The mention of specific products, trademarks, or brand
names is for purposes of identification only. Such mention is not to be
interpreted in any way as an endorsement or certification of such products or
brands by the National Institute of Standards and Technology, nor does it
imply that the products so identified are necessarily the best available for
the purpose. All trademarks mentioned herein belong to their respective
owners.

Common applications of the Ferrers functions include spherical or
hyperspherical and hyperspheroidal harmonics (e.g., the quantum theory of
angular momentum), conical functions, the Mehler–Fock integral transforms,
opposite antipodal fundamental solutions of the Laplace–Beltrami operator and
Helmholtz operators on hyperspheres, and many other applications. The Ferrers
functions appear whenever one performs harmonic analysis on spheres or
hyperspheres with dimension greater than or equal to one because they arise
from the method of separation of variables for the Laplace–Beltrami operator
on these constant positive curvature Riemannian manifolds. In fact, the
orthogonal Gegenbauer (or ultraspherical) polynomials provide a basis for
functions on hyperspheres (as well as on Euclidean space, hyperbolic geometry
and other isotropic manifolds) and are fundamentally connected to the Ferrers
functions of the first kind (see [11, equation (14.3.21)].

The following relations expressing the Ferrers function of the second kind
$\mathsf{Q}_{\nu}^{\mu}$ in terms of the associated Legendre functions
$P_{\nu}^{\mu}$ and $Q_{\nu}^{\mu}$ will be useful. Equations (2.9) and (2.12)
can be found in [11, equation (14.23.6)], and the other equations follow from
the connection relations [11, equations (14.9.12) and (14.9.15)].

######  Theorem 2.1.

  1. $(a)$

For $\mathop{\rm Im}x>0$, we have

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)$ $\displaystyle={\rm
e}^{-\frac{3}{2}\pi{\rm i}\mu}Q_{\nu}^{\mu}(x)+\frac{\pi{\rm i}}{2}{\rm
e}^{\frac{1}{2}\pi{\rm i}\mu}P_{\nu}^{\mu}(x)$ (2.9)
$\displaystyle=\frac{1}{2}\pi\cot(\pi\mu){\rm e}^{\frac{1}{2}\pi{\rm
i}\mu}P_{\nu}^{\mu}(x)-\frac{\pi{{\rm e}^{-\frac{1}{2}\pi{\rm
i}\mu}}}{2\sin(\pi\mu)}\frac{\Gamma(\nu+\mu+1)}{\Gamma(\nu-\mu+1)}P_{\nu}^{-\mu}(x)$
(2.10) $\displaystyle={\rm e}^{-\frac{1}{2}\pi{\rm
i}\mu}\left(\cos(\pi\mu)-{\rm
i}\frac{\sin(\pi(\mu-\nu))}{2\cos(\pi\nu)}\right)Q_{\nu}^{\mu}(x)$
$\displaystyle\phantom{=}+{\rm i}{\rm e}^{-\frac{1}{2}\pi{\rm
i}\mu}\frac{\sin(\pi(\mu-\nu))}{2\cos(\pi\nu)}Q_{-\nu-1}^{\mu}(x).$ (2.11)

  2. $(b)$

For $\mathop{\rm Im}x<0$, we have

$\displaystyle\allowdisplaybreaks\mathsf{Q}_{\nu}^{\mu}(x)={\rm
e}^{-\frac{1}{2}\pi{\rm i}\mu}Q_{\nu}^{\mu}(x)-\frac{\pi{\rm i}}{2}{\rm
e}^{-\frac{1}{2}\pi{\rm i}\mu}P_{\nu}^{\mu}(x)$ (2.12)
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)}{}=\frac{1}{2}\pi\cot(\pi\mu){\rm
e}^{-\frac{1}{2}\pi{\rm i}\mu}P_{\nu}^{\mu}(x)-\frac{\pi{{\rm
e}^{\frac{1}{2}\pi{\rm
i}\mu}}}{2\sin(\pi\mu)}\frac{\Gamma(\nu+\mu+1)}{\Gamma(\nu-\mu+1)}P_{\nu}^{-\mu}(x)$
(2.13) $\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)}{}={\rm
e}^{-\frac{3}{2}\pi{\rm i}\mu}\left(\\!\cos(\pi\mu)+{\rm
i}\frac{\sin(\pi(\mu-\nu))}{2\cos(\pi\nu)}\right)Q_{\nu}^{\mu}(x)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=}{}-{\rm i}{\rm
e}^{-\frac{3}{2}\pi{\rm
i}\mu}\frac{\sin(\pi(\mu-\nu))}{2\cos(\pi\nu)}Q_{-\nu-1}^{\mu}(x).$ (2.14)

##  3 Group I hypergeometric representations for
$\bm{\mathsf{Q}_{\nu}^{\mu}(x)}$

The following result appears in [9, p. 167] in a slightly different form.
However, it is claimed therein that the result is only valid for $x\in(0,1)$,
where in fact it is valid for $x\in(-1,1)$. On the other hand, the same
formula is reproduced in terms of trigonometric functions in [9, p. 169],
where the full range $x\in(-1,1)$ is indicated. This result is also stated in
[11, equation (14.3.2)].

######  Theorem 3.1.

Let $x\in D_{1}$, $\nu\in\mathbb{C}$, $\mu\in\mathbb{C}\setminus\mathbb{Z}$,
$\nu+\mu\not\in-\mathbb{N}$. Then

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=\frac{\pi}{2\sin(\pi\mu)}\Bigg{[}\frac{\cos(\pi\mu)}{\Gamma(1-\mu)}{\left(\frac{1+x}{1-x}\right)}^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1-\mu};\frac{1-x}{2}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=\frac{\pi}{2\sin(\pi\mu)}\Bigg{[}}{}-\frac{\Gamma(\nu+\mu+1)}{\Gamma(\mu+1)\Gamma(\nu-\mu+1)}{\left(\frac{1-x}{1+x}\right)}^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1+\mu};\frac{1-x}{2}\right)\\!\Bigg{]}.$

###### Proof.

According to [11, equation (14.3.20)] and [9, Entry 19, p. 160], the
associated Legendre function of the second kind satisfies

$\displaystyle{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(z)=\frac{\Gamma(-\mu)\Gamma(\nu+\mu+1)}{2\Gamma(\nu-\mu+1)}\left(\frac{z-1}{z+1}\right)^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1+\mu};\frac{1-z}{2}\right)$
$\displaystyle\hphantom{{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(z)=}{}+\frac{1}{2}\Gamma(\mu)\left(\frac{z+1}{z-1}\right)^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1-\mu};\frac{1-z}{2}\right)\\!,$

where $z\in D_{2}$. This can be rewritten as

$\displaystyle{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(z)=\frac{-\Gamma(\nu+\mu+1)}{2\Gamma(\nu-\mu+1)\sin(\pi\mu)\Gamma(\mu+1)}{\left(\frac{z-1}{z+1}\right)}^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1+\mu};\frac{1-z}{2}\right)$
$\displaystyle\hphantom{{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(z)=}{}+\frac{\pi}{2\sin(\pi\mu)\Gamma(1-\mu)}{\left(\frac{z+1}{z-1}\right)}^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1-\mu};\frac{1-z}{2}\right)\\!,$

using the reflection formula (2.1). Using (2.8), we derive from this equation
an expression for $\mathsf{Q}_{\nu}^{\mu}{(x)}$, namely

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=\frac{\pi}{4\sin(\pi\mu)}\Bigg{[}\frac{-{2}\Gamma(\nu+\mu+1)}{\Gamma(\nu-\mu+1)\Gamma(\mu+1)}{\left(\frac{1-x}{1+x}\right)}^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1+\mu};\frac{1-x}{2}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=\frac{\pi}{4\sin(\pi\mu)}\Bigg{[}}{}+\frac{{\rm
e}^{-{\rm i}\pi\mu}+{\rm e}^{{\rm
i}\pi\mu}}{\Gamma(1-\mu)}{\left(\frac{1+x}{1-x}\right)}^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1-\mu};\frac{1-x}{2}\right)\Bigg{]}$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)}{}=\frac{\pi}{2\sin(\pi\mu)}\Bigg{[}\frac{\cos(\pi\mu)}{\Gamma(1-\mu)}{\left(\frac{1+x}{1-x}\right)}^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1-\mu};\frac{1-x}{2}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=\frac{\pi}{2\sin(\pi\mu)}\Bigg{[}}{}-\frac{\Gamma(\nu+\mu+1)}{\Gamma(\nu-\mu+1)\Gamma(\mu+1)}{\left(\frac{1-x}{1+x}\right)}^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1+\mu};\frac{1-x}{2}\right)\Bigg{]},$

which is true for all $x\in(-1,1)$. Since both sides of the claimed identity
are analytic functions on $D_{1}$, the identity is true for all $x\in D_{1}$.
∎

######  Theorem 3.2.

Let $x\in D_{1}$, $\nu\in\mathbb{C}$, $\mu\in\mathbb{C}\setminus\mathbb{Z}$,
such that $\nu+\mu\notin-\mathbb{N}$. Then

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=-\frac{1}{2}\Bigg{[}\cos(\pi\nu)\Gamma(\mu){\left(\frac{1-x}{1+x}\right)}^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1-\mu};\frac{1+x}{2}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=}{}+\cos(\pi(\nu+\mu))\Gamma(-\mu)\frac{\Gamma(\nu+\mu+1)}{\Gamma(\nu-\mu+1)}{\left(\frac{1+x}{1-x}\right)}^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1+\mu};\frac{1+x}{2}\right)\Bigg{]}.$

###### Proof.

According to [9, Entry 20, p. 160], the associated Legendre function of the
second kind satisfies

$\displaystyle{\rm e}^{-{\rm i}\pi\mu}Q_{\nu}^{\mu}(z)=-\frac{1}{2}{\rm
e}^{\mp{\rm
i}\pi\nu}\Gamma(\mu){\left(\frac{z-1}{z+1}\right)}^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1-\mu};\frac{1+z}{2}\right)$
$\displaystyle\hphantom{{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(z)=}{}-\frac{1}{2}{\rm e}^{\mp{\rm
i}\pi\nu}\frac{\Gamma(\nu+\mu+1)\Gamma(-\mu)}{\Gamma(\nu-\mu+1)}{\left(\frac{z+1}{z-1}\right)}^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1+\mu};\frac{1+z}{2}\right)\\!,$

when $\pm\mathop{\rm Im}z>0$. Using (2.8), we derive from this equation an
expression for $\mathsf{Q}_{\nu}^{\mu}{(x)}$, namely

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=-\frac{1}{4}\Bigg{[}{\rm e}^{-{\rm
i}\pi\nu}\Gamma(\mu){\left(\frac{1-x}{1+x}\right)}^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1-\mu};\frac{1+x}{2}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=-\frac{1}{4}\Bigg{[}}{}+{{\rm
e}^{-{\rm
i}\pi(\nu+\mu)}}\frac{\Gamma(\nu+\mu+1)\Gamma(-\mu)}{\Gamma(\nu-\mu+1)}{\left(\frac{1+x}{1-x}\right)}^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1+\mu};\frac{1+x}{2}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=-\frac{1}{4}\Bigg{[}}{}+{\rm
e}^{{\rm
i}\pi\nu}\Gamma(\mu){\left(\frac{1-x}{1+x}\right)}^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1-\mu};\frac{1+x}{2}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=-\frac{1}{4}\Bigg{[}}{}+{{\rm
e}^{{\rm
i}\pi(\nu+\mu)}}\frac{\Gamma(\nu+\mu+1)\Gamma(-\mu)}{\Gamma(\nu-\mu+1)}{\left(\frac{1+x}{1-x}\right)}^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1+\mu};\frac{1+x}{2}\right)\Bigg{]}$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)}{}=-\frac{1}{2}\Bigg{[}\cos(\pi\nu)\Gamma(\mu){\left(\frac{1-x}{1+x}\right)}^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1-\mu};\frac{1+x}{2}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=-\frac{1}{2}\Bigg{[}}{}+\cos(\pi(\nu+\mu))\frac{\Gamma(\nu+\mu+1)\Gamma(-\mu)}{\Gamma(\nu-\mu+1)}\left(\frac{1+x}{1-x}\right)^{\frac{\mu}{2}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1+\mu};\frac{1+x}{2}\right)\Bigg{]}.$

This completes the proof. ∎

######  Theorem 3.3.

Let $x\in D_{1}$, $\nu\in\mathbb{C}$, $\mu\in\mathbb{C}\setminus\mathbb{Z}$,
such that $\nu+\mu\notin-\mathbb{N}$. Then

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=\frac{(1+x)^{\nu}}{2^{\nu+1}}\Bigg{[}\cos(\pi\mu)\Gamma(\mu){\left(\frac{1+x}{1-x}\right)}^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,-\nu-\mu}{1-\mu};\frac{x-1}{x+1}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=\frac{(1+x)^{\nu}}{2^{\nu+1}}\Bigg{[}}{}+\frac{\Gamma(\nu+\mu+1)\Gamma(-\mu)}{\Gamma(\nu-\mu+1)}{\left(\frac{1-x}{1+x}\right)}^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\mu-\nu}{1+\mu};\frac{x-1}{x+1}\right)\Bigg{]}.$

###### Proof.

This follows from [9, Entry 21, p. 160]. ∎

######  Theorem 3.4.

Let $x\in D_{1}$, $\nu\in\mathbb{C}$, $\mu\in\mathbb{C}\setminus\mathbb{Z}$,
such that $\nu+\mu\notin-\mathbb{N}$. Then

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=\frac{-2^{\nu}}{(1-x)^{\nu+1}}\Bigg{[}\Gamma(\mu)\cos(\pi\nu)\left(\frac{1-x}{1+x}\right)^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\nu+1,\nu-\mu+1}{1-\mu};\frac{x+1}{x-1}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=}{}+\frac{\Gamma(-\mu)\cos(\pi(\nu+\mu))\Gamma(\nu+\mu+1)}{\Gamma(\nu-\mu+1)}\left(\frac{1+x}{1-x}\right)^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\nu+1,\nu+\mu+1}{1+\mu};\frac{x+1}{x-1}\right)\Bigg{]}.$

###### Proof.

This follows from [9, Entry 22, p. 160]. ∎

######  Theorem 3.5.

Let $\nu,\mu\in\mathbb{C}$ such that $\nu+\mu\notin-\mathbb{N}_{0}$,
$2\nu\notin\mathbb{Z}$. If $\pm\mathop{\rm Im}x>0$ then

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=2^{\nu}\left(\cos(\pi\mu)\mp{\rm
i}\frac{\sin(\pi(\mu-\nu))}{2\cos(\pi\nu)}\right)\frac{\Gamma(\nu+1)\Gamma(\nu+\mu+1)}{\Gamma(2\nu+2)}$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=2^{\nu}}{}\times(1+x)^{-1+\frac{1}{2}\mu-\nu}(1-x)^{-\frac{1}{2}\mu}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\nu-\mu+1,\nu+1}{2\nu+2};\frac{2}{1+x}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=}{}\pm{\rm i}\pi
2^{-\nu-2}\sec(\pi\nu)\frac{\Gamma(-\nu)}{\Gamma(-2\nu)\Gamma(\nu-\mu+1)}$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=2^{\nu}}{}\times(1+x)^{\nu+\frac{1}{2}\mu}(1-x)^{-\frac{1}{2}\mu}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,-\nu-\mu}{-2\nu};\frac{2}{1+x}\right)\\!.$

###### Proof.

We represent $\mathsf{Q}_{\nu}^{\mu}(x)$ by (2.11) if $\mathop{\rm Im}x>0$ and
by (2.14) if $\mathop{\rm Im}x<0$. Then we use [9, Entry 23, p. 161], namely

$\displaystyle{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(x)=2^{\nu}\frac{\Gamma(\nu+1)\Gamma(\nu+\mu+1)}{\Gamma(2\nu+2)}\frac{(x+1)^{{\frac{1}{2}\mu}-\nu-1}}{(x-1)^{{\frac{1}{2}\mu}}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\nu+1,\nu-\mu+1}{2\nu+2};\frac{2}{1+x}\right)\\!,$

for $x\in D_{2}$ and the desired result follows. ∎

######  Theorem 3.6.

Let $\nu,\mu\in\mathbb{C}$ such that $\nu+\mu\notin-\mathbb{N}_{0}$,
$2\nu\notin\mathbb{Z}$. If $\pm\mathop{\rm Im}x>0$ then

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=2^{\nu}{\rm e}^{\mp\pi{\rm
i}(\nu+1)}\left(\cos(\pi\mu)\mp{\rm
i}\frac{\sin(\pi(\mu-\nu))}{2\cos(\pi\nu)}\right)\frac{\Gamma(\nu+1)\Gamma(\nu+\mu+1)}{\Gamma(2\nu+2)}$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=2^{\nu}}{}\times(1+x)^{\frac{1}{2}\mu}(1-x)^{{-\nu-\frac{1}{2}\mu-1}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\nu+\mu+1,\nu+1}{2\nu+2};\frac{2}{1-x}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=}\pm{\rm i}\pi
2^{-\nu-2}{\rm e}^{\pm\pi{\rm
i}\nu}\sec(\pi\nu)\frac{\Gamma(-\nu)}{\Gamma(-2\nu)\Gamma(\nu-\mu+1)}$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=2^{\nu}}{}\times(1+x)^{\frac{1}{2}\mu}(1-x)^{\nu-\frac{1}{2}\mu}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\mu-\nu}{-2\nu};\frac{2}{1-x}\right)\\!.$

###### Proof.

We represent $\mathsf{Q}_{\nu}^{\mu}(x)$ by (2.11) if $\mathop{\rm Im}x>0$ and
by (2.14) if $\mathop{\rm Im}x<0$. Then we use [9, Entry 24, p. 161], namely

$\displaystyle{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(x)=2^{\nu}\frac{\Gamma(\nu+1)\Gamma(\nu+\mu+1)}{\Gamma(2\nu+2)}\frac{(x+1)^{\frac{1}{2}\mu}}{(x-1)^{\nu+\frac{1}{2}\mu+1}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\nu+1,\nu+\mu+1}{2\mu+2};\frac{2}{1-x}\right)\\!,$

for $x\in D_{2}$ and the desired result follows. ∎

######  Remark 3.7.

The hypergeometric representations of $\mathsf{Q}_{\nu}^{\mu}(x)$ given in the
previous six theorems can be written in slightly different forms by applying
Euler’s transformation (2.3). For instance, the result of Theorem 3.1 can be
written as

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=\frac{\pi}{2\sin(\pi\mu)}\Bigg{[}2^{\mu}\frac{\cos(\pi\mu)}{\Gamma(1-\mu)}\big{(}1-x^{2}\big{)}^{-\frac{\mu}{2}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\nu-\mu+1,-\mu-\nu}{1-\mu};\frac{1-x}{2}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=\frac{\pi}{2\sin(\pi\mu)}\Bigg{[}}{}-2^{-\mu}\frac{\Gamma(\nu+\mu+1)}{\Gamma(\mu+1)\Gamma(\nu-\mu+1)}\big{(}1-x^{2}\big{)}^{\frac{\mu}{2}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\nu\\!+\mu\\!+1,\mu\\!-\nu}{1+\mu};\frac{1\\!-x}{2}\right)\Bigg{]}.$

We observe that by applying Pfaff’s transformation (2.4) and (2.5), the
results of Theorems 3.3, 3.4 and 3.6 follow immediately from those of Theorems
3.1, 3.2 and 3.5, respectively.

