# An Alternative Approach to Computing $\beta(2k+1)$

Naomi Tanabe  Department of Mathematics, Bowdoin College, Brunswick, ME 04011
<EMAIL_ADDRESS>and  Nawapan Wattanawanichkul  University of Illinois
Urbana-Champaign, Champaign, IL 04011<EMAIL_ADDRESS>
###### Abstract.

This paper presents a new approach to evaluating the special values of the
Dirichlet beta function, $\beta(2k+1)$, where $k$ is any nonnegative integer.
Our approach relies on some properties of the Euler numbers and polynomials,
and uses basic calculus and telescoping series. By a similar procedure, we
also yield an integral representation of $\beta(2k)$. The idea of our proof
adapts from a previous study by Ciaurri et al., where the authors introduced a
new proof of Euler's formula for $\zeta(2k)$.

##  1\. Introduction

It is well known that the value of the Riemann $\zeta$-function at a positive
even integer $2k$ can be expressed as

$\zeta(2k)\ =\ \sum_{n=1}^{\infty}\frac{1}{n^{2k}}\ =\
\cfrac{(-1)^{k-1}2^{2k-1}\pi^{2k}}{(2k)!}B_{2k},$ (1.1)

where $B_{k}$ is the $k$-th Bernoulli number. One of the classical proofs of
this formula is attributed to Euler, which involves considering the expansion
of $\pi z\cot(\pi z)$ in two different ways. However, over time, numerous
other proofs have been developed utilizing a variety of techniques and
approaches, including notable examples such as [1], [3]–[9], [11], and
[13]–[15]. The multitude of proofs reflects the fundamental importance of this
formula and the richness of the mathematical concepts it connects.

On the other hand, the Riemann $\zeta$-function has been generalized in many
ways, including the Dirichlet $L$-functions. In a similar manner to Equation
(1.1), formulas for the special values of the Dirichlet $L$-functions have
been established as follows (for details see, for example, [10, Section 7-2,
Corollary 2.10]).

######  Theorem 1 ([10]).

Let $\chi$ be a primitive character of conductor $N$ and $k$ be a positive
integer satisfying $\chi(-1)=(-1)^{k}$. Then we have

$L(k,\chi)=(-1)^{k-1}\frac{\tau(\chi)}{2}\left(\frac{2\pi
i}{N}\right)^{k}\frac{B_{k,\overline{\chi}}}{k!},$ (1.2)

where $B_{k,\overline{\chi}}$ is the generalized Bernoulli number associated
with the conjugate of the character $\chi$, and $\tau(\chi)$ is the Gauss sum
of the character defined as

$\tau(\chi)\ =\ \sum_{a=1}^{N}\chi(a)e^{\frac{2\pi ia}{N}}.$

These formulas play a critical role in number theory, particularly in the
study of primes in arithmetic progressions, and have many connections with
various mathematical objects such as modular forms, automorphic
representations, and Galois representations.

In this paper, we focus on Equation (1.2) particularly for the $L$-function
associated with the primitive Dirichlet character $\chi_{4}$ modulo $4$, also
known as the Dirichlet $\beta$-function

$\beta(s)=\sum_{m=0}^{\infty}\frac{(-1)^{m}}{(2m+1)^{s}}.$

More precisely, it is our aim to present an alternative approach to computing
$L(2k+1,\chi_{4})=\beta(2k+1)$, whose formula can be stated as in the
following theorem, according to Equation (1.2).

######  Theorem 2.

Let $\chi_{4}$ be a primitive character of conductor $4$ and $k$ be a
nonnegative integer. We have

$\beta(2k+1)=(-1)^{k+1}\left(\frac{\pi}{2}\right)^{2k+1}\frac{B_{2k+1,\chi_{4}}}{(2k+1)!},$
(1.3)

where $B_{2k+1,\chi_{4}}$ is the $2k+1$-th generalized Bernoulli number
associated with $\chi_{4}$.