Each of the previous six theorems represents $\mathsf{Q}_{\nu}^{\mu}(x)$ as a
sum of two functions each of which is itself a solution of the associated
Legendre equation (1.1). In the following we identify these solutions.

######  Remark 3.8.

In Theorems 3.1 and 3.3, $\mathsf{Q}_{\nu}^{\mu}(x)$ is expressed as a linear
combination of ${\sf P}_{\nu}^{\mu}(x)$ and ${\sf P}_{\nu}^{-\mu}(x)$. This
follows from the connection formula [11, cf. equation (14.9.2)]

$\displaystyle\frac{2}{\pi}\sin(\pi\mu)\mathsf{Q}_{\nu}^{\mu}(x)=\cos(\pi\mu){\sf
P}_{\nu}^{\mu}(x)-\frac{\Gamma(\nu+\mu+1)}{\Gamma(\nu-\mu+1)}{\sf
P}_{\nu}^{-\mu}(x),$ (3.1)

and the hypergeometric representation [11, equation (14.3.1)] of the Ferrers
function of the first kind

$\displaystyle{\sf
P}_{\nu}^{\mu}(x)=\left(\frac{1+x}{1-x}\right)^{\frac{1}{2}\mu}\frac{1}{\Gamma(1-\mu)}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1-\mu};\frac{1-x}{2}\right)\\!.$
(3.2)

Analogously, in Theorems 3.2 and 3.4, $\mathsf{Q}_{\nu}^{\mu}(x)$ is written
as a linear combination of ${\sf P}_{\nu}^{\mu}(-x)$ and ${\sf
P}_{\nu}^{-\mu}(-x)$. This follows from the connection relation

$\displaystyle\frac{2}{\pi}\sin(\pi\mu)\mathsf{Q}_{\nu}^{\mu}(x)=-\cos(\pi\nu){\sf
P}_{\nu}^{\mu}(-x)+\cos(\pi(\nu+\mu))\frac{\Gamma(\nu+\mu+1)}{\Gamma(\nu-\mu+1)}{\sf
P}_{\nu}^{-\mu}(-x){,}$

which is a consequence of [11, equations (14.9.7) and (14.9.10)]. We will use
this observation to derive additional hypergeometric representations of
$\mathsf{Q}_{\nu}^{\mu}(x)$.

The connection relation [11, equation (14.9.10)]

$\displaystyle\frac{2}{\pi}\sin(\pi(\nu+\mu))\mathsf{Q}_{\nu}^{\mu}(x)=\cos(\pi(\nu+\mu)){\sf
P}_{\nu}^{\mu}(x)-{\sf P}_{\nu}^{\mu}(-x),$

and (3.2) lead to the following result.

######  Theorem 3.9.

Let $x\in D_{1}$, $\mu,\nu\in\mathbb{C}$ such that $\nu+\mu\notin-\mathbb{Z}$.
Then

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=\frac{\pi}{2}\cot(\pi(\nu+\mu))\left(\frac{1+x}{1-x}\right)^{\frac{1}{2}\mu}\frac{1}{\Gamma(1-\mu)}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1-\mu};\frac{1-x}{2}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=}{}-\frac{\pi}{2}\csc(\pi(\nu+\mu))\left(\frac{1-x}{1+x}\right)^{\frac{1}{2}\mu}\frac{1}{\Gamma(1-\mu)}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu,\nu+1}{1-\mu};\frac{1+x}{2}\right)\\!.$

######  Remark 3.10.

The hypergeometric representation in Theorem 3.9 involves two hypergeometric
functions with the same parameters $a=\nu+1$, $b=-\nu$, $c=1-\mu$ but two
different arguments $(1\pm x)/2$ while the previous results in this section
involve two hypergeometric functions with two different set of parameters but
the same argument. By representing $\mathsf{Q}_{\nu}^{\mu}(x)$ in terms of
$\big{\\{}{\sf P}_{\nu}^{\mu}(-x),{\sf P}_{\nu}^{-\mu}(x)\big{\\}}$,
$\big{\\{}{\sf P}_{\nu}^{-\mu}(-x),{\sf P}_{\nu}^{\mu}(x)\big{\\}}$,
$\big{\\{}{\sf P}_{\nu}^{-\mu}(-x),{\sf P}_{\nu}^{-\mu}(x)\big{\\}}$,
respectively, we can obtain three additional hypergeometric representations of
$\mathsf{Q}_{\nu}^{\mu}(x)$.

######  Remark 3.11.

In Theorem 3.5, $\mathsf{Q}_{\nu}^{\mu}(x)$ is expressed as a linear
combination of $Q_{\nu}^{\mu}(x)$ and $Q_{-\nu-1}^{\mu}(x)$ in the upper and
lower half-plane. The argument of the hypergeometric function is
$w_{5}=2/(1+x)$. If one considers $x\in(-1,1)$, then this function maps to
$(1,\infty)$, where the hypergeometric function takes two values depending on
whether the value is approached from above or below the ray $[1,\infty)$.
These values can be computed by Theorems B.1–B.4 which are given in Appendix
B. We map $w_{5}(x\pm{\rm i}0)$ to $(1+x)/2$, $(x-1)/(x+1)$, $(1-x)/2$,
$(x+1)/(x-1)$, which have already been encountered in Theorems 3.2, 3.3, 3.1
and 3.4, respectively. Similar remarks apply to Theorem 3.6.

##  4 Group II hypergeometric representations for
$\bm{\mathsf{Q}_{\nu}^{\mu}(x)}$

######  Theorem 4.1.

Let $x\in D_{1}^{+}$, $\nu\in\mathbb{C}$,
$\mu\in\mathbb{C}\setminus\mathbb{Z}$,
$\nu+\mu\in\mathbb{C}\setminus-\mathbb{N}$. Then

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=\frac{2^{\mu-1}\Gamma(\mu)\cos(\pi\mu)}{\big{(}1-x^{2}\big{)}^{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{\nu-\mu+1}{2},\frac{-\nu-\mu}{2}}{1-\mu};1-x^{2}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=}{}+\frac{\Gamma(-\mu)\Gamma(\nu+\mu+1)\big{(}1-x^{2}\big{)}^{\frac{1}{2}\mu}}{2^{1+\mu}\Gamma(\nu-\mu+1)}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{\nu+\mu+1}{2},\frac{\mu-\nu}{2}}{1+\mu};1-x^{2}\right)\\!.$

###### Proof.

Use the Gauss hypergeometric representation of the associated Legendre
function of the second kind [9, Entry 25, p. 161],

$\displaystyle{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(z)=2^{\mu-1}\Gamma(\mu)(z^{2}-1)^{-{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{\nu-\mu+1}{2},\frac{-\nu-\mu}{2}}{1-\mu};1-z^{2}\right)$
$\displaystyle\hphantom{{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(z)=}{}+\frac{\Gamma(\nu+\mu+1)\Gamma(-\mu)}{2^{1+\mu}\Gamma(\nu-\mu+1)}(z^{2}-1)^{{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{\nu+\mu+1}{2},\frac{\mu-\nu}{2}}{1+\mu};1-z^{2}\right)\\!,$

valid for $z\in D_{2}$ with $\mathop{\rm Re}z>0$. Now (2.8) gives the desired
representation for $x\in(0,1)$. Since both sides of the equation are analytic
for $x\in D_{1}$, $\mathop{\rm Re}x>0$. The full statement follows. ∎

######  Theorem 4.2.

Let $\nu,\mu\in\mathbb{C}$ such that $\nu+\mu\notin-\mathbb{N}$ and
$\nu+\frac{1}{2}\notin\mathbb{Z}$. If $\pm\mathop{\rm Im}x>0$ then

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=\sqrt{\pi}\,2^{-\nu-1}{\rm
e}^{\pm\frac{1}{2}{\rm i}\pi(-\nu+\mu-1)}\left(\cos(\pi\mu)\mp{\rm
i}\frac{\sin(\pi(\mu-\nu))}{2\cos(\pi\nu)}\right)\frac{\Gamma(\nu+\mu+1)}{\Gamma\big{(}\nu+\frac{3}{2}\big{)}}$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=+}{}\times\big{(}1-x^{2}\big{)}^{-\frac{1}{2}\nu-\frac{1}{2}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{\nu-\mu+1}{2},\frac{\nu+\mu+1}{2}}{\nu+\frac{3}{2}};\frac{1}{1-x^{2}}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=}{}+\pi^{\frac{3}{2}}2^{\nu-1}\frac{{\rm
e}^{\pm\frac{1}{2}{\rm
i}\pi(\nu+\mu+1)}\sec(\pi\nu)}{\Gamma(\nu-\mu+1)\Gamma\big{(}\frac{1}{2}-\nu\big{)}}\big{(}1-x^{2}\big{)}^{\frac{1}{2}\nu}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\frac{\nu+\mu}{2},\frac{\mu-\nu}{2}}{\frac{1}{2}-\nu};\frac{1}{1-x^{2}}\right)\\!.$

###### Proof.

We represent $\mathsf{Q}_{\nu}^{\mu}(x)$ by (2.11) if $\mathop{\rm Im}x>0$ and
by (2.14) if $\mathop{\rm Im}x<0$. Then we use [9, Entry 26, p. 161]

$\displaystyle{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(z)=\sqrt{\pi}\,2^{-\nu-1}\frac{\Gamma(\nu+\mu+1)}{\Gamma\big{(}\nu+\frac{3}{2}\big{)}}(z^{2}-1)^{\tfrac{1}{2}(\nu+1)}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{\nu-\mu+1}{2},\frac{\nu+\mu+1}{2}}{\nu+\frac{3}{2}};\frac{1}{1-z^{2}}\right)\\!,$

for $z\in D_{2}$ and the desired result follows. ∎

######  Theorem 4.3 (equation (14.3.12) in [11]).

Let $x\in D_{1}$ and $\nu,\mu\in\mathbb{C}$, such that
$\nu+\mu\notin-\mathbb{N}$. Then

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=\frac{\sqrt{\pi}\,2^{\mu-1}}{(1-x^{2})^{\frac{1}{2}\mu}}\Bigg{[}\frac{-\sin\big{(}\frac{\pi}{2}(\nu+\mu)\big{)}\Gamma\big{(}\frac{\nu+\mu+1}{2}\big{)}}{\Gamma\big{(}\frac{\nu-\mu+2}{2}\big{)}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\frac{\nu+\mu}{2},\frac{\nu-\mu+1}{2}}{\frac{1}{2}};x^{2}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=\frac{\sqrt{\pi}\,2^{\mu-1}}{(1-x^{2})^{\frac{1}{2}\mu}}\Bigg{[}}{}+\frac{2\cos\big{(}\tfrac{\pi}{2}(\nu+\mu)\big{)}\Gamma\big{(}\frac{\nu+\mu+2}{2}\big{)}\,x}{\Gamma\big{(}\frac{\nu-\mu+1}{2}\big{)}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{-\nu-\mu+1}{2},\frac{\nu-\mu+2}{2}}{\frac{3}{2}};x^{2}\right)\\!\Bigg{]}.$

###### Proof.

By [9, Entry 27, p. 161], the associated Legendre function of the second kind
satisfies

$\displaystyle{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(z)=\frac{\pi^{\frac{1}{2}}2^{\mu}\Gamma\big{(}\frac{\nu+\mu+2}{2}\big{)}z}{\Gamma\big{(}\frac{\nu-\mu+1}{2}\big{)}(z^{2}-1)^{{\frac{1}{2}\mu}}}{\rm
e}^{\pm\frac{1}{2}{\rm
i}\pi(\mu-\nu)}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{-\nu-\mu+1}{2},\frac{\nu-\mu+2}{2}}{\frac{3}{2}};z^{2}\right)$
$\displaystyle\hphantom{{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(z)=}{}+\frac{\pi^{\frac{1}{2}}2^{\mu-1}\Gamma\big{(}\frac{\nu+\mu+1}{2}\big{)}}{\Gamma\big{(}\frac{\nu-\mu+1}{2}\big{)}(z^{2}-1)^{{\frac{1}{2}\mu}}}{\rm
e}^{\pm\frac{1}{2}{\rm
i}\pi(\mu-\nu-1)}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{-\nu-\mu}{2},\frac{\nu-\mu+1}{2}}{\frac{1}{2}};z^{2}\right)\\!,$

where the upper and lower sign holds according to $\pm\mathop{\rm Im}z>0$.
Using this equation with (2.8), one derives

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=\frac{\pi^{\frac{1}{2}}2^{\mu-2}}{(1-x^{2})^{{\frac{1}{2}\mu}}}\Bigg{[}2x\big{(}{\rm
e}^{\frac{1}{2}{\rm i}\pi(\nu+\mu)}+{\rm e}^{-\frac{1}{2}{\rm
i}\pi(\nu+\mu)}\big{)}\frac{\Gamma\big{(}\frac{\nu+\mu+2}{2}\big{)}}{\Gamma\big{(}\frac{\nu-\mu+1}{2}\big{)}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{-\nu-\mu+1}{2},\frac{\nu-\mu+2}{2}}{\frac{3}{2}};x^{2}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=\frac{\pi^{\frac{1}{2}}2^{\mu-2}}{(1-x^{2})^{{\frac{1}{2}\mu}}}\Bigg{[}}{}+\big{(}{\rm
e}^{-\frac{1}{2}{\rm i}\pi(\nu+\mu+1)}+{\rm e}^{\frac{1}{2}{\rm
i}\pi(\nu+\mu+1)}\big{)}\frac{\Gamma\big{(}\frac{\nu+\mu+1}{2}\big{)}}{\Gamma\big{(}\frac{\nu-\mu+2}{2}\big{)}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{-\nu-\mu}{2},\frac{\nu-\mu+1}{2}}{\frac{1}{2}};x^{2}\right)\\!\Bigg{]}.$

Since this is equivalent to the claimed result, we have completed the proof. ∎

######  Theorem 4.4.

Let $\nu,\mu\in\mathbb{C}$ such that $\nu+\mu\notin-\mathbb{N}$ and
$\nu+\frac{1}{2}\notin\mathbb{Z}$. If $\pm\mathop{\rm Im}x>0$ then

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=\sqrt{\pi}\,2^{-\nu-1}{\rm e}^{\pm{\rm
i}\pi\mu}\left(\cos(\pi\mu)\mp{\rm
i}\frac{\sin(\pi(\mu-\nu))}{2\cos(\pi\nu)}\right)\frac{\Gamma(\nu+\mu+1)}{\Gamma\big{(}\nu+\frac{3}{2}\big{)}}$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=\sqrt{\pi}}{}\times
x^{{-\nu-\mu-1}}\big{(}1-x^{2}\big{)}^{\frac{1}{2}\mu}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{\nu+\mu+1}{2},\frac{\nu+\mu+2}{2}}{\nu+\frac{3}{2}};\frac{1}{x^{2}}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=}{}+\pi^{\frac{3}{2}}2^{\nu-1}\frac{{\rm
e}^{\pm\pi{\rm
i}\left(\frac{1}{2}+\mu\right)}\sec(\pi\nu)}{\Gamma(\nu-\mu+1)\Gamma\big{(}\frac{1}{2}-\nu\big{)}}x^{\nu-\mu}\big{(}1-x^{2}\big{)}^{\frac{1}{2}\mu}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{\mu-\nu}{2},\frac{\mu-\nu+1}{2}}{\frac{1}{2}-\nu};\frac{1}{x^{2}}\right)\\!.$