To prove the above theorem, we use the techniques analogous to those
introduced in [5]. The method utilizes basic calculus and telescoping series
to derive the desired formula. Applying the same techniques, we also obtain an
integral representation for $\beta(2k)$, when $k$ is a positive integer.

######  Theorem 3.

For any positive integer $k$, we have

$\beta(2k)\ =\
\frac{(-1)^{k-1}\pi^{2k}}{2(2k-1)!}\int_{0}^{1/2}E_{2k-1}(t)\sec({\pi t})dt.$
(1.4)

In the next section, we recall definitions and important properties of Euler
polynomials $E_{k}(x)$, which are crucial to our work. Then Section 3 is
devoted to our proof of Theorem 2, a formula for $\beta(2k+1)$. Lastly, we
derive Theorem 3, an integral representation for $\beta(2k)$, in Section 4.

##  2\. Euler Numbers and Polynomials

In this section, we introduce some useful properties of the Euler numbers and
polynomials, which will be used repeatedly in Sections 3 and 4.

######  Definition 1.

The $k$-th Euler number $E_{k}$ is defined by the generating function

$\sum_{k=0}^{\infty}E_{k}\frac{t^{k}}{k!}\ =\ \frac{2e^{t}}{e^{2t}+1}.$ (2.1)

By expanding the right-hand side of the equation above, the first few Euler
numbers can be observed as

$E_{0}=1,\ E_{1}=0,\ E_{2}=-1,\ E_{3}=0,\ E_{4}=5,\ \dots.$

######  Definition 2.

The $k$-th Euler polynomial $E_{k}(x)$ is defined by the generating function

$\sum_{k=0}^{\infty}E_{k}(x)\frac{t^{k}}{k!}\ =\
\frac{2e^{xt}}{e^{t}+1},\text{ where }|t|\leq\pi,x\in\mathbb{R}.$ (2.2)

Again, by expanding the right-hand side of the above equation, we see that the
first few Euler polynomials are

$E_{0}(x)=1,\ E_{1}(x)=x-\frac{1}{2},\ E_{2}(x)=x^{2}-x,\
E_{3}=x^{3}-\frac{3}{2}x^{2}+\frac{1}{4},\ \dots.$

Using Definitions 1 and 2, we observe the following noteworthy proposition.

######  Proposition 1.

For the Euler polynomials and $k\in\mathbb{Z}_{\geq 0}$, the followings are
true:

  1. (1.1)

$E_{k}(1-x)=(-1)^{k}E_{k}(x)$ and, in particular, $\displaystyle
E_{2k+1}\left(1/2\right)=0,$

  2. (1.2)

$\displaystyle E_{k}=2^{k}E_{k}\left(1/2\right),$

  3. (1.3)

$E_{k}(x+1)+E_{k}(x)=2x^{k},$

  4. (1.4)

$E_{2k}(1)=E_{2k}(0)=0,$

  5. (1.5)

$E^{\prime}_{0}(x)=0$ and $E^{\prime}_{k}(x)=kE_{k-1}(x)$ when $k\geq 1,$

  6. (1.6)

$\displaystyle
E_{2k}\left(1/2\right)=-\frac{B_{2k+1,\chi_{4}}}{(2k+1)2^{2k-1}}$, where
$B_{2k+1,\chi_{4}}$ is the generalized Bernoulli number associated with the
primitive character modulo 4.

###### Proof.

To prove the first statement, we substitute $x$ with $1-x$ in Equation (2.2)
and get

$\quad\sum_{k=0}^{\infty}E_{k}(1-x)\frac{t^{k}}{k!}\ =\
\frac{2e^{(1-x)t}}{e^{t}+1}\ =\ \frac{2e^{-xt}}{e^{-t}+1}\ =\
\sum_{k=0}^{\infty}(-1)^{k}E_{k}(x)\frac{t^{k}}{k!}.$

Comparing the coefficients of $t^{k}$ term on both sides of the equation gives
us the desired result. The second part in (1.1) then follows by evaluating the
equation at $x=1/2$ when $k$ is an odd integer. The second and third
statements are obtained similarly, by evaluating Equation (2.2) at $x=1/2$ and
at $x-1,$ respectively.

The statement (1.4) follows from substituting $x=0$ to equations in (1.1) and
(1.3), which yields $E_{2k}(1)=E_{2k}(0)$ and $E_{2k}(1)+E_{2k}(0)=0$,
respectively.

To verify (1.5), we differentiate Equation (2.2) with respect to $x$;

$\hskip 21.68121pt\sum_{k=0}^{\infty}E^{\prime}_{k}(x)\frac{t^{k}}{k!}\ =\
\frac{d}{dx}\frac{2e^{xt}}{e^{t}+1}\ =\ t\cdot\frac{2e^{xt}}{e^{t}+1}.$

The right-hand side of the equation is then the product of $t$ and the
generating function of the the Euler polynomials. Therefore, we see that

$\qquad\sum_{k=0}^{\infty}E^{\prime}_{k}(x)\frac{t^{k}}{k!}\ =\
\sum_{k=0}^{\infty}E_{k}(x)\frac{t^{k+1}}{k!}\ =\
\sum_{k=1}^{\infty}E_{k-1}(x)\frac{t^{k}}{(k-1)!},$ (2.3)

which means $E^{\prime}_{k}(x)=kE^{\prime}_{k-1}(x)$ when $k\geq 1$. Since the
constant term of the right-hand side of Equation (2.3) is 0, we conclude that
$E^{\prime}_{0}(x)=0$.

Lastly, for (1.6), we recall the relation between the $k$-th generalized
Bernoulli number $B_{k,\chi}$ associated with $\chi$ and the $k$-th Bernoulli
polynomial $B_{k}(x)$ given by

$\displaystyle B_{k,\chi}\ =\
N^{k-1}\sum_{a=1}^{N}\chi(a)B_{k}\left(\frac{a}{N}\right).$

Here, $N$ is the conductor of the character $\chi$. See, for example, [2,
Section 4.3]. In particular, when $\chi=\chi_{4}$,

$B_{k,\chi_{4}}\ =\
4^{k-1}\left(B_{k}\left(\frac{1}{4}\right)-B_{k}\left(\frac{3}{4}\right)\right).$
(2.4)

Likewise, the Euler polynomials can be related to the Bernoulli polynomials as

$E_{k-1}(x)\ =\
\frac{2^{k}}{k}\left(B_{k}\left(\frac{x+1}{2}\right)-B_{k}\left(\frac{x}{2}\right)\right),$

(see [12] for details). In particular, when $x=1/2$, we have

$E_{k-1}\left(\frac{1}{2}\right)\ =\
\frac{2^{k}}{k}\left(B_{k}\left(\frac{3}{4}\right)-B_{k}\left(\frac{1}{4}\right)\right).$
(2.5)

Comparing Equations (2.4) and (2.5) gives us

$E_{k-1}\left(\frac{1}{2}\right)\ =\
\frac{2^{k}}{k}\cdot\frac{-B_{k,\chi_{4}}}{4^{k-1}}\ =\
-\frac{B_{k,\chi_{4}}}{2^{k-2}k}.$

A proof is completed by replacing $k$ with $2k+1$. ∎

##  3\. Computing $\beta(2k+1)$

In this section, we prove the formula for $\beta(2k+1)$ as stated in Theorem
2.

###### Proof of Theorem 2.

We will use the following auxiliary integral

$I(k,m)\ =\ \int_{0}^{1/2}E_{2k}(t)\sin((2m+1)\pi t)\ dt,$ (3.1)

for integers $k,m\geq 0$. For clarity, we split the proof into three main
steps.