###### Proof.

We represent $\mathsf{Q}_{\nu}^{\mu}(x)$ by (2.11) if $\mathop{\rm Im}x>0$ and
by (2.14) if $\mathop{\rm Im}x<0$. Then we use [9, Entry 28, p. 162] which
agrees with (2.6), and the desired result follows. ∎

######  Theorem 4.5.

Let $x\in D_{1}^{+}$, $\nu\in\mathbb{C}$,
$\mu\in\mathbb{C}\setminus\mathbb{Z}$ such that $\nu+\mu\notin-\mathbb{N}$.
Then

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=\frac{2^{\mu-1}\cos(\pi\mu)\Gamma(\mu)}{\big{(}1-x^{2}\big{)}^{{\frac{1}{2}\mu}}}x^{\nu+\mu}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{-\nu-\mu}{2},\frac{-\nu-\mu+1}{2}}{1-\mu};\frac{x^{2}-1}{x^{2}}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=}{}+\frac{\Gamma(\nu+\mu+1)\Gamma(-\mu)}{2^{\mu+1}\Gamma(\nu-\mu+1)}\big{(}1-x^{2}\big{)}^{{\frac{1}{2}\mu}}x^{\nu-\mu}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{\mu-\nu}{2},\frac{\mu-\nu+1}{2}}{1+\mu};\frac{x^{2}-1}{x^{2}}\right)\\!.$

###### Proof.

The hypergeometric representation [9, Entry 29, p. 162],

$\displaystyle{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(z)=\frac{2^{\mu-1}\Gamma(\mu)z^{\nu+\mu}}{(z^{2}-1)^{\frac{1}{2}\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{-\nu-\mu}{2},\frac{1-\nu-\mu}{2}}{1-\mu};1-\frac{1}{z^{2}}\right)$
$\displaystyle\hphantom{{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(z)=}{}+\frac{\Gamma(\nu+\mu+1)\Gamma(-\mu)z^{\nu-\mu}(z^{2}-1)^{\frac{1}{2}\mu}}{2^{\mu+1}\Gamma(\nu-\mu+1)}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{\mu-\nu}{2},\frac{\mu-\nu+1}{2}}{1+\mu};1-\frac{1}{z^{2}}\right)\\!,$

is valid for $\mathop{\rm Re}z\\!>\\!0$, $z\not\in(0,1]$ $\big{(}$then
$1-\frac{1}{z^{2}}\notin[1,\infty)\big{)}$. Using (2.8), the result follows. ∎

######  Theorem 4.6.

Let $x\in D_{1}$, $\mu,\nu\in\mathbb{C}$ such that $\nu+\mu\notin\mathbb{N}$.
Then

$\displaystyle{\sf
Q}_{\nu}^{\mu}(x)=\sqrt{\pi}\,2^{\mu}\Bigg{[}\frac{-\Gamma\big{(}\frac{\nu+\mu+1}{2}\big{)}\sin\big{(}\frac{\pi}{2}(\nu+\mu)\big{)}}{2\Gamma\big{(}\frac{\nu-\mu+2}{2}\big{)}\big{(}1-x^{2}\big{)}^{\tfrac{1}{2}(\nu+1)}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{\nu-\mu+1}{2},\frac{\nu+\mu+1}{2}}{\frac{1}{2}};\frac{x^{2}}{x^{2}-1}\right)$
$\displaystyle\hphantom{{\sf
Q}_{\nu}^{\mu}(x)=}{}+\frac{\Gamma\big{(}\frac{\nu+\mu+2}{2}\big{)}\cos\big{(}\frac{\pi}{2}(\nu+\mu)\big{)}x\big{(}1-x^{2}\big{)}^{{\frac{1}{2}(\nu-1)}}}{\Gamma\big{(}\frac{\nu-\mu+1}{2}\big{)}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{\mu-\nu+1}{2},\frac{-\nu-\mu+1}{2}}{\frac{3}{2}};\frac{x^{2}}{x^{2}-1}\right)\\!\Bigg{]}.$

###### Proof.

The hypergeometric representation of the associated Legendre function of the
second kind [9, Entry 30, p. 162],

$\displaystyle{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(z)=\frac{\sqrt{\pi}\,2^{\mu}\Gamma\big{(}\frac{\nu+\mu+2}{2}\big{)}}{\Gamma\big{(}\frac{\nu-\mu+1}{2}\big{)}}{\rm
e}^{\mp\frac{1}{2}{\rm
i}\pi\left(\nu-\frac{1}{2}\right)}z(z^{2}-1)^{{\frac{1}{2}(\nu-1)}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{-\nu-\mu+1}{2},\frac{\mu-\nu+1}{2}}{\frac{3}{2}};\frac{z^{2}}{z^{2}-1}\right)$
$\displaystyle\hphantom{{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(z)=}{}+\frac{\sqrt{\pi}\,2^{\mu-1}\Gamma\big{(}\frac{\nu+\mu+1}{2}\big{)}}{\Gamma\big{(}\frac{\nu-\mu+1}{2}\big{)}}{\rm
e}^{\mp\frac{1}{2}{\rm
i}\pi\left(\nu+\frac{1}{2}\right)}(z^{2}-1)^{{\frac{1}{2}\nu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{-\nu-\mu}{2},\frac{\mu-\nu}{2}}{\frac{1}{2}};\frac{z^{2}}{z^{2}-1}\right)\\!,$

holds with the upper and lower sign chosen according to $\pm\mathop{\rm
Im}z>0$. Then (2.8) yields the desired result. ∎

######  Remark 4.7.

Theorem 4.3 (or 4.6) writes $\mathsf{Q}_{\nu}^{\mu}(x)$ as a sum of an even
and an odd solution of the associated Legendre equation.

######  Remark 4.8.

The hypergeometric representations of $\mathsf{Q}_{\nu}^{\mu}(x)$ given in the
previous six theorems can be written in slightly different forms by applying
Euler’s transformation (2.3). For instance, the result of Theorem 4.3 can be
written as

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=\sqrt{\pi}\,2^{\mu-1}\big{(}1-x^{2}\big{)}^{{\frac{1}{2}\mu}}\Biggl{[}\frac{-\sin\big{(}\tfrac{\pi}{2}(\nu+\mu)\big{)}\Gamma\big{(}\frac{\nu+\mu+1}{2}\big{)}}{\Gamma\big{(}\frac{\nu-\mu+2}{2}\big{)}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{\nu+\mu+1}{2},\frac{\mu-\nu}{2}}{\frac{1}{2}};x^{2}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=\sqrt{\pi}\,2^{\mu-1}\big{(}1-x^{2}\big{)}^{{\frac{1}{2}\mu}}\Biggl{[}}{}+\frac{2\cos\big{(}\tfrac{\pi}{2}(\nu+\mu)\big{)}\Gamma\big{(}\frac{\nu+\mu+2}{2}\big{)}}{\Gamma\big{(}\frac{\nu-\mu+1}{2}\big{)}}x{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{\nu+\mu+2}{2},\frac{\mu-\nu+1}{2}}{\frac{3}{2}};x^{2}\\!\right)\Biggr{]}.$

######  Remark 4.9.

We observe that by applying Pfaff’s transformation (2.4) and (2.5), the
results of Theorems 4.5, 4.4 and 4.6 follow immediately from those of Theorems
4.1, 4.2 and 4.3, respectively. The hypergeometric representation of
$\mathsf{Q}_{\nu}^{\mu}(x)$ stated in Theorem 4.1 (or 4.5) is of the form
(3.1) with the Ferrers function ${\sf P}_{\nu}^{\mu}(x)$ replaced by the
hypergeometric representation

$\displaystyle{\sf
P}_{\nu}^{\mu}(x)=\frac{2^{\mu}}{\Gamma(1-\mu)}\big{(}1-x^{2}\big{)}^{-\frac{1}{2}\mu}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\frac{1}{2}\mu-\frac{1}{2}\nu,\frac{1}{2}-\frac{1}{2}\mu+\frac{1}{2}\nu}{1-\mu};1-x^{2}\right)\\!.$

######  Remark 4.10.

Theorem 4.2 expresses $\mathsf{Q}_{\nu}^{\mu}(x)$ in terms of the associated
Legendre functions $Q_{\nu}^{\mu}(x)$ and $Q_{-\nu-1}^{\mu}(x)$. The argument
of the hypergeometric function is $w_{8}=1/\big{(}1-x^{2}\big{)}$. If one
considers $x\in(-1,1)$, then this function maps to $(1,\infty)$, where the
Gauss hypergeometric function takes two values depending on whether the value
is approached from above or below the ray $[1,\infty)$. Hence, to compute
these values, one must use Theorems B.1–B.4. Then the arguments of the
hypergeometric functions are transformed to $1-x^{2}$,
$x^{2}/\big{(}x^{2}-1\big{)}$, $x^{2}$, $\big{(}x^{2}-1\big{)}/x^{2}$, and we
obtain the results that are already contained in Theorems 4.1, 4.6, 4.3 and
4.5, respectively. Similar remarks apply to Theorem 4.4.

##  5 Group III hypergeometric representations for
$\bm{\mathsf{Q}_{\nu}^{\mu}(x)}$

In this section we focus on hypergeometric functions listed in Section 1, with
arguments $w_{j}$, $j=13,\dots,18$. Since we want to represent the Ferrers
functions (defined on $D_{1}$), we replace $\sqrt{x^{2}-1}$ by ${\rm
i}\sqrt{1-x^{2}}$. The latter is an analytic function on $D_{1}$. We note that
${\rm i}\sqrt{1-x^{2}}=\pm\sqrt{x^{2}-1}$ if $\pm\mathop{\rm Im}x>0$.

The following result in terms of trigonometric functions appears in [9, p.
168] in a different form.

######  Theorem 5.1.

Let $x\in D_{1}$, $\nu,\mu\in\mathbb{C}$, $\nu+\mu\notin-\mathbb{N}$,
$\nu+\frac{1}{2}\notin\mathbb{Z}$. Then for the upper sign chosen everywhere
as well as with the lower sign chosen everywhere,

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=\frac{\sqrt{\pi}}{2^{\frac{3}{2}}\big{(}1-x^{2}\big{)}^{\frac{1}{4}}}$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=}{}\times\Biggl{[}{\rm
e}^{\pm\frac{{\rm
i}\pi}{2}\left(\mu+\frac{1}{2}\right)}\frac{\Gamma\big{(}\nu+\frac{1}{2}\big{)}}{\Gamma(\nu-\mu+1)}\big{(}x\pm{\rm
i}\sqrt{1\\!-x^{2}}\big{)}^{\nu+\frac{1}{2}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{1}{2}\\!+\mu,\frac{1}{2}\\!-\mu}{\frac{1}{2}-\nu};\frac{\mp
x+{\rm i}\sqrt{1\\!-x^{2}}}{2{\rm i}\sqrt{1-x^{2}}}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=\times\Biggl{[}}{}+{\rm
e}^{\mp\frac{{\rm
i}\pi}{2}\left(\mu+\frac{1}{2}\right)}\frac{\Gamma(\nu+\mu+1)}{\Gamma\big{(}\nu+\frac{3}{2}\big{)}}\left(1+{\rm
e}^{\pm{\rm
i}\pi(\nu+\mu)}\frac{\cos(\pi\mu)}{\cos(\pi\nu)}\right)\big{(}x\mp{\rm
i}\sqrt{1-x^{2}}\big{)}^{\nu+\frac{1}{2}}$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=\times\Biggl{[}+}{}\times{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{1}{2}+\mu,\frac{1}{2}-\mu}{\nu+\frac{3}{2}};\frac{\mp
x+{\rm i}\sqrt{1-x^{2}}}{2{\rm i}\sqrt{1-x^{2}}}\right)\\!\Bigg{]},$

or equivalently with $\mathop{\rm Re}\theta\in(0,\pi)$,

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(\cos\theta)=\frac{\sqrt{\pi}}{2^{\frac{3}{2}}\sqrt{\sin\theta}}\Biggl{[}{\rm
e}^{\pm\frac{{\rm
i}\pi}{2}\left(\mu+\frac{1}{2}\right)}\frac{\Gamma\big{(}\nu+\frac{1}{2}\big{)}}{\Gamma(\nu-\mu+1)}{\rm
e}^{\pm{\rm
i}\left(\nu+\frac{1}{2}\right)\theta}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{1}{2}+\mu,\frac{1}{2}-\mu}{\frac{1}{2}-\nu};\frac{1}{2}\pm\frac{\rm
i}{2}\cot\theta\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(\cos\theta)=}{}+{\rm
e}^{\mp\frac{{\rm
i}\pi}{2}\left(\mu+\frac{1}{2}\right)}\frac{\Gamma(\nu+\mu+1)}{\Gamma\big{(}\nu+\frac{3}{2}\big{)}}\left(1+{\rm
e}^{\mp{\rm i}\pi(\nu+\mu)}\frac{\cos(\pi\mu)}{\cos(\pi\nu)}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(\cos\theta)=+}{}\times{\rm
e}^{\mp{\rm
i}\left(\nu+\frac{1}{2}\right)\theta}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{1}{2}+\mu,\frac{1}{2}-\mu}{\nu+\frac{3}{2}};\frac{1}{2}\pm\frac{\rm
i}{2}\cot\theta\right)\Biggr{]}.$

###### Proof.

We first prove the result with the upper sign. We claim that

$\displaystyle w_{13}=\frac{-x+{\rm i}\sqrt{1-x^{2}}}{2{\rm i}\sqrt{1-x^{2}}}$

is a conformal map from $D_{1}$ to the complex plane cut along the rays
$(-\infty,0]$ and $[1,\infty)$. To see this let $x=\cos\theta$. This is a
conformal map from the strip $S=\\{\theta\colon\mathop{\rm Re}\theta<\pi\\}$
onto $D_{1}$. Then

$\displaystyle w_{13}=\frac{1}{2}+\frac{\rm i}{2}\cot\theta.$

Now $\cot\theta$ is a conformal map from $S$ to $\mathbb{C}\setminus((-{\rm
i}\infty,-{\rm i}]\cup[{\rm i},{\rm i}\infty))$ which establishes the claim.
Hence, using the principal value of the hypergeometric function, the
composition of the hypergeometric function with $w_{13}$ is an analytic
function on $D_{1}$. Further, we note that $x\pm{\rm
i}\sqrt{1-x^{2}}\notin(-\infty,0]$ for $x\in D_{1}$. Therefore, the right-hand
side of the stated identity is an analytic function on $D_{1}$, so it is
sufficient to prove this identity for $\mathop{\rm Im}x>0$. For $\mathop{\rm
Im}z>0$, we insert the hypergeometric representation of the associated
Legendre function [9, Entry 31, p. 162]

$\displaystyle{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(z)=\frac{\sqrt{\pi}\,\Gamma(\nu\\!+\mu\\!+1)\big{(}z\\!-\sqrt{z^{2}\\!-1}\big{)}^{\nu+\frac{1}{2}}}{\sqrt{2}\Gamma\big{(}\nu+\frac{3}{2}\big{)}(z^{2}\\!-1)^{\frac{1}{4}}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{1}{2}+\mu,\frac{1}{2}\\!-\mu}{\nu+\frac{3}{2}};\frac{-z+\sqrt{z^{2}-1}}{2\sqrt{z^{2}-1}}\right)\\!,$
(5.1)

in (2.11), We obtain the desired result by using (2.1) and the identity

$\displaystyle 2{\rm e}^{{\rm i}\pi\mu}\left(\sin(\pi\nu)-{\rm
i}\frac{\sin(\pi(\mu-\nu))}{2\cos(\pi\nu)}\right)=1+{\rm e}^{{\rm
i}\pi(\nu+\mu)}\frac{\cos(\pi\mu)}{\cos(\pi\nu)}.$

The proof of the result with the lower sign is similar. The function

$\displaystyle w_{17}=\frac{x+{\rm i}\sqrt{1-x^{2}}}{2{\rm
i}\sqrt{1-x^{2}}}{,}$

is also a conformal map from $D_{1}$ to the complex plane cut along the rays
$(-\infty,0]$ and $[1,\infty)$. Hence the right-hand side of the claimed
identity is again an analytic function on $D_{1}$, so it is sufficient to
prove it for $\mathop{\rm Im}x>0$ or $\mathop{\rm Im}x<0$. The desired result
follows from (2.11) and the hypergeometric representation of the associated
Legendre function cf. [9, Entry 35, p. 162] in the half-plane $\mathop{\rm
Im}x>0$. However, it is easier to use (2.14) and (5.1) in the half-plane
$\mathop{\rm Im}x<0$. In this half-plane $\sqrt{x^{2}-1}=-{\rm
i}\sqrt{1-x^{2}}$ and the desired representation follows. In order to obtain
the trigonometric form of the representation set $x=\cos\theta$ with
$\mathop{\rm Re}\theta\in(0,\pi)$ and note that ${\rm
i}\sqrt{1-x^{2}}=\sin\theta$. This completes the proof. ∎

The following result in terms of trigonometric functions appears in [9, p.
168] in a different form.

######  Theorem 5.2.

Let $x\in D_{1}$, $\nu,\mu\in\mathbb{C}$, $\nu+\mu\notin-\mathbb{N}$,
$\nu+\frac{1}{2}\notin\mathbb{Z}$. Then for the upper sign chosen everywhere
as well as with the lower sign chosen everywhere,

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=\sqrt{\pi}\,2^{\mu-1}\big{(}1-x^{2}\big{)}^{{\frac{1}{2}\mu}}$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=}{}\times\Biggl{[}{\rm
e}^{\pm{\rm
i}\pi\left(\mu+\frac{1}{2}\right)}\frac{\Gamma\big{(}\nu+\frac{1}{2}\big{)}}{\Gamma(\nu-\mu+1)}\big{(}x\pm{\rm
i}\sqrt{1-x^{2}}\big{)}^{\nu-\mu}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{1}{2}+\mu,\mu-\nu}{\frac{1}{2}-\nu};\frac{x\mp{\rm
i}\sqrt{1-x^{2}}}{x\pm{\rm i}\sqrt{1-x^{2}}}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=\times\Biggl{[}}{}+\frac{\Gamma(\nu+\mu+1)}{\Gamma\big{(}\nu+\frac{3}{2}\big{)}}\left(1+{\rm
e}^{\pm{\rm
i}\pi(\nu+\mu)}\frac{\cos(\pi\mu)}{\cos(\pi\nu)}\right)\big{(}x\mp{\rm
i}\sqrt{1-x^{2}}\big{)}^{\nu+\mu+1}$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=\times\Biggl{[}+}{}\times{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{1}{2}+\mu,\nu+\mu+1}{\nu+\frac{3}{2}};\frac{x\mp{\rm
i}\sqrt{1-x^{2}}}{x\pm{\rm i}\sqrt{1-x^{2}}}\right)\\!\Biggr{]},$

or equivalently with $\mathop{\rm Re}\theta\in(0,\pi)$,

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(\cos\theta)=\sqrt{\pi}\,2^{\mu-1}(\sin\theta)^{\mu}\Biggl{[}{\rm
e}^{\pm{\rm
i}\pi\left(\mu+\frac{1}{2}\right)}\frac{\Gamma\big{(}\nu+\frac{1}{2}\big{)}}{\Gamma(\nu-\mu+1)}{\rm
e}^{\pm{\rm
i}(\nu-\mu)\theta}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{1}{2}+\mu,\mu-\nu}{\frac{1}{2}-\nu};{\rm
e}^{\mp 2{\rm i}\theta}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(\cos\theta)=}{}+\\!\frac{\Gamma(\nu\\!+\\!\mu\\!+\\!1)}{\Gamma\big{(}\nu+\frac{3}{2}\big{)}}\left(\\!1\\!+\\!{\rm
e}^{\pm{\rm i}\pi(\nu+\mu)}\frac{\cos(\pi\mu)}{\cos(\pi\nu)}\right){\rm
e}^{\mp{\rm
i}(\nu+\mu+1)\theta}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{1}{2}\\!+\\!\mu,\nu\\!+\\!\mu\\!+\\!1}{\nu+\frac{3}{2}};{\rm
e}^{\mp 2{\rm i}\theta}\right)\\!\Biggr{]}.$

###### Proof.

We first prove the result with the upper sign. We note that the function

$\displaystyle w_{14}=\frac{x-{\rm i}\sqrt{1-x^{2}}}{x+{\rm
i}\sqrt{1-x^{2}}}=\frac{w_{13}}{w_{13}-1}$

is a conformal map from $D_{1}$ to the complex plane cut along the ray
$[0,\infty)$. Hence the right-hand side of the stated identity is an analytic
function on $D_{1}$, so it is sufficient to prove it for $\mathop{\rm Im}x>0$.
Now we insert the hypergeometric representation [9, Entry 32, p. 162]

$\displaystyle{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(z)=\sqrt{\pi}\,2^{\mu}\frac{\Gamma(\nu+\mu+1)}{\Gamma\big{(}\mu+\frac{3}{2}\big{)}}(z^{2}-1)^{\frac{1}{2}\mu}\big{(}z+\sqrt{z^{2}-1}\big{)}^{\nu+\mu+1}$
$\displaystyle\hphantom{{\rm e}^{-{\rm
i}\pi\mu}Q_{\nu}^{\mu}(z)=}{}\times{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{1}{2}+\mu,\nu+\mu+1}{\nu+\frac{3}{2}};\frac{z-\sqrt{z^{2}-1}}{z+\sqrt{z^{2}-1}}\right)$
(5.2)

in (2.11), and obtain the desired result. To prove the result with the lower
sign we can either use the hypergeometric representation [9, Entry 36, p. 163]
of $Q_{\nu}^{\mu}(z)$ in the half-plane $\mathop{\rm Im}z>0$ or (5.2) in the
half-plane $\mathop{\rm Im}z<0$. This completes the proof. ∎

######  Theorem 5.3.

Let $x\in D_{1}^{+}$, $\nu,\mu\in\mathbb{C}$, $2\mu\not\in\mathbb{Z}$,
$\nu+\mu\notin-\mathbb{N}$. Then for the upper sign chosen everywhere as well
as with the lower sign chosen everywhere,

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=\frac{\Gamma(-\mu)}{2^{\mu+1}}\frac{\Gamma(\nu\\!+\\!\mu\\!+\\!1)}{\Gamma(\nu\\!-\\!\mu\\!+\\!1)}\big{(}1-x^{2}\big{)}^{\frac{1}{2}\mu}\big{(}x\pm{\rm
i}\sqrt{1\\!-\\!x^{2}}\big{)}^{\nu-\mu}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{1}{2}\\!+\\!\mu,\mu\\!-\\!\nu}{1+2\mu};\frac{2{\rm
i}\sqrt{1-x^{2}}}{\pm\,x\\!+\\!{\rm i}\sqrt{1\\!-\\!x^{2}}}\right)\\!$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=}{}+2^{\mu-1}\Gamma(\mu)\frac{\cos(\pi\mu)\big{(}x\pm{\rm
i}\sqrt{1-x^{2}}\big{)}^{\nu+\mu}}{\big{(}1-x^{2}\big{)}^{{\frac{1}{2}\mu}}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{1}{2}-\mu,-\nu-\mu}{1-2\mu};\frac{2{\rm
i}\sqrt{1-x^{2}}}{\pm\,x+{\rm i}\sqrt{1-x^{2}}}\right)\\!,$

or equivalently with $\mathop{\rm Re}\theta\in(0,\pi)$,

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(\cos\theta)=\frac{\Gamma(-\mu)}{2^{\mu+1}}\frac{\Gamma(\nu+\mu+1)}{\Gamma(\nu-\mu+1)}(\sin\theta)^{\mu}{\rm
e}^{\pm{\rm
i}(\nu-\mu)\theta}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{1}{2}+\mu,\mu-\nu}{1+2\mu};1-{\rm
e}^{\mp 2{\rm i}\theta}\right)$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(\cos\theta)=}{}+2^{\mu-1}\Gamma(\mu)\frac{\cos(\pi\mu){\rm
e}^{\pm{\rm
i}(\nu+\mu)\theta}}{(\sin\theta)^{\mu}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{1}{2}-\mu,-\nu-\mu}{1-2\mu};1-{\rm
e}^{\mp 2{\rm i}\theta}\right)\\!.$

###### Proof.

We first prove the result with the upper sign. We observe that the function

$\displaystyle w_{15}=\frac{2{\rm i}\sqrt{1-x^{2}}}{x+{\rm
i}\sqrt{1-x^{2}}}=\frac{1}{w_{17}}{,}$

maps the imaginary axis to the branch cut $[1,\infty)$ of the hypergeometric
function. Then $w_{15}$ is a conformal map from $D_{1}^{+}$ to the upper half-
plane. Therefore, the right-hand side of the stated identity is analytic on
$D_{1}^{+}$ so it is sufficient to prove it for $x\in(0,1)$. We obtain the
desired result by using (2.8) and the hypergeometric representation of
$Q_{\nu}^{\mu}(x)$ given in [9, Entry 33, p. 163]. However, it is much simpler
to use (2.10) and the hypergeometric representation [9, Entry 15, p. 158]

$\displaystyle
P_{\nu}^{\mu}(z)={\frac{2^{\mu}\big{(}z+\sqrt{z^{2}-1}\big{)}^{\nu+\mu}}{\Gamma(1-\mu)(z^{2}-1)^{\frac{1}{2}\mu}}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{-\nu-\mu,\frac{1}{2}-\mu}{1-2\mu};\frac{2\sqrt{z^{2}-1}}{z+\sqrt{z^{2}-1}}\right)\\!.$

In order to derive the result with the lower sign, use [9, Entry 34, p. 163]
or (2.13) and (5.2) in $\mathop{\rm Re}x>0$, $\mathop{\rm Im}x<0$,
respectively. This completes the proof. ∎

######  Remark 5.4.

The hypergeometric representations of $\mathsf{Q}_{\nu}^{\mu}(x)$ given in the
previous three theorems can be written in slightly different forms by using
Euler’s transformation (2.3). We observe that by applying Pfaff’s
transformations (2.4) and (2.5), Theorem 5.2 follows from Theorem 5.1.
Moreover, the result of Theorem 5.3 with the lower sign follows from the same
theorem with the upper sign.

######  Remark 5.5.

Theorem 5.1 with the upper sign represents $\mathsf{Q}_{\nu}^{\mu}(x)$ in the
upper half-plane as a linear combination of $Q_{\nu}^{\mu}(x)$ and
$Q_{-\nu-1}^{\mu}(x)$ (cf. (2.11)). In particular, if $x\in(-1,1)$,
$\mathsf{Q}_{\nu}^{\mu}(x)$ is given by a linear combination of
$Q_{\nu}^{\mu}(x+{\rm i}0)$ and $Q_{-\nu-1}^{\mu}(x+{\rm i}0)$. The argument
$w_{13}$, $x\in(-1,1)$, of the hypergeometric functions lies on the line
$\mathop{\rm Re}w=\frac{1}{2}$. Theorem 5.1 with the lower sign represents
$\mathsf{Q}_{\nu}^{\mu}(x)$ in the lower half-plane as a linear combination of
$Q_{\nu}^{\mu}(x)$ and $Q_{-\nu-1}^{\mu}(x)$. In particular, if $x\in(-1,1)$,
$\mathsf{Q}_{\nu}^{\mu}(x)$ is given by a linear combination of
$Q_{\nu}^{\mu}(x-{\rm i}0)$ and $Q_{-\nu-1}^{\mu}(x-{\rm i}0)$. Again the
argument $w_{17}$, $x\in(-1,1)$, of the hypergeometric functions lies on the
line $\mathop{\rm Re}w=\frac{1}{2}$. Similar results hold regarding Theorem
5.2. In this case the argument $w_{14}$, $x\in(-1,1)$, of the hypergeometric
functions lies on the unit circle $|w|=1$ excluding $w=1$. Theorem 5.3
represents $\mathsf{Q}_{\nu}^{\mu}(x)$ as a linear combination (3.1) of ${\sf
P}_{\nu}^{-\mu}(x)$ and ${\sf P}_{\nu}^{\mu}(x)$ for $x\in D_{1}^{+}$, using
an appropriate hypergeometric representation of the Ferrers functions of the
first kind. The argument $w_{15}$, $x\in(0,1)$, lies on the circle $|w-1|=1$
excluding $w=2$. The range of values of $w_{j}$, $j=13,\dots,18$,
$x\in(-1,1)$, is depicted in Figure 1.

Figure 1: This figure represents a plot in the complex plane of the complex
arguments which appear in Theorems 5.1, 5.2 and 5.3 when $x\in(-1,1)$. The
symbol $O$ represents the origin of the complex $w$-plane, with the dashed
horizontal and vertical lines emanating from it being the real and imaginary
axes respectively. The symbol $A$ represents the line parallel to the
imaginary axis $w=\frac{1}{2}\pm\frac{\rm i}{2}\cot\theta$ in Theorem 5.1,
where $\theta\in(0,\pi)$. The symbol $B$ represents the unit circle centered
at the origin excluding the point located at $w=1$ in Theorem 5.2. The symbol
$C$ represents the unit circle centered at the point $w=1$ excluding the point
located at $w=0$ in Theorem 5.3. The point $w=2$ ($D$) corresponding to $x=0$
is excluded in Theorem 5.3. The symbol $E$ represents the branch cut of the
hypergeometric function located at $w\in[1,\infty)$. The symbols $F$ and $G$
represent the two points where all three curves intersect each other at
$w=\frac{1}{2}\pm\frac{\sqrt{3}}{2}{\rm i}$ respectively.

##  6 A Fourier series representation

A Fourier series for a function $f(x)$ is given by the infinite sum

$\displaystyle f(x)=\sum_{n=-\infty}^{\infty}a_{n}{\rm e}^{{\rm i}n\theta},$

where the coefficients $a_{n}$ must satisfy certain specific conditions to
guarantee convergence, see below. In this section we perform analysis for a
Fourier series representation of the Ferrers function of the second kind.

######  Theorem 6.1.

Let $x\in D_{1}$ and $\nu,\mu\in\mathbb{C}$ such that
$\nu+\mu\not\in-\mathbb{N}$. Then

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(x)=\sqrt{\pi}\,2^{\mu-1}\big{(}1-x^{2}\big{)}^{\frac{1}{2}\mu}\frac{\Gamma(\nu+\mu+1)}{\Gamma\big{(}\nu+\frac{3}{2}\big{)}}$
$\displaystyle\hphantom{\mathsf{Q}_{\nu}^{\mu}(x)=}{}\times\Bigg{[}u^{\nu+\mu+1}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\mu+\frac{1}{2},\nu+\mu+1}{\nu+\frac{3}{2}};\frac{u}{v}\right)+v^{\nu+\mu+1}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\mu+\frac{1}{2},\nu+\mu+1}{\nu+\frac{3}{2}};\frac{v}{u}\right)\\!\Bigg{]},$

where

$\displaystyle u:=x+{\rm i}\sqrt{1-x^{2}},\qquad v:=x-{\rm
i}\sqrt{1-x^{2}}=\frac{1}{u}.$

###### Proof.

Both sides of the stated identity are analytic functions on $D_{1}$, so it is
sufficient to prove it for $x\in(-1,1)$. If $x\in(-1,1)$ we use (2.8) with
$Q_{\nu}^{\mu}(x+{\rm i}0)$ and $Q_{\nu}^{\mu}(x-{\rm i}0)$ both expressed
through (5.2). Then the desired identity follows noting that $\sqrt{(x\pm{\rm
i}0)^{2}-1}=\pm{\rm i}\sqrt{1-x^{2}}$. ∎

If we set $x=\cos\theta$, $\mathop{\rm Re}\theta\in(0,\pi)$, then Theorem 6.1
gives

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(\cos\theta)=\sqrt{\pi}\,2^{\mu-1}(\sin\theta)^{\mu}\frac{\Gamma(\nu+\mu+1)}{\Gamma\big{(}\nu+\frac{3}{2}\big{)}}$
$\displaystyle\qquad{}\times\Bigg{[}{\rm e}^{{\rm
i}(\nu\\!+\mu\\!+1)\theta}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\mu\\!+\frac{1}{2},\nu\\!+\mu\\!+1}{\nu+\frac{3}{2}};{\rm
e}^{2{\rm i}\theta}\right)\\!+{\rm e}^{-{\rm
i}(\nu+\mu+1)\theta}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\mu+\frac{1}{2},\nu+\mu+1}{\nu+\frac{3}{2}};{\rm
e}^{-2{\rm i}\theta}\right)\\!\Bigg{]}.$

If $\theta\in(0,\pi)$ then the arguments $w={\rm e}^{\pm 2{\rm i}\theta}$ of
the hypergeometric function lie on the unit circle $|w|=1$. Provided the
hypergeometric series converges at ${\rm e}^{\pm 2{\rm i}\theta}$ we obtain
(using Abel’s theorem on power series) [11, equation (14.13.2)]

$\displaystyle\mathsf{Q}_{\nu}^{\mu}(\cos\theta)=\sqrt{\pi}\,2^{\mu}(\sin\theta)^{\mu}\sum_{k=0}^{\infty}\frac{\Gamma(\nu+\mu+k+1)}{\Gamma\big{(}\nu+k+\frac{3}{2}\big{)}}\frac{\big{(}\mu+\frac{1}{2}\big{)}_{k}}{k!}\cos((\nu+\mu+2k+1)\theta).$
(6.1)

Regarding the convergence of the series in (6.1) we have the following result.
Statement $(b)$ of Theorem 6.2 is more precise than the corresponding
statement in [11, Section 14.13].

######  Theorem 6.2.

Let $\theta\in(0,\pi)$, $\nu,\mu\in\mathbb{C}$ such that
$\nu+\mu\in\mathbb{C}\setminus-\mathbb{N}$.