1) Summing auxiliary functions. First, we find the recurrence relation among
the auxiliary functions $I(k,m)$ and derive the closed form solution. We begin
with the simplest case when $k=0$. Using the fact that $E_{0}(t)=1$ for any
real $t$, we have that

$I(0,m)\ =\ \int_{0}^{1/2}\sin((2m+1)\pi t)\,dt\ =\ \frac{1}{(2m+1)\pi}.$
(3.2)

For $k\geq 1$, we integrate Equation (3.1) by parts and obtain

$\begin{split}I(k,m)&=-\left[E_{2k}(t)\frac{\cos((2m+1)\pi
t)}{(2m+1)\pi}\right]_{t=0}^{t=1/2}+\int_{0}^{1/2}E^{\prime}_{2k}(t)\frac{\cos((2m+1)\pi
t)}{(2m+1)\pi}\,dt\\\
&=\frac{1}{(2m+1)\pi}\int_{0}^{1/2}E^{\prime}_{2k}(t)\cos((2m+1)\pi
t)\,dt,\end{split}$

where the last equality follows from (1.4).

Now, applying (1.5) and integrating by parts again, we get

$\begin{split}I(k,m)&=-\frac{2k(2k-1)}{(2m+1)^{2}\pi^{2}}\int_{0}^{1/2}E_{2k-2}(t)\sin((2m+1)\pi
t)\,dt\\\ &=\frac{-2k(2k-1)}{(2m+1)^{2}\pi^{2}}I(k-1,m).\end{split}$

Applying this recurrence relation repeatedly, together with the value of
$I(0,m)$ from Equation $\eqref{eqn:I(0,m)}$, we obtain the closed form of our
auxiliary functions

$\begin{split}I(k,m)&=\frac{-2k(2k-1)}{(2m+1)^{2}\pi^{2}}\cdot\frac{-(2k-2)(2k-3)}{(2m+1)^{2}\pi^{2}}\dots\frac{-2\cdot
1}{(2m+1)^{2}\pi^{2}}\cdot\frac{1}{(2m+1)\pi}\\\
&=\frac{(-1)^{k}(2k)!}{(2m+1)^{2k+1}\pi^{2k+1}},\end{split}$

for any nonnegative integers $k$ and $m$. Multiplying each $I(k,m)$ by
$(-1)^{m}$ and summing up over nonnegative integers $m$ relate $I(k,m)$ to
$\beta(2k+1)$ as

$\sum_{m=0}^{\infty}(-1)^{m}I(k,m)\ =\
\sum_{m=0}^{\infty}\frac{(-1)^{m}(-1)^{k}(2k)!}{(2m+1)^{2k+1}\pi^{2k+1}}\ =\
\frac{(-1)^{k}(2k)!}{\pi^{2k+1}}\beta(2k+1).$ (3.3)

2) Modifying auxiliary functions. We now modify our auxiliary functions
$I(k,m)$ as follows:

$I^{*}(k,m)\ :=\ \int_{0}^{1/2}E^{*}_{2k}(t)\sin((2m+1)\pi t)\,dt,$

where

$E^{*}_{2k}(t)\ :=\ E_{2k}(t)-\frac{E_{2k}}{2^{2k}}\sin(\pi t).$

This can also be written as

$\begin{split}I^{*}(k,m)&=\int_{0}^{1/2}\left(E_{2k}(t)-\frac{E_{2k}}{2^{2k}}\sin(\pi
t)\right)\sin((2m+1)\pi t)\,dt\\\
&=I(k,m)-\int_{0}^{1/2}\frac{E_{2k}}{2^{2k}}\sin(\pi t)\sin((2m+1)\pi
t)\,dt,\end{split}$

and therefore,

$I(k,m)\ =\ I^{*}(k,m)+\frac{E_{2k}}{2^{2k}}\int_{0}^{1/2}\sin(\pi
t)\sin((2m+1)\pi t)\,dt.$ (3.4)

Furthermore, applying the following trigonometric identity

$\sin(\alpha)\sin(\beta)\ =\ \frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{2}$

to the integrand in Equation (3.4) yields that

$\begin{split}\frac{E_{2k}}{2^{2k}}\int_{0}^{1/2}\sin(\pi t)\sin((2m+1)\pi
t)\,dt&=\frac{E_{2k}}{2^{2k}}\int_{0}^{1/2}\frac{\cos(2m\pi t)-\cos((2m+2)\pi
t)}{2}\,dt\\\ &=\begin{cases}\displaystyle\frac{E_{2k}}{2^{2k+2}}\hskip
17.34518pt\text{if }m=0,\\\ 0\hskip
38.30301pt\text{otherwise.}\end{cases}\end{split}$

We also note that (1.2) and (1.6) give

$\frac{E_{2k}}{2^{2k+2}}\ =\ -\cfrac{B_{2k+1,\chi_{4}}}{(2k+1)2^{2k+1}}.$

Hence, Equation (3.4) can be written as

$I(k,m)\ =\
\begin{cases}I^{*}(k,m)-\cfrac{B_{2k+1,\chi_{4}}}{(2k+1)2^{2k+1}},&\text{if
}m=0,\\\ I^{*}(k,m),&\text{if }m\geq 1.\end{cases}$

Thus,

$\displaystyle\sum_{m=0}^{\infty}(-1)^{m}I(k,m)$
$\displaystyle=\left(I^{*}(k,0)-\cfrac{B_{2k+1,\chi_{4}}}{(2k+1)2^{2k+1}}\right)+\sum_{m=1}^{\infty}(-1)^{m}I^{*}(k,m)$
$\displaystyle=\sum_{m=0}^{\infty}(-1)^{m}I^{*}(k,m)-\cfrac{B_{2k+1,\chi}}{(2k+1)2^{2k+1}}.$
(3.5)

Comparing this with Equation (3.3), it boils down to simplify Equation (3.5)
to obtain the desired result. Indeed, we will show that
$\sum_{m=0}^{\infty}(-1)^{m}I^{*}(k,m)=0$ in the following subsection.

3) Computing telescoping series. We now show that the infinite series
$\sum_{m=0}^{\infty}(-1)^{m}I^{*}(k,m),$ which is defined as
$\lim_{N\to\infty}\sum_{m=0}^{N}(-1)^{m}I^{*}(k,m)$, converges to 0 by using
trigonometric identities and telescoping sums. Consider

$\begin{split}&\lim_{N\to\infty}\sum_{m=0}^{N}(-1)^{m}I^{*}(k,m)\\\
&=\lim_{N\to\infty}\left(I^{*}(k,0)-I^{*}(k,1)+\cdots+(-1)^{N-1}I^{*}(k,N-1)+(-1)^{N}I^{*}(k,N)\right)\\\
&=\lim_{N\to\infty}\int_{0}^{1/2}\Big{(}E^{*}_{2k}(t)\sin(\pi
t)-E^{*}_{2k}(t)\sin(3\pi t)+\cdots\\\ &\hskip
28.90755pt+(-1)^{N-1}E^{*}_{2k}(t)\sin((2N-1)\pi
t)+(-1)^{N}E^{*}_{2k}(t)\sin((2N+1)\pi t)\Big{)}\,dt.\end{split}$

Applying the following trigonometric identity

$\sin((2m+1)x)\ =\ \cfrac{\cos((2m-1)x)-\cos((2m+3)x)}{2\sin(2x)},$

we obtain the telescoping series

$\displaystyle\displaystyle\lim_{N\to\infty}\int_{0}^{1/2}\Big{(}E^{*}_{2k}(t)\cdot\cfrac{\cos(-\pi
t)-\cos(3\pi t)}{2\sin(2\pi t)}-E^{*}_{2k}(t)\cdot\cfrac{\cos(\pi t)-\cos(5\pi
t)}{2\sin(2\pi t)}$ $\displaystyle\hskip
36.135pt+E^{*}_{2k}(t)\cdot\cfrac{\cos(3\pi t)-\cos(7\pi t)}{2\sin(2\pi
t)}-E^{*}_{2k}(t)\cdot\cfrac{\cos(5\pi t)-\cos(9\pi t)}{2\sin(2\pi t)}+\cdots$
$\displaystyle\hskip 36.135pt+(-1)^{N}E^{*}_{2k}(t)\cdot\cfrac{\cos((2N-1)\pi
t)-\cos((2N+3)\pi t)}{2\sin(2\pi t)}\Big{)}\,dt.$ (3.6)