  1. $(a)$

If $\mathop{\rm Re}\mu<0$ then the series in (6.1) converges absolutely.

  2. $(b)$

If $0\leq\mathop{\rm Re}\mu<\frac{1}{2}$ then the series in (6.1) converges,
but, if $\theta\neq\frac{1}{2}\pi$, it does not converge absolutely.

  3. $(c)$

If $\mathop{\rm Re}\mu\geq\frac{1}{2}$ and $\theta\neq\frac{1}{2}\pi$, then
the series in (6.1) diverges.

###### Proof.

$(a)$ It is known [11, Section 15.2(i)] that the Gauss hypergeometric series
${}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,b}{c};w\right)$ converges absolutely
on the unit circle $|w|=1$ if $\mathop{\rm Re}(c-a-b)>0$. In our case
$a=\mu+\frac{1}{2}$, $b=\nu+\mu+1$, $c=\nu+\frac{3}{2}$ so $c-a-b=-2\mu$. If
$\mathop{\rm Re}\mu<0$ it follows that the series in (6.1) is the sum of two
absolutely convergent series and so is itself absolutely convergent.

$(b)$ Suppose that $0\leq\mathop{\rm Re}\mu<\frac{1}{2}$. If $-1<\mathop{\rm
Re}(c-a-b)\leq 0$ then the Gauss hypergeometric series converges conditionally
at $|w|=1$, $w\neq 1$ [11, Section 15.2(i)]. It follows that the series in
(6.1) is the sum of two convergent series and so is itself convergent.
However, it is not true that the sum of two conditionally convergent series is
conditionally convergent. We still have to show that the series in (6.1) does
not converge absolutely if $\theta\neq\frac{1}{2}\pi$. According to [11,
equation (5.11.12)],

$\displaystyle\frac{\Gamma(a+z)}{\Gamma(b+z)}\sim z^{a-b}$

as $z\to+\infty$. Therefore, as $k\to\infty$,

$\displaystyle\frac{\Gamma(\nu\\!+\mu\\!+k\\!+1)}{\Gamma\big{(}\nu+k+\frac{3}{2}\big{)}}\frac{\big{(}\mu\\!+\\!\frac{1}{2}\big{)}_{k}}{k!}=\frac{\Gamma(\nu\\!+\mu\\!+k\\!+\\!1)}{\Gamma\big{(}\nu+k+\frac{3}{2}\big{)}}\frac{\Gamma\big{(}\mu\\!+\frac{1}{2}\\!+k\big{)}}{\Gamma(k+1)\Gamma\big{(}\mu+\frac{1}{2}\big{)}}\sim\frac{k^{\,\mu-\frac{1}{2}}k^{\,\mu-\frac{1}{2}}}{\Gamma\big{(}\mu+\frac{1}{2}\big{)}}=\frac{k^{2\mu-1}}{\Gamma\big{(}\mu+\frac{1}{2}\big{)}}.$

Since $1/\Gamma\big{(}\mu+\frac{1}{2}\big{)}\neq 0$, there are positive
constants $\kappa$ and $K$ such that

$\displaystyle\left|\frac{\Gamma(\nu+\mu+k+1)}{\Gamma\big{(}\nu+k+\frac{3}{2}\big{)}}\frac{\big{(}\mu+\frac{1}{2}\big{)}_{k}}{k!}\right|\geq\frac{\kappa}{k},$

for $k\geq K$. The second part of statement $(b)$ now follows from Lemma 6.3.

$(c)$ Suppose that $\mathop{\rm Re}\mu\geq\frac{1}{2}$ and the series in (6.1)
converges. Then the terms of the series must converge to $0$. It follows that
$\cos((\nu+\mu+2k+1)\theta)\to 0$ as $k\to\infty$. By Lemma 6.4 this is
impossible unless $\theta=\frac{1}{2}\pi$. Therefore, the series in (6.1) will
diverge for $\theta\neq\frac{1}{2}\pi$. ∎

######  Lemma 6.3.

Let $a\in\mathbb{C}$ and $\theta\in(0,\pi)$. Then

$\displaystyle\sum_{k=1}^{\infty}\frac{1}{k}|\cos((a+2k)\theta)|=\infty$

unless $\cos a=0$ and $\theta=\frac{1}{2}\pi$.

###### Proof.

If $\theta=\frac{1}{2}\pi$ then $|\cos((a+2k)\theta)|$ is independent of $k$
and the assertion follows. Now suppose that $\theta\neq\frac{1}{2}\pi$. Since
$|\cos z|\geq|\cos(\mathop{\rm Re}z)|$ for all $z\in\mathbb{C}$ it is enough
to consider $a\in\mathbb{R}$. Then
$|\cos((a+2k)\theta)|\geq\cos^{2}((a+2k)\theta)$. Therefore, it is enough to
show that

$\displaystyle\sum_{k=1}^{\infty}\frac{1}{k}\cos^{2}((a+2k)\theta)=\infty.$

Now

$\displaystyle\frac{1}{k}\cos^{2}((a+2k)\theta)=\frac{1}{2k}+\frac{1}{2k}\cos(2(a+2k)\theta).$

The series $\sum_{k=1}^{\infty}1/(2k)$ diverges. The series
$\sum_{k=1}^{\infty}\cos(2(a+2k)\theta)/(2k)$ converges by the Dirichlet test
provided the partial sums $\sum_{k=1}^{n}\cos(2(a+2k)\theta)$ form a bounded
sequence. This is true for $\theta\in(0,\pi)$, $\theta\neq\frac{1}{2}\pi$. ∎

######  Lemma 6.4.

Let $a,b\in\mathbb{C}$ and $\sin b\neq 0$. Then the sequence $\cos(a+bn)$ does
not converge to $0$ as $n\to\infty$.

###### Proof.

Suppose that $\cos(a+bn)\to 0$ as $n\to\infty$. We have

$\displaystyle\cos(a+b(n+1))=\cos(a+bn)\cos b-\sin(a+bn)\sin b.$

If we let $n\to\infty$ and use $\sin b\neq 0$ we obtain that $\sin(a+bn)\to
0$. Since $\cos^{2}x+\sin^{2}x=1$ this is a contradiction. ∎

##  7 Convergence regions of the Gauss hypergeometric series

In the hypergeometric representations of $\mathsf{Q}_{\nu}^{\mu}(x)$ derived
in Sections 3–5 we worked with the principal value of the hypergeometric
function defined on $\mathbb{C}\setminus[1,\infty)$. If we wish to use the
hypergeometric series (2.2) we have to add the condition $|w_{j}|<1$.
Therefore, it is of interest to determine the regions in the $x$-plane on
which $|w_{j}|<1$. In most cases these regions are obvious but not in all of
them.

###  7.1 Group I

  * •

The hypergeometric functions in Theorem 3.1 have the argument $w_{1}=(1-x)/2$.
Now $|w_{1}|<1$ gives the condition $|1-x|<2$ describing a disk centered at
$x=1$ with radius $2$.

  * •

The hypergeometric functions in Theorem 3.2 have the argument $w_{2}=(1+x)/2$.
Then $|w_{2}|<1$ is satisfied if and only if $|1+x|<2$.

  * •

The hypergeometric functions in Theorem 3.3 have the argument
$w_{3}=(x-1)/(x+1)$. Then $|w_{3}|<1$ is satisfied if and only if $\mathop{\rm
Re}x>0$.

  * •

The hypergeometric functions in Theorem 3.4 have the argument
$w_{4}=(x+1)/(x-1)$. Then $|w_{4}|<1$ is satisfied if and only if $\mathop{\rm
Re}x<0$.

  * •

The hypergeometric functions in Theorem 3.5 have the argument $w_{5}=2/(1+x)$.
Then $|w_{5}|<1$ is satisfied if and only if $|1+x|>2$.

  * •

The hypergeometric function in Theorem 3.6 have the argument $w_{6}=2/(1-x)$.
Then $|w_{6}|<1$ is satisfied if and only if $|1-x|>2$.

###  7.2 Group II

  * •

The hypergeometric functions in Theorem 4.1 have the argument $w_{7}=1-x^{2}$.
Then $|w_{7}|<1$ is satisfied if and only if $|1-x||1+x|<1$. This is the
shaded region in Figure 2 bounded by a lemniscate.

Figure 2: The region $|w_{7}|<1$.

  * •

The hypergeometric functions in Theorem 4.2 have the argument
$w_{8}=1/\big{(}1-x^{2}\big{)}$. Then $|w_{8}|<1$ is satisfied if and only if
$|1+x||1-x|>1$, that is, if $x$ lies in the unshaded region in Figure 2.

  * •

The hypergeometric functions in Theorem 4.3 have the argument $w_{9}=x^{2}$.
Then $|w_{9}|<1$ is satisfied if and only if $|x|<1$.

  * •

The hypergeometric functions in Theorem 4.4 have the argument
$w_{10}=1/x^{2}$. Then $|w_{10}|<1$ is satisfied if and only if $|x|>1$.

  * •

The hypergeometric functions in Theorem 4.5 have the argument
$w_{11}=\big{(}x^{2}-1\big{)}/x^{2}$. Then $|w_{11}|<1$ is equivalent to
$\mathop{\rm Re}x^{2}>\frac{1}{2}$ or $(\mathop{\rm Re}x)^{2}-(\mathop{\rm
Im}x)^{2}>\frac{1}{2}$. This determines the shaded region bounded by a
hyperbola depicted in Figure 3.

Figure 3: The region $|w_{11}|<1$.

  * •

The hypergeometric functions in Theorem 4.6 have the argument
$w_{12}=x^{2}/\big{(}x^{2}-1\big{)}$. Then $|w_{12}|<1$ is satisfied if and
only if $(\mathop{\rm Re}x)^{2}-(\mathop{\rm Im}x)^{2}<\frac{1}{2}$. This
determines the unshaded region in the $x$-plane in Figure 3.

###  7.3 Group III

For abbreviation, let

$\displaystyle y={\rm i}\sqrt{1-x^{2}}$

for $x\in D_{1}$.

  * •

The hypergeometric functions in Theorem 5.1 (upper sign) have the argument

$\displaystyle
w_{13}=\frac{-x+y}{2y}=\frac{w_{14}}{w_{14}-1},\qquad\text{where}\quad
w_{14}=\frac{x-y}{x+y}.$

Therefore, $|w_{13}|<1$ is equivalent to $\mathop{\rm Re}w_{14}<\frac{1}{2}$.
Let $x\in D_{1}$. Then $x=\cos\theta$, $\theta=\alpha+{\rm i}\beta$, where
$\alpha\in(0,\pi)$ and $\beta\in\mathbb{R}$, so

$\displaystyle w_{14}=\frac{\cos\theta-{\rm i}\sin\theta}{\cos\theta+{\rm
i}\sin\theta}={\rm e}^{-2{\rm i}\theta}={\rm e}^{-2{\rm i}\alpha}{\rm
e}^{2\beta}.$

Hence, $|w_{13}|{<}1$ is equivalent to ${{\rm
e}^{2\beta}\cos(2\alpha){<}\frac{1}{2}}$. Then
$\frac{1}{4}\pi\leq\alpha\leq\frac{3}{4}\pi$ or
$\beta{<}-\frac{1}{2}\ln(2\cos(2\alpha))$. In the $x$-plane the curve
$\beta=-\frac{1}{2}\ln(2\cos(2\alpha))$, for $0<\alpha<\frac{1}{4}\pi$ is
given by

$\displaystyle x=\cos(\alpha+{\rm
i}\beta)=\frac{1}{2}\big{(}t+t^{-1}\big{)}\cos\alpha+\frac{1}{2}{\rm
i}\big{(}t-t^{-1}\big{)}\sin\alpha,$ (7.1)

where $t=\sqrt{2\cos(2\alpha)}$. This curve and the corresponding curve for
$\frac{3}{4}\pi<\alpha<\pi$ are shown in Figure 4. Therefore, for $x\in
D_{1}$, $|w_{13}|<1$ is satisfied if and only if $x$ lies in the unshaded
region in Figure 4.

Figure 4: The region $|w_{16}|<1$ ($|w_{13}|>1$).

  * •

The hypergeometric functions in Theorem 5.1 (lower sign) have the argument

$\displaystyle w_{17}=\frac{x+y}{2y}.$

The condition $|w_{17}|\\!<\\!1$ is satisfied if and only if $x$ lies in the
unshaded region of Figure 5.

Figure 5: The region $|w_{15}|<1$ ($|w_{17}|>1$).

  * •

The hypergeometric functions in Theorem 5.2 (upper sign) have the argument

$\displaystyle w_{14}=\frac{x-y}{x+y}.$

We take $\sqrt{x^{2}-1}={\rm i}\sqrt{1-x^{2}}$ for $x\in D_{1}$. Then using
$x=\cos\theta$, $\mathop{\rm Re}\theta\in(0,\pi)$ gives $w_{14}={\rm
e}^{-2{\rm i}\theta}$ so $|w_{14}|={\rm e}^{2\mathop{\rm Im}\theta}$. Then
$|w_{14}|<1$ means $\mathop{\rm Im}\theta<0$ which therefore implies that
$\mathop{\rm Im}x>0$. Then $|w_{14}|={\rm e}^{2\beta}$ so $|w_{14}|<1$ is
satisfied if and only if $\mathop{\rm Im}x>0$.

  * •

The hypergeometric functions in Theorem 5.2 (lower sign) have the argument

$\displaystyle w_{18}=\frac{x+y}{x-y}.$

Since $w_{18}=1/w_{14}$, then $|w_{18}|<1$ if and only if $\mathop{\rm
Im}x<0$. Of course, these results depend on the choice of the sign of the root
$\sqrt{x^{2}-1}$.

  * •

The hypergeometric functions in Theorem 5.3 (upper sign) have the argument

$\displaystyle w_{15}=\frac{2y}{x+y}=\frac{1}{w_{17}}.$

Then $|w_{15}|<1$ is equivalent to $|w_{17}|>1$ and this means that $x$ lies
in the shaded region of Figure 5.

  * •

The hypergeometric functions in Theorem 5.3 (lower sign) have the argument

$\displaystyle w_{16}=\frac{2y}{-x+y}=\frac{1}{w_{13}}.$

Then $|w_{16}|<1$ is equivalent to $|w_{13}|>1$ and this means that $x$ lies
in the shaded region of Figure 5.

We close this paper with the following interesting observations. In the
arguments $w_{j}$, $j=13,\dots,18$, we replaced $\sqrt{x^{2}-1}$ by ${\rm
i}\sqrt{1-x^{2}}$ because Ferrers functions are naturally defined on the
domain $D_{1}$. However, if we are interested in hypergeometric
representations of the associated Legendre function $Q_{\nu}^{\mu}(x)$ we will
use $\sqrt{x^{2}-1}:=\sqrt{x-1}\sqrt{x+1}$ which changes the domains on which
$|w_{j}|<1$. Let us denote these modifications of $w_{j}$ by $w_{j}^{\ast}$.
Since $w_{13}(x)=w_{13}^{\ast}(x)$ for $\mathop{\rm Im}x>0$ the part of the
curve $|w^{\ast}_{13}|=1$ lying in the region $\mathop{\rm Re}x>0$,
$\mathop{\rm Im}x>0$ is given by (7.1) for $0<\alpha<\frac{1}{6}\pi$. The
parts of the curve in the other quadrants are reflections of this arc at the
real and imaginary axis. The curve surrounding $1$ is shown in Figure 6. This
means that the hypergeometric representation (5.1) with the Gauss
hypergeometric series in place of ${}_{2}F_{1}$ holds for all $z$ that lie
outside these curves. Moreover, we obtain that $|w^{\ast}_{14}|<1$ for all
$x\in D_{2}$. Therefore, the hypergeometric representation (5.2) with the
Gauss hypergeometric series in place of ${}_{2}F_{1}$ holds for all $z\in
D_{2}$. This appears to be the only representation of $Q_{\nu}^{\mu}$ valid on
$D_{2}$ by the Gauss hypergeometric series.

Figure 6: The curve $|w^{\ast}_{13}|=1$. Note that by analyzing the equation
of the curve one can see that the innermost points occur at
$x=\pm\frac{\sqrt{3}}{2}\approx\pm 0.866025$ and the outermost points occur at
$x=\pm\frac{3\sqrt{2}}{4}\approx\pm 1.06066$. Also, the points with the
largest absolute value of their imaginary parts on the curve occur at
$\mathop{\rm Re}x=\pm\sqrt{\frac{15}{32}+\frac{7\sqrt{5}}{32}}\approx\pm
0.978718$ and $\mathop{\rm
Im}x=\pm\sqrt{-\frac{11}{32}+\frac{5\sqrt{5}}{32}}\approx\pm 0.0750708$.

##  Appendix A Olbricht’s 72 solutions of the associated Legendre equation

Richard Olbricht [10, pp. 17–20] gave a list of 72 solutions of the associated
Legendre equation (1.1) in terms of the Gauss hypergeometric function
${}_{2}F_{1}(a,b;c;z)$. He does not state the domains of definition of these
solutions and does not identify them in terms of known solutions. Therefore,
it might be useful to go through the list and add a few remarks to each entry.

We will regularly use Euler’s and Pfaff’s transformations (2.3)–(2.5) and the
domains (1.2)–(1.6). Olbricht’s solutions will be denoted by $L^{{\rm
I}}_{j}$, $L^{{\rm II}}_{j}$, $L^{{\rm III}}_{j}$, $j=1,2,\dots,24$.

###  A.1 The Group I arguments

Consider

$\displaystyle L^{{\rm
I}}_{1}(x):=\left(\frac{1-x}{2}\right)^{\frac{1}{2}\mu}\left(\frac{1+x}{2}\right)^{\frac{1}{2}\mu}{}_{2}F_{1}\left(\mu-\nu,\nu+\mu+1;1+\mu;\frac{1-x}{2}\right).$
(A.1)

If we use principal values of the powers and also of the hypergeometric
function ${}_{2}F_{1}$ then $L^{{\rm I}}_{1}$ is analytic on the domain
$D_{1}$. Using (2.3) and [11, equation (14.3.1)] Olbricht has

$\displaystyle L^{{\rm I}}_{1}(x)=\Gamma(1+\mu){\sf P}_{\nu}^{-\mu}(x)$

for $x\in D_{1}$. If $\mathbf{L}^{I}_{1}$ denotes the right-hand side of (A.1)
with ${}_{2}F_{1}$ replaced by Olver’s
$\mathbf{F}(a,b;c;z):=\frac{1}{\Gamma(c)}{{}_{2}F_{1}}(a,b;c;z)$ [11, equation
(15.2.2)] then

$\displaystyle\mathbf{L}^{I}_{1}(x)={\sf P}_{\nu}^{-\mu}(x)$

and this equation is true for all $\nu,\mu\in\mathbb{C}$. The function

$\displaystyle L^{{\rm
I}}_{2}(x):=\left(\frac{1-x}{2}\right)^{-\frac{1}{2}\mu}\left(\frac{1+x}{2}\right)^{\frac{1}{2}\mu}{}_{2}F_{1}\left({-}\nu,\nu+1;1-\mu;\frac{1-x}{2}\right)$

is analytic on $D_{1}$. By applying (2.3) to $L^{{\rm I}}_{1}$ and changing
$\mu$ to $-\mu$,

$\displaystyle L^{{\rm I}}_{2}(x)=\Gamma(1-\mu){\sf P}_{\nu}^{\mu}(x),$

for $x\in D_{1}$. Olbricht has

$\displaystyle L^{{\rm
I}}_{3}(x):=\left(\frac{1-x}{2}\right)^{\frac{1}{2}\mu}\left(\frac{1+x}{2}\right)^{-\frac{1}{2}\mu}{}_{2}F_{1}\left({-}\nu,\nu+1;1+\mu;\frac{1-x}{2}\right),$

and by (2.3), $L^{{\rm I}}_{1}=L^{{\rm I}}_{3}$. Also,

$\displaystyle L^{{\rm
I}}_{4}(x):=\left(\frac{1-x}{2}\right)^{-\frac{1}{2}\mu}\left(\frac{1+x}{2}\right)^{-\frac{1}{2}\mu}{}_{2}F_{1}\left({-}\nu-\mu,\nu-\mu+1;1-\mu;\frac{1-x}{2}\right),$

and by (2.3), $L^{{\rm I}}_{2}=L^{{\rm I}}_{4}$. One also has

$\displaystyle L^{{\rm I}}_{5}(x):=L^{{\rm I}}_{1}(-x),$ $\displaystyle
L^{{\rm I}}_{6}(x):=L^{{\rm I}}_{2}(-x),$ $\displaystyle L^{{\rm
I}}_{7}(x):=L^{{\rm I}}_{3}(-x),$ $\displaystyle L^{{\rm I}}_{8}(x):=L^{{\rm
I}}_{4}(-x),$ $\displaystyle L^{{\rm
I}}_{9}(x):=\left(\frac{2}{1-x}\right)^{-\nu}\left(\frac{x+1}{x-1}\right)^{\frac{1}{2}\mu}{}_{2}F_{1}\left({-}\nu,\mu-\nu;-2\nu;\frac{2}{1-x}\right).$

Using principal values, $L^{{\rm I}}_{9}$ is analytic on $D_{3}$. Using [9,
Entry 24, p. 161] and (2.3) with [11, equation (14.9.14)]

$\displaystyle\bm{Q}_{\nu}^{\mu}(x)=\bm{Q}_{\nu}^{-\mu}(x)$

and [11, equation (14.3.10)]

$\displaystyle\bm{Q}_{\nu}^{\mu}(x):={\rm e}^{-\mu\pi{\rm
i}}\frac{Q_{\nu}^{\mu}(x)}{\Gamma(\nu+\mu+1)},$

we find

$\displaystyle L^{{\rm
I}}_{9}(x)=\frac{4^{-\nu}}{\sqrt{\pi}}\Gamma\left(\frac{1}{2}-\nu\right)\bm{Q}_{-\nu-1}^{\mu}(-x)$

for $x\in D_{3}$. We also have

$\displaystyle L^{{\rm
I}}_{10}(x):=\left(\frac{2}{1-x}\right)^{\nu+1}\left(\frac{x+1}{x-1}\right)^{\frac{1}{2}\mu}{}_{2}F_{1}\left(\nu+\mu+1,\nu+1;2\nu+2;\frac{2}{1-x}\right),$

and $L^{{\rm I}}_{10}$ is obtained from $L^{{\rm I}}_{9}$ by replacing
$\nu\mapsto-\nu-1$. Hence

$\displaystyle L^{{\rm
I}}_{10}(x)=\frac{4^{\nu+1}}{\sqrt{\pi}}\Gamma\left(\nu+\frac{3}{2}\right)\bm{Q}_{\nu}^{\mu}(-x)$

for $x\in D_{3}$. Also,

$\displaystyle L^{{\rm
I}}_{11}(x):=\left(\frac{2}{1-x}\right)^{-\nu}\left(\frac{x+1}{x-1}\right)^{-\frac{1}{2}\mu}{}_{2}F_{1}\left({-}\nu,-\nu-\mu;-2\nu;\frac{2}{1-x}\right),$

and $L^{{\rm I}}_{11}$ is obtained from $L^{{\rm I}}_{9}$ by replacing
$\mu\mapsto-\mu$. Hence $L^{{\rm I}}_{11}=L^{{\rm I}}_{9}$. One also has

$\displaystyle L^{{\rm
I}}_{12}(x):=\left(\frac{2}{1-x}\right)^{\nu+1}\left(\frac{x+1}{x-1}\right)^{-\frac{1}{2}\mu}{}_{2}F_{1}\left(\nu+1,\nu-\mu+1;2\nu+2;\frac{2}{1-x}\right),$

where $L^{{\rm I}}_{12}$ is obtained from $L^{{\rm I}}_{10}$ by replacing
$\mu\mapsto-\mu$, so $L^{{\rm I}}_{12}=L^{{\rm I}}_{10}$. Also,

$\displaystyle L^{{\rm I}}_{13}(x):=L^{{\rm I}}_{9}(-x),$

and using principal values, $L^{{\rm I}}_{13}(x)$ is analytic on $D_{2}$, and
therefore

$\displaystyle L^{{\rm
I}}_{13}(x)=\frac{4^{-\nu}}{\sqrt{\pi}}\Gamma\left(\frac{1}{2}-\nu\right)\bm{Q}_{-\nu-1}^{\mu}(x)$

for $x\in D_{2}$. For $x\in D_{2}$, we also have

$\displaystyle\allowdisplaybreaks L^{{\rm I}}_{14}(x):=L^{{\rm I}}_{10}(-x),$
$\displaystyle L^{{\rm I}}_{15}(x):=L^{{\rm I}}_{11}(-x),$ $\displaystyle
L^{{\rm I}}_{16}(x):=L^{{\rm I}}_{12}(-x),$ $\displaystyle L^{{\rm
I}}_{17}(x):=\left(\frac{x-1}{x+1}\right)^{\frac{1}{2}\mu}\left(\frac{2}{1+x}\right)^{-\nu}{}_{2}F_{1}\left({-}\nu,\mu-\nu;1+\mu;\frac{x-1}{x+1}\right).$

This function is analytic on $D_{2}$. By [9, Section 4.1.2, Entry 3]

$\displaystyle L^{{\rm I}}_{17}(x)=\Gamma(1+\mu)P_{\nu}^{-\mu}(x)$

for $x\in D_{2}$. One also has

$\displaystyle L^{{\rm
I}}_{18}(x):=\left(\frac{x-1}{x+1}\right)^{-\frac{1}{2}\mu}\left(\frac{2}{1+x}\right)^{-\nu}{}_{2}F_{1}\left({-}\nu,-\mu-\nu;1-\mu;\frac{x-1}{x+1}\right),$

where $L^{{\rm I}}_{18}$ is obtained from $L^{{\rm I}}_{17}$ by replacing
$\mu\mapsto-\mu$. Hence

$\displaystyle L^{{\rm I}}_{18}(x):=\Gamma(1-\mu)P_{\nu}^{\mu}(x)$

for $x\in D_{2}$. Also,

$\displaystyle L^{{\rm
I}}_{19}(x):=\left(\frac{x-1}{x+1}\right)^{\frac{1}{2}\mu}\left(\frac{2}{1-x}\right)^{\nu+1}{}_{2}F_{1}\left(\nu+\mu+1,\nu+1;1+\mu;\frac{x-1}{x+1}\right),$

and by (2.3), $L^{{\rm I}}_{19}=L^{{\rm I}}_{17}$. Also,

$\displaystyle L^{{\rm
I}}_{20}(x):=\left(\frac{x-1}{x+1}\right)^{-\frac{1}{2}\mu}\left(\frac{2}{1+x}\right)^{\nu+1}{}_{2}F_{1}\left(\nu+1,\nu-\mu+1;1-\mu;\frac{x-1}{x+1}\right),$

and by (2.3), $L^{{\rm I}}_{20}=L^{{\rm I}}_{18}$. Also,

$\displaystyle L^{{\rm I}}_{21}(x):=L^{{\rm I}}_{17}(-x),$ $\displaystyle
L^{{\rm I}}_{22}(x):=L^{{\rm I}}_{18}(-x),$ $\displaystyle L^{{\rm
I}}_{23}(x):=L^{{\rm I}}_{19}(-x),$ $\displaystyle L^{{\rm
I}}_{24}(x):=L^{{\rm I}}_{21}(-x).$

###  A.2 The Group II arguments

The first function in Group II of Olbricht’s list is

$\displaystyle L^{{\rm
II}}_{1}(x):=\big{(}1-x^{2}\big{)}^{\frac{1}{2}\mu}{}_{2}F_{1}\left(\frac{1}{2}\mu-\frac{1}{2}\nu,\frac{1}{2}\nu+\frac{1}{2}\mu+\frac{1}{2};\frac{1}{2};x^{2}\right).$

It is an even solution of (1.1) on $D_{1}$ uniquely determined by the initial
conditions $y(0)=1$, $y^{\prime}(0)=0$. By [11, equation (14.5.1)],

$\displaystyle L^{{\rm
II}}_{1}(x)=2^{-\mu-1}\pi^{{-\frac{1}{2}}}\Gamma\left(\frac{1}{2}\nu-\frac{1}{2}\mu+1\right)\Gamma\left({-}\frac{1}{2}\nu-\frac{1}{2}\mu+\frac{1}{2}\right)\big{(}{\sf
P}_{\nu}^{\mu}(x)+{\sf P}_{\nu}^{\mu}(-x)\big{)}$

for $x\in D_{1}$. We also have

$\displaystyle L^{{\rm
II}}_{2}(x):=x\big{(}1-x^{2}\big{)}^{\frac{1}{2}\mu}{}_{2}F_{1}\left(\frac{1}{2}\mu-\frac{1}{2}\nu+\frac{1}{2},\frac{1}{2}\nu+\frac{1}{2}\mu+1;\frac{3}{2};x^{2}\right).$

This function is analytic on $D_{1}$. It is an odd solution of (1.1) on
$D_{1}$ uniquely determined by the initial conditions $y(0)=0$,
$y^{\prime}(0)=1$. Using [11, equation (14.5.2)],

$\displaystyle L^{{\rm
II}}_{2}(x):=2^{-\mu-2}\pi^{{-\frac{1}{2}}}\Gamma\left(\frac{1}{2}\nu-\frac{1}{2}\mu+\frac{1}{2}\right)\Gamma\left({-}\frac{1}{2}\nu-\frac{1}{2}\mu\right)\big{(}{\sf
P}_{\nu}^{\mu}(-x)-{\sf P}_{\nu}^{\mu}(x)\big{)},$

for $x\in D_{1}$. Also,

$\displaystyle L^{{\rm
II}}_{3}(x):=\big{(}1-x^{2}\big{)}^{-\frac{1}{2}\mu}{}_{2}F_{1}\left({-}\frac{1}{2}\nu-\frac{1}{2}\mu,\frac{1}{2}\nu-\frac{1}{2}\mu+\frac{1}{2};\frac{1}{2};x^{2}\right),$

and by (2.3), $L^{{\rm II}}_{3}=L^{{\rm II}}_{1}$. Furthermore,

$\displaystyle L^{{\rm
II}}_{4}(x):=x\big{(}1-x^{2}\big{)}^{-\frac{1}{2}\mu}{}_{2}F_{1}\left({-}\frac{1}{2}\nu-\frac{1}{2}\mu+\frac{1}{2},\frac{1}{2}\nu-\frac{1}{2}\mu+1;\frac{3}{2};x^{2}\right),$

and by (2.3), $L^{{\rm II}}_{4}=L^{{\rm II}}_{2}$. Also,

$\displaystyle L^{{\rm
II}}_{5}(x):=\big{(}1-x^{2}\big{)}^{\frac{1}{2}\mu}{}_{2}F_{1}\left(\frac{1}{2}\mu-\frac{1}{2}\nu,\frac{1}{2}\nu+\frac{1}{2}\mu+\frac{1}{2};1+\mu;1-x^{2}\right).$

This function is analytic on $D_{1}\setminus{\rm i}\mathbb{R}$. By [9, Section
4.1.2, Entry 5],

$\displaystyle L^{{\rm II}}_{5}(x)=2^{\mu}\Gamma(1+\mu){\sf
P}_{\nu}^{-\mu}(x),$

for $x\in D_{1}^{+}$. One also has

$\displaystyle L^{{\rm
II}}_{6}(x):=x\big{(}1-x^{2}\big{)}^{\frac{1}{2}\mu}{}_{2}F_{1}\left(\frac{1}{2}\mu-\frac{1}{2}\nu+\frac{1}{2},\frac{1}{2}\nu+\frac{1}{2}\mu+1;1+\mu;1-x^{2}\right),$

and by (2.3), $L^{{\rm II}}_{6}=L^{{\rm II}}_{5}$ for $x\in D_{1}^{+}$. Also,

$\displaystyle L^{{\rm
II}}_{7}(x):=\big{(}1-x^{2}\big{)}^{-\frac{1}{2}\mu}{}_{2}F_{1}\left({-}\frac{1}{2}\nu-\frac{1}{2}\mu,\frac{1}{2}\nu-\frac{1}{2}\mu+\frac{1}{2};1-\mu;1-x^{2}\right),$

and $L^{{\rm II}}_{7}$ is obtained from $L^{{\rm II}}_{5}$ by replacing
$\mu\mapsto-\mu$. Therefore,

$\displaystyle L^{{\rm II}}_{7}(x)=2^{-\mu}\Gamma(1-\mu){\sf
P}_{\nu}^{\mu}(x)$

for $x\in D_{1}^{+}$. Also,

$\displaystyle L^{{\rm
II}}_{8}(x):=x\big{(}1-x^{2}\big{)}^{-\frac{1}{2}\mu}{}_{2}F_{1}\left(\frac{1}{2}\nu-\frac{1}{2}\mu+1,-\frac{1}{2}\nu-\frac{1}{2}\mu+\frac{1}{2};1-\mu;1-x^{2}\right),$

and by (2.3), $L^{{\rm II}}_{8}(x)=L^{{\rm II}}_{7}(x)$ for $x\in D_{1}^{+}$.
Also,

$\displaystyle L^{{\rm
II}}_{9}(x):={x^{\nu-\mu}}\big{(}x^{2}-1\big{)}^{\frac{1}{2}\mu}{}_{2}F_{1}\left(\frac{1}{2}\mu-\frac{1}{2}\nu,\frac{1}{2}\mu-\frac{1}{2}\nu+\frac{1}{2};\frac{1}{2}-\nu;\frac{1}{x^{2}}\right).$

We use $\big{(}x^{2}-1\big{)}^{\alpha}$ as an abbreviation for
$(x-1)^{\alpha}(x+1)^{\alpha}$, so $\big{(}x^{2}-1\big{)}^{\alpha}$ is
analytic on $D_{2}$. Olbricht has $\big{(}1-x^{2}\big{)}^{\frac{1}{2}\mu}$ in
place of $\big{(}x^{2}-1\big{)}^{\frac{1}{2}\mu}$ but this has the
disadvantage that when using principal values the function is not analytic for
any real $x$. $L^{{\rm II}}_{9}$ is analytic on $D_{2}$. By [9, Section 4.1.2,
Entry 28],