To cancel out repetitive terms in Equation (3), we need to extend the function

$f(t)\ =\ \cfrac{E^{*}_{2k}(t)}{\sin(2\pi t)},\qquad\text{ for }t\in(0,1/2),$

to $t=0$ and $1/2$.

When $t=0$, we note that
$E^{*}_{2k}(0)=E_{2k}(0)-\frac{E_{2k}}{2^{2k}}\cdot\sin(0)=0$ by (1.4). We
then evaluate the limit of $f(t)$ when $t$ approaches $0$ using L'Hôpital's
rule as follows

$\lim_{t\to 0}\frac{E_{2k}(t)-\cfrac{E_{2k}}{2^{2k}}\sin(\pi t)}{\sin(2\pi
t)}\ =\ \frac{2kE_{2k-1}(0)-\cfrac{E_{2k}}{2^{2k}}\pi}{2\pi},$

which is some constant.

As for $t=1/2$, notice that
$E^{*}_{2k}(1/2)=E_{2k}(1/2)-\frac{E_{2k}}{2^{2k}}\sin(\pi/2)=0$ by (1.2).
Then the limit of $f(t)$ as $t$ approaches $1/2$ can be evaluated as

$\lim_{t\to 1/2}\frac{E_{2k}(t)-\cfrac{E_{2k}}{2^{2k}}\sin(\pi t)}{\sin(2\pi
t)}\ =\ \frac{2k\cdot
E_{2k-1}(1/2)-\cfrac{E_{2k}}{2^{2k}}\pi\cos(\pi/2)}{2\pi\cos(\pi)},$

which equals $0$ by using (1.1). Thus, $f(t)$ is well-defined on $[0,1/2]$,
and, hence, most of the terms in Equation (3) get cancelled. Moreover, since
the first two terms $f(t)\cos(-\pi t)$ and $f(t)\cos(\pi t)$ are equal, we are
left with

$\displaystyle\sum_{m=0}^{\infty}(-1)^{m}I^{*}(k,m)$
$\displaystyle=\displaystyle\lim_{N\to\infty}(-1)^{N-1}\int_{0}^{1/2}\frac{E^{*}_{2k}(t)}{2\sin(2\pi
t)}\left(\cos((2N+1)\pi t)-\cos((2N+3)\pi t)\right)\,dt$
$\displaystyle=\displaystyle\lim_{N\to\infty}(-1)^{N-1}\int_{0}^{1/2}\frac{E^{*}_{2k}(t)}{2\sin(2\pi
t)}\left(-2\sin((2N+2)\pi t)\sin(-\pi t)\right)\,dt$
$\displaystyle=\displaystyle\lim_{N\to\infty}(-1)^{N-1}\int_{0}^{1/2}\frac{E^{*}_{2k}(t)}{2\cos(\pi
t)}\left(\sin((2N+2)\pi t)\right)\,dt.$ (3.7)

To proceed further, we justify that the function
$\frac{E^{*}_{2k}(t)}{2\cos(\pi t)}\sin((2N+2)\pi t)$ is differentiable on
$[0,1/2]$ with continuous derivative. Similar to the case of $f(t)$, we extend
the function

$g(t)\ =\ \cfrac{E^{*}_{2k}(t)}{\cos(\pi t)},\hskip 14.45377pt\text{ for
}t\in[0,1/2),$

to $t=1/2$, which