$\displaystyle L^{{\rm
II}}_{9}(x)=2^{-\nu}\pi^{{-\frac{1}{2}}}\Gamma\left(\frac{1}{2}-\nu\right)\bm{Q}_{-\nu-1}^{\mu}(x)$

for $x\in D_{2}$. Also,

$\displaystyle L^{{\rm
II}}_{10}(x):=\left(\frac{1}{x}\right)^{\nu+\mu+1}\big{(}x^{2}-1\big{)}^{\frac{1}{2}\mu}{}_{2}F_{1}\left(\frac{1}{2}\nu+\frac{1}{2}\mu+\frac{1}{2},\frac{1}{2}\nu+\frac{1}{2}\mu+1;\nu+\frac{3}{2};\frac{1}{x^{2}}\right).$

Olbricht has $\big{(}1-x^{2}\big{)}^{\frac{1}{2}\mu}$ in place of
$\big{(}x^{2}-1\big{)}^{\frac{1}{2}\mu}$, and $L^{{\rm II}}_{10}$ is obtained
from $L^{{\rm II}}_{9}$ by replacing $\nu\mapsto-\nu-1$. Hence

$\displaystyle L^{{\rm
II}}_{10}(x)=2^{\nu+1}\pi^{{-\frac{1}{2}}}\Gamma\left(\nu+\frac{3}{2}\right)\bm{Q}_{\nu}^{\mu}(x)$

for $x\in D_{2}$. Also,

$\displaystyle L^{{\rm
II}}_{11}(x):={x^{{{\mu-\nu-1}}}}\big{(}x^{2}-1\big{)}^{-\frac{1}{2}\mu}{}_{2}F_{1}\left(\frac{1}{2}\nu-\frac{1}{2}\mu+\frac{1}{2},\frac{1}{2}\nu-\frac{1}{2}\mu+1;\nu+\frac{3}{2};\frac{1}{x^{2}}\right).$

Olbricht has $\big{(}1-x^{2}\big{)}^{\frac{1}{2}\mu}$ in place of
$\big{(}x^{2}-1\big{)}^{\frac{1}{2}\mu}$, and $L^{{\rm II}}_{11}$ is obtained
from $L^{{\rm II}}_{10}$ by replacing $\mu\mapsto-\mu$. Hence $L^{{\rm
II}}_{11}=L^{{\rm II}}_{10}$. Also,

$\displaystyle L^{{\rm
II}}_{12}(x):={x^{\nu+\mu}}\big{(}x^{2}-1\big{)}^{-\frac{1}{2}\mu}{}_{2}F_{1}\left({-}\frac{1}{2}\nu-\frac{1}{2}\mu,-\frac{1}{2}\nu-\frac{1}{2}\mu+\frac{1}{2};\frac{1}{2}-\nu;\frac{1}{x^{2}}\right).$

Olbricht has $\big{(}1-x^{2}\big{)}^{\frac{1}{2}\mu}$ in place of
$\big{(}x^{2}-1\big{)}^{\frac{1}{2}\mu}$, and $L^{{\rm II}}_{12}$ is obtained
from $L^{{\rm II}}_{9}$ by replacing $\mu\mapsto-\mu$. Hence $L^{{\rm
II}}_{12}=L^{{\rm II}}_{9}$. Also,

$\displaystyle L^{{\rm
II}}_{13}(x):=\big{(}x^{2}-1\big{)}^{\frac{1}{2}\nu}{}_{2}F_{1}\left(\frac{1}{2}\mu-\frac{1}{2}\nu,-\frac{1}{2}\mu-\frac{1}{2}\nu;\frac{1}{2}-\nu;\frac{1}{1-x^{2}}\right).$

Olbricht has $\big{(}1-x^{2}\big{)}^{\frac{1}{2}\nu}$ in place of
$\big{(}x^{2}-1\big{)}^{\frac{1}{2}\nu}$, and $L^{{\rm II}}_{13}$ is analytic
on $D_{2}$. By (2.4), $L^{{\rm II}}_{13}=L^{{\rm II}}_{9}$. More precisely, we
use (2.4) for $x>1$ and then apply the identity theorem for analytic
functions. Also,

$\displaystyle L^{{\rm
II}}_{14}(x):=\big{(}x^{2}-1\big{)}^{-\frac{1}{2}\nu-\frac{1}{2}}{}_{2}F_{1}\left(\frac{1}{2}\nu+\frac{1}{2}\mu+\frac{1}{2},\frac{1}{2}\nu-\frac{1}{2}\mu+\frac{1}{2};\nu+\frac{3}{2};\frac{1}{1-x^{2}}\right).$

Olbricht has $\big{(}1-x^{2}\big{)}^{-\frac{1}{2}\nu-\frac{1}{2}}$ in place of
$\big{(}x^{2}-1\big{)}^{-\frac{1}{2}\nu-\frac{1}{2}}$, and by (2.4), $L^{{\rm
II}}_{14}=L^{{\rm II}}_{10}$. Also,

$\displaystyle L^{{\rm
II}}_{15}(x):=x\big{(}x^{2}-1\big{)}^{\frac{1}{2}\nu-\frac{1}{2}}{}_{2}F_{1}\left(\frac{1}{2}\mu-\frac{1}{2}\nu+\frac{1}{2},-\frac{1}{2}\nu-\frac{1}{2}\mu+\frac{1}{2};\frac{1}{2}-\nu;\frac{1}{1-x^{2}}\right).$

Olbricht has $\big{(}1-x^{2}\big{)}^{\frac{1}{2}\nu-\frac{1}{2}}$ in place of
$\big{(}x^{2}-1\big{)}^{\frac{1}{2}\nu-\frac{1}{2}}$, and by (2.4), $L^{{\rm
II}}_{15}=L^{{\rm II}}_{12}$. Also,

$\displaystyle L^{{\rm
II}}_{16}(x):=x\big{(}x^{2}-1\big{)}^{\frac{1}{2}\nu-1}{}_{2}F_{1}\left(\frac{1}{2}\nu+\frac{1}{2}\mu+1,\frac{1}{2}\nu-\frac{1}{2}\mu+1;\nu+\frac{3}{2};\frac{1}{1-x^{2}}\right).$

Olbricht has $\big{(}1-x^{2}\big{)}^{\frac{1}{2}\nu-1}$ in place of
$\big{(}x^{2}-1\big{)}^{\frac{1}{2}\nu-1}$, and by (2.4), $L^{{\rm
II}}_{16}=L^{{\rm II}}_{11}$. Also,

$\displaystyle L^{{\rm
II}}_{17}(x):=\big{(}1-x^{2}\big{)}^{\frac{1}{2}\nu}{}_{2}F_{1}\left(\frac{1}{2}\mu-\frac{1}{2}\nu,-\frac{1}{2}\mu-\frac{1}{2}\nu;\frac{1}{2};\frac{x^{2}}{x^{2}-1}\right),$

and by (2.4), $L^{{\rm II}}_{17}=L^{{\rm II}}_{1}$. Also,

$\displaystyle L^{{\rm
II}}_{18}(x):=\big{(}1-x^{2}\big{)}^{-\frac{1}{2}\nu-\frac{1}{2}}{}_{2}F_{1}\left(\frac{1}{2}\nu+\frac{1}{2}\mu+\frac{1}{2},\frac{1}{2}\nu-\frac{1}{2}\mu+\frac{1}{2};\frac{1}{2};\frac{x^{2}}{x^{2}-1}\right),$

and by (2.4), $L^{{\rm II}}_{18}=L^{{\rm II}}_{3}$. Also,

$\displaystyle L^{{\rm
II}}_{19}(x):=x\big{(}1-x^{2}\big{)}^{\frac{1}{2}\nu-\frac{1}{2}}{}_{2}F_{1}\left(\frac{1}{2}\mu-\frac{1}{2}\nu+\frac{1}{2},-\frac{1}{2}\nu-\frac{1}{2}\mu+\frac{1}{2};\frac{3}{2};\frac{x^{2}}{x^{2}-1}\right),$

and by (2.4), $L^{{\rm II}}_{19}=L^{{\rm II}}_{2}$. Also,

$\displaystyle L^{{\rm
II}}_{20}(x):=x\big{(}1-x^{2}\big{)}^{-\frac{1}{2}\nu-1}{}_{2}F_{1}\left(\frac{1}{2}\nu+\frac{1}{2}\mu+1,\frac{1}{2}\nu-\frac{1}{2}\mu+1;\frac{3}{2};\frac{x^{2}}{x^{2}-1}\right),$

and by (2.4), $L^{{\rm II}}_{20}=L^{{\rm II}}_{4}$. Also,

$\displaystyle L^{{\rm
II}}_{21}(x):=x^{{\nu-\mu}}\big{(}1-x^{2}\big{)}^{\frac{1}{2}\mu}{}_{2}F_{1}\left(\frac{1}{2}\mu-\frac{1}{2}\nu,\frac{1}{2}\mu-\frac{1}{2}\nu+\frac{1}{2};1+\mu;\frac{x^{2}-1}{x^{2}}\right),$

and by (2.4), $L^{{\rm II}}_{21}=L^{{\rm II}}_{5}$. Also,

$\displaystyle L^{{\rm
II}}_{22}(x):=x^{{-\nu-\mu-1}}\big{(}1-x^{2}\big{)}^{\frac{1}{2}\mu}{}_{2}F_{1}\left(\frac{1}{2}\nu+\frac{1}{2}\mu+\frac{1}{2},\frac{1}{2}\nu+\frac{1}{2}\mu+1;1+\mu;\frac{x^{2}-1}{x^{2}}\right),$

and by (2.4), $L^{{\rm II}}_{22}=L^{{\rm II}}_{6}$. Also,

$\displaystyle L^{{\rm
II}}_{23}(x):=x^{{\mu-\nu-1}}\big{(}1-x^{2}\big{)}^{-\frac{1}{2}\mu}{}_{2}F_{1}\left(\frac{1}{2}\nu-\frac{1}{2}\mu+1,\frac{1}{2}\nu-\frac{1}{2}\mu+\frac{1}{2};1-\mu;\frac{x^{2}-1}{x^{2}}\right),$

and by (2.4), $L^{{\rm II}}_{23}=L^{{\rm II}}_{7}$. Also,

$\displaystyle L^{{\rm
II}}_{24}(x):=x^{\mu+\nu}\big{(}1-x^{2}\big{)}^{-\frac{1}{2}\mu}{}_{2}F_{1}\left({-}\frac{1}{2}\nu-\frac{1}{2}\mu+\frac{1}{2},-\frac{1}{2}\mu-\frac{1}{2}\nu;1-\mu;\frac{x^{2}-1}{x^{2}}\right),$

and by (2.4), $L^{{\rm II}}_{24}=L^{{\rm II}}_{8}$.

###  A.3 The Group III arguments

In this section there appears the function $\sqrt{x^{2}-1}$. There are two
versions of this function:

$\displaystyle y_{1}:={\rm i}\sqrt{1-x^{2}},\qquad\text{for}\quad x\in D_{1},$
$\displaystyle y_{2}:=x\sqrt{1-x^{-2}},\qquad\text{for}\quad
x\in\mathbb{C}\setminus[-1,1],$

where the root denotes its principal value. If $\pm\mathop{\rm Im}x>0$ then
$y_{1}=\pm y_{2}$. Each entry in Olbricht’s list has two versions depending on
whether $\sqrt{x^{2}-1}$ has the meaning $y_{k}$, $k=1,2$. We will denote
these functions by $L^{{\rm III}}_{j,k}$, where $1\leq j\leq 24$, $1\leq k\leq
2$.

In Entries 1, 2, 3, 4 of Olbricht’s list the hypergeometric function has the
argument

$\displaystyle w_{17}=\frac{\sqrt{x^{2}-1}+x}{2\sqrt{x^{2}-1}}.$

If we use $y_{1}$ for $\sqrt{x^{2}-1}$ then the function

$\displaystyle w_{17,1}=\frac{y_{1}+x}{2y_{1}}$

is a conformal map from $D_{1}$ to the complex plane cut along the rays
$(-\infty,0]$ and $[1,\infty)$. To see this let $x=\cos\theta$. This is a
conformal map from the strip $S=\\{\theta\colon\mathop{\rm
Re}\theta\in(0,\pi)\\}$ onto $D_{1}$. Then

$\displaystyle\frac{y_{1}+x}{2y_{1}}=\frac{1}{2}-\frac{\rm i}{2}\cot\theta.$

Now $\cot\theta$ is a conformal map from $S$ to $\mathbb{C}\setminus((-{\rm
i}\infty,-{\rm i}]\cup[{\rm i},{\rm i}\infty))$. Therefore, using the
principal value of the hypergeometric function, the function
${}_{2}F_{1}(a,b;c;w_{17,1})$ the function is analytic on $D_{1}$. On the
other hand, if we use $y_{2}$ for $\sqrt{x^{2}-1}$ then the values of

$\displaystyle w_{17,2}=\frac{y_{2}+x}{2y_{2}}$

lie on the ray $[1,\infty)$ for $x>1$, so ${}_{2}F_{1}(a,b;c;w_{17,2})$ cannot
be defined for $x>1$ when using the principal value of the hypergeometric
function. Therefore we will not allow $y_{2}$ in Entries $1,2,3,4$ of
Olbrichts’s list.

In Entries 5, 6, 7, 8 the hypergeometric function has argument

$\displaystyle w_{13}=\frac{\sqrt{x^{2}-1}-x}{2\sqrt{x^{2}-1}}.$

Since $y_{1}$ is an even function of $x$ ($y_{2}$ is odd) we have

$\displaystyle w_{13,1}(x)=\frac{y_{1}-x}{2y_{1}}=w_{17,1}(-x)$

so we can define ${}_{2}F_{1}(a,b;c;w_{13,1})$ again on $D_{1}$. However, the
situation is now different for $y_{2}$. If we use $y_{2}$ for $\sqrt{x^{2}-1}$
then the function

$\displaystyle w_{13,2}=\frac{y_{2}-x}{2y_{2}}$

is analytic on $D_{2}$ and its range is
$\big{\\{}z\in\mathbb{C}\colon\mathop{\rm
Re}z<\frac{1}{2}\big{\\}}\setminus\big{(}{-}\infty,-\frac{1}{2}\big{)}$.
Therefore, ${}_{2}F_{1}(a,b;c;w_{13,2})$ is analytic on $D_{2}$. This follows
as in the proof of Theorem 5.1.

In the following list of 24 solutions of Olbricht the factors in front of the
hypergeometric function are rewritten. This is necessary in order to obtain
reasonable branch cuts. For example, solution 7 of Olbricht is

$\displaystyle\left(\frac{\sqrt{x^{2}-1}-x}{2\sqrt{x^{2}-1}}\right)^{\frac{1}{2}\nu+\frac{1}{2}}\left(\frac{\sqrt{x^{2}-1}+x}{2\sqrt{x^{2}-1}}\right)^{-\frac{1}{2}\nu}{}_{2}F_{1}\left(\frac{1}{2}-\mu,\mu+\frac{1}{2};\nu+\frac{3}{2};\frac{\sqrt{x^{2}-1}-x}{2\sqrt{x^{2}-1}}\right).$

If we choose $y_{2}$ for $\sqrt{x^{2}-1}$ then, up to a constant factor, this
solution should agree with the associated Legendre function
$Q_{\nu}^{\mu}(x)$. However, the basis of the first power is a negative number
for $x>1$. Using that

$\displaystyle\Big{(}x-\sqrt{x^{2}-1}\Big{)}\Big{(}x+\sqrt{x^{2}-1}\Big{)}=1,$

we rewrite the solution as

$\displaystyle(2y_{2})^{-\frac{1}{2}}(x+y_{2})^{-\nu-\frac{1}{2}}{}_{2}F_{1}\left(\frac{1}{2}-\mu,\mu+\frac{1}{2};\nu+\frac{3}{2};\frac{y_{2}-x}{2y_{2}}\right).$

This function is now analytic on $D_{2}$ as it should be.

The first entry in Olbricht’s list is

$\displaystyle L^{{\rm
III}}_{1,1}(x):=(2y_{1})^{\nu}{}_{2}F_{1}\left(\mu-\nu,-\mu-\nu;\frac{1}{2}-\nu;\frac{y_{1}+x}{2y_{1}}\right)=L^{{\rm
III}}_{5,1}(-x).$

The function $L^{{\rm III}}_{5,1}$ will be identified below. Also,

$\displaystyle L^{{\rm
III}}_{2,1}(x):=(2y_{1})^{-\frac{1}{2}}(y_{1}-x)^{\nu+\frac{1}{2}}{}_{2}F_{1}\left(\frac{1}{2}-\mu,\mu+\frac{1}{2};\frac{1}{2}-\nu;\frac{y_{1}+x}{2y_{1}}\right),$

and by (2.3), $L^{{\rm III}}_{2,1}=L^{{\rm III}}_{1,1}$. Also,

$\displaystyle L^{{\rm
III}}_{3,1}(x):=(2y_{1})^{-\frac{1}{2}}(y_{1}-x)^{-\nu-\frac{1}{2}}{}_{2}F_{1}\left(\frac{1}{2}-\mu,\mu+\frac{1}{2};\nu+\frac{3}{2};\frac{y_{1}+x}{2y_{1}}\right),$

and $L^{{\rm III}}_{3,1}$ is obtained from $L^{{\rm III}}_{2,1}$ by replacing
$\nu\mapsto-\nu-1$. Also,

$\displaystyle L^{{\rm
III}}_{4,1}(x):=(2y_{1})^{-\nu-1}{}_{2}F_{1}\left(\nu-\mu+1,\nu+\mu+1;\nu+\frac{3}{2};\frac{y_{1}+x}{2y_{1}}\right),$

and by (2.3), $L^{{\rm III}}_{4,1}=L^{{\rm III}}_{3,1}$.

The fifth entry in Olbricht’s list is

$\displaystyle L^{{\rm
III}}_{5,k}(x):=(2y_{k})^{\nu}{}_{2}F_{1}\left(\mu-\nu,-\mu-\nu;\frac{1}{2}-\nu;\frac{y_{k}-x}{2y_{k}}\right),$

and this function is analytic on $D_{k}$ for $k=1,2$. By [9, Section 4.1.2,
Entry 31] and (2.3),

$\displaystyle L^{{\rm
III}}_{5,2}(x)=\pi^{-1/2}\Gamma\left(\frac{1}{2}-\nu\right)\bm{Q}_{-\nu-1}^{\mu}(x)\qquad\text{for}\quad
x\in D_{2}.$

Since $y_{1}=y_{2}$ for $\mathop{\rm Im}x>0$,

$\displaystyle L^{{\rm
III}}_{5,1}(x)=\pi^{-1/2}\Gamma\left(\frac{1}{2}-\nu\right)\bm{Q}_{-\nu-1}^{\mu}(x)\qquad\text{if}\quad\mathop{\rm
Im}x>0.$

We know from [12, p. 147, equation (21)] that

$\displaystyle\bm{Q}_{-\nu-1}^{\mu}\big{(}1+(x-1){\rm e}^{2\pi{\rm
i}}\big{)}={\rm e}^{-\pi{\rm i}\mu}\bm{Q}_{-\nu-1}^{\mu}(x)-\frac{\pi{\rm
i}}{\Gamma(-\nu-\mu)}P_{-\nu-1}^{-\mu}(x).$

Therefore,

$\displaystyle L^{{\rm
III}}_{5,1}(x)=\pi^{-1/2}\Gamma\left(\frac{1}{2}-\nu\right)\left({\rm
e}^{-\pi{\rm i}\mu}\bm{Q}_{-\nu-1}^{\mu}(x)-\frac{\pi{\rm
i}}{\Gamma(-\nu-\mu)}P_{\nu}^{-\mu}(x)\right)\qquad\text{if}\quad\mathop{\rm
Im}x<0.$

Entries 6, 7, 8 in Olbricht’s list are

$\displaystyle L^{{\rm
III}}_{6,k}(x):=(2y_{k})^{-\frac{1}{2}}(x+y_{k})^{\nu+\frac{1}{2}}{}_{2}F_{1}\left(\frac{1}{2}-\mu,\mu+\frac{1}{2};\frac{1}{2}-\nu;\frac{y_{k}-x}{2y_{k}}\right),$
$\displaystyle L^{{\rm
III}}_{7,k}(x):=(2y_{k})^{-\frac{1}{2}}(x+y_{k})^{-\nu-\frac{1}{2}}{}_{2}F_{1}\left(\frac{1}{2}-\mu,\mu+\frac{1}{2};\nu+\frac{3}{2};\frac{y_{k}-x}{2y_{k}}\right),$
$\displaystyle L^{{\rm
III}}_{8,k}(x):=(2y_{k})^{-\nu-1}{}_{2}F_{1}\left(\nu-\mu+1,\nu+\mu+1;\nu+\frac{3}{2};\frac{y_{k}-x}{2y_{k}}\right).$

By (2.3), $L^{{\rm III}}_{6,k}=L^{{\rm III}}_{5,k}$ and $L^{{\rm
III}}_{8,k}=L^{{\rm III}}_{7,k}$. $L^{{\rm III}}_{7,k}$ is obtained from
$L^{{\rm III}}_{6,k}$ by replacing $\nu\mapsto-\nu-1$.

In Entries 9, 10, 11, 12 the hypergeometric function has the argument

$\displaystyle w_{15}=\frac{2\sqrt{x^{2}-1}}{\sqrt{x^{2}-1}+x}.$

If $\sqrt{x^{2}-1}$ is replaced by $y_{1}$ or $y_{2}$ then purely imaginary
$x$ are mapped to the branch cut $[1,\infty)$ of the hypergeometric function.
Therefore, we will work on $D_{k}^{+}$. The function

$\displaystyle L^{{\rm
III}}_{9,k}(x):=(2y_{k})^{\mu}(x+y_{k})^{\nu-\mu}{}_{2}F_{1}\left(\mu-\nu,\mu+\frac{1}{2};1+2\mu;\frac{2y_{k}}{y_{k}+x}\right)$

is analytic on $D_{k}^{+}$ for $k=1,2$. By [9, Section 4.1.2, Entry 15],

$\displaystyle L^{{\rm
III}}_{9,2}(x)=4^{\mu}\Gamma(1+\mu)P_{\nu}^{-\mu}(x)\qquad\text{for}\quad x\in
D_{2}^{+}.$

This implies

$\displaystyle L^{{\rm III}}_{9,1}(x)={\rm e}^{\frac{1}{2}{\rm
i}\pi\mu}4^{\mu}\Gamma(1+\mu){\sf P}_{\nu}^{-\mu}(x)\qquad\text{for}\quad x\in
D_{1}^{+}.$

Also,

$\displaystyle L^{{\rm
III}}_{10,k}(x):=(2y_{k})^{-\mu}(x+y_{k})^{\mu+\nu}{}_{2}F_{1}\left({-}\nu-\mu,\frac{1}{2}-\mu;1-2\mu;\frac{2y_{k}}{y_{k}+x}\right)$

so $L^{{\rm III}}_{10,k}$ is obtained from $L^{{\rm III}}_{9,k}$ by replacing
$\mu\mapsto-\mu$. Hence

$\displaystyle L^{{\rm
III}}_{10,2}(x)=4^{-\mu}\Gamma(1-\mu)P_{\nu}^{\mu}(x)\qquad\text{for}\quad
x\in D_{2}^{+}.$

Also,

$\displaystyle L^{{\rm
III}}_{11,k}(x):=(2y_{k})^{-\mu}(y_{k}+x)^{-1+\mu-\nu}{}_{2}F_{1}\left(\nu-\mu+1,\frac{1}{2}-\mu;1-2\mu;\frac{2y_{k}}{y_{k}+x}\right),$

and $L^{{\rm III}}_{11,k}$ is obtained from $L^{{\rm III}}_{10,k}$ by
replacing $\nu\mapsto-\nu-1$. By (2.3), $L^{{\rm III}}_{11,k}=L^{{\rm
III}}_{10,k}$ on $D_{k}^{+}$. Also,

$\displaystyle L^{{\rm
III}}_{12,k}(x):=(2y_{k})^{\mu}(y_{k}+x)^{-1-\mu-\nu}{}_{2}F_{1}\left(\nu+\mu+1,\mu+\frac{1}{2};1+2\mu;\frac{2y_{k}}{y_{k}+x}\right).$

$L^{{\rm III}}_{12,k}$ is obtained from $L^{{\rm III}}_{9,k}$ by replacing
$\nu\mapsto-\nu-1$. By (2.3), $L^{{\rm III}}_{12,k}=L^{{\rm III}}_{9,k}$ on
$D_{k}^{+}$.

Entries 13, 14, 15, 16 in Olbricht’s list use the argument

$\displaystyle w_{16}=\frac{2\sqrt{x^{2}-1}}{\sqrt{x^{2}-1}-x}.$

Again purely imaginary $x$ are mapped to the branch cut $[1,\infty)$ of the
hypergeometric function. Olbricht has

$\displaystyle L^{{\rm
III}}_{13,k}(x):=(2y_{k})^{\mu}(x-y_{k})^{\nu-\mu}{}_{2}F_{1}\left(\mu-\nu,\mu+\frac{1}{2};1+2\mu;\frac{2y_{k}}{y_{k}-x}\right),$
$\displaystyle L^{{\rm
III}}_{14,k}(x):=(2y_{k})^{-\mu}(x-y_{k})^{\mu+\nu}{}_{2}F_{1}\left({-}\nu-\mu,\frac{1}{2}-\mu;1-2\mu;\frac{2y_{k}}{y_{k}-x}\right),$
$\displaystyle L^{{\rm
III}}_{15}(x):=(2y_{k})^{-\mu}(x-y_{k})^{-1+\mu-\nu}{}_{2}F_{1}\left(\nu-\mu+1,\frac{1}{2}-\mu;1-2\mu;\frac{2y_{k}}{y_{k}-x}\right),$
$\displaystyle L^{{\rm
III}}_{16}(x):=(2y_{k})^{\mu}(x-y_{k})^{-1-\mu-\nu}{}_{2}F_{1}\left(\nu+\mu+1,\mu+\frac{1}{2};1+2\mu;\frac{2y_{k}}{y_{k}-x}\right).$

Using (2.4) these functions can be reduced to Entries 9, 10, 11, 12 as follows

$\displaystyle L^{{\rm III}}_{j,k}=L^{{\rm III}}_{j-4,k}\qquad\text{for}\quad
j=13,14,15,16.$

Entries 17 to 24 in Olbricht’s list are

$\displaystyle\allowdisplaybreaks L^{{\rm
III}}_{17,k}(x):=(2y_{k})^{\mu}(y_{k}-x)^{\nu-\mu}{}_{2}F_{1}\left(\mu-\nu,\mu+\frac{1}{2};\frac{1}{2}-\nu;\frac{x+y_{k}}{x-y_{k}}\right),$
$\displaystyle L^{{\rm
III}}_{18,k}(x):=(2y_{k})^{-\mu}(y_{k}-x)^{\mu+\nu}{}_{2}F_{1}\left({-}\nu-\mu,\frac{1}{2}-\mu;\frac{1}{2}-\nu;\frac{x+y_{k}}{x-y_{k}}\right),$
$\displaystyle L^{{\rm
III}}_{19,k}(x):=(2y_{k})^{\mu}(y_{k}-x)^{-\nu-\mu-1}{}_{2}F_{1}\left(\frac{1}{2}+\mu,\nu+\mu+1;\nu+\frac{3}{2};\frac{x+y_{k}}{x-y_{k}}\right),$
$\displaystyle L^{{\rm
III}}_{20,k}(x):=(2y_{k})^{-\mu}(y_{k}-x)^{\mu-\nu-1}{}_{2}F_{1}\left(\frac{1}{2}-\mu,\nu-\mu+1;\nu+\frac{3}{2};\frac{x+y_{k}}{x-y_{k}}\right),$
$\displaystyle L^{{\rm
III}}_{21,k}(x):=(2y_{k})^{\mu}(x+y_{k})^{\nu-\mu}{}_{2}F_{1}\left(\mu-\nu,\mu+\frac{1}{2};\frac{1}{2}-\nu;\frac{x-y_{k}}{x+y_{k}}\right),$
$\displaystyle L^{{\rm
III}}_{22,k}(x):=(2y_{k})^{-\mu}(x+y_{k})^{\mu+\nu}{}_{2}F_{1}\left({-}\nu-\mu,\frac{1}{2}-\mu;\frac{1}{2}-\nu;\frac{x-y_{k}}{x+y_{k}}\right),$
$\displaystyle L^{{\rm
III}}_{23,k}(x):=(2y_{k})^{\mu}(x+y_{k})^{-1-\mu-\nu}{}_{2}F_{1}\left(\frac{1}{2}+\mu,\nu+\mu+1;\nu+\frac{3}{2};\frac{x-y_{k}}{x+y_{k}}\right),$
$\displaystyle L^{{\rm
III}}_{24,k}(x):=(2y_{k})^{-\mu}(x+y_{k})^{-1+\mu-\nu}{}_{2}F_{1}\left(\frac{1}{2}-\mu,\nu-\mu+1;\nu+\frac{3}{2};\frac{x-y_{k}}{x+y_{k}}\right).$

Using (2.4), these functions can be reduced to Entries 1 to 8 as follows

$\displaystyle L^{{\rm III}}_{j,k}=L^{{\rm III}}_{j-16,k}\qquad\text{for}\quad
j=17,18,\dots,24.$

##  Appendix B Values on the cut $\bm{[1,\infty)}$ of the Gauss
hypergeometric function

The hypergeometric function has a branch cut along the ray $[1,\infty)$. Along
this branch cut it has limiting values
${}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,b}{c};x\pm{\rm i}0\right)$, $x>1$,
that are usually different when we approach the branch cut from above or
below. Since these limiting values play a role in this paper, we note the
following results.

######  Theorem B.1.

Let $a,b\in\mathbb{C}$, $c\in\mathbb{C}\setminus-\mathbb{N}_{0}$,
$x\in(1,\infty)$. Then

$\displaystyle{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,b}{c};x\pm{\rm
i}0\right)=\frac{\Gamma(c)\Gamma(b-a)}{\Gamma(b)\Gamma(c-a)}{\rm e}^{\pm{\rm
i}\pi
a}x^{-a}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,a-c+1}{a-b+1};\frac{1}{x}\right)$
$\displaystyle\hphantom{{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,b}{c};x\pm{\rm
i}0\right)=}{}+\frac{\Gamma(c)\Gamma(a-b)}{\Gamma(a)\Gamma(c-b)}{\rm
e}^{\pm{\rm i}\pi
b}x^{-b}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{b,b-c+1}{b-a+1};\frac{1}{x}\right)\\!.$

###### Proof.

The connection relation [11, equation (15.8.2)] and (2.1) give

$\displaystyle{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,b}{c};z\right)=\frac{\Gamma(c)\Gamma(b-a)}{\Gamma(b)\Gamma(c-a)}(-z)^{-a}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,a-c+1}{a-b+1};\frac{1}{z}\right)$
$\displaystyle\hphantom{{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,b}{c};z\right)=}{}+\frac{\Gamma(c)\Gamma(a-b)}{\Gamma(a)\Gamma(c-b)}(-z)^{-b}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{b,b-c+1}{b-a+1};\frac{1}{z}\right)\\!.$

This identity holds for all $z\in\mathbb{C}\setminus[0,\infty)$. Setting
$z=x+i\epsilon$, $x>1$, $\epsilon\in\mathbb{R}\setminus\\{0\\}$, and letting
$\epsilon\to 0^{+}$ and $\epsilon\to 0^{-}$, we obtain the stated result. ∎

We note that the limiting values of the generalized hypergeometric function on
the branch cut have recently been investigated by Karp and Prilepkina [7].
Using [11, equations (15.8.3), (15.8.4) and (15.8.5)], one may also obtain
alternative representations of the limiting values of the hypergeometric
function on the branch cut.

######  Theorem B.2.

Let $a,b\in\mathbb{C}$, $c\in\mathbb{C}\setminus-\mathbb{N}_{0}$,
$x\in(1,\infty)$. Then

$\displaystyle{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,b}{c};x\pm{\rm
i}0\right)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}x^{1-c}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a-c+1,b-c+1}{a+b-c+1};1-x\right)$
$\displaystyle\hphantom{{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,b}{c};x\pm{\rm
i}0\right)=}{}+{\rm e}^{\pm{\rm
i}\pi(a+b-c)}\frac{\Gamma(c)\Gamma(a\\!+b\\!-c)}{\Gamma(a)\Gamma(b)}(x\\!-1)^{c-a-b}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{c-a,c-b}{c\\!-a\\!-b\\!+1};1\\!-x\right)\\!.$

######  Theorem B.3.

Let $x\in(1,\infty)$. Then

$\displaystyle{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,b}{c};x\pm{\rm
i}0\right)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}x^{-a}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,a-c+1}{a+b-c+1};1-\frac{1}{x}\right)$
$\displaystyle\hphantom{{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,b}{c};x\pm{\rm
i}0\right)=}{}+{\rm e}^{\pm{\rm
i}\pi(a+b-c)}\frac{\Gamma(c)\Gamma(a+b-c)}{\Gamma(a)\Gamma(b)}(x-1)^{c-a-b}x^{a-c}$
$\displaystyle\hphantom{{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,b}{c};x\pm{\rm
i}0\right)=+}{}\times{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{1-a,c-a}{c-a-b+1};1-\frac{1}{x}\right)\\!.$

######  Theorem B.4.

Let $x\in(1,\infty)$. Then

$\displaystyle{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,b}{c};x\pm{\rm
i}0\right)={\rm e}^{\pm{\rm i}\pi
a}\frac{\Gamma(c)\Gamma(b-a)}{\Gamma(b)\Gamma(c-a)}(x-1)^{-a}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,c-b}{a-b+1};\frac{1}{1-x}\right)$
$\displaystyle\hphantom{{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,b}{c};x\pm{\rm
i}0\right)=}{}+{\rm e}^{\pm{\rm i}\pi
b}\frac{\Gamma(c)\Gamma(a-b)}{\Gamma(a)\Gamma(c-b)}(x-1)^{b}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{b,c-a}{b-a+1};\frac{1}{1-x}\right)\\!.$

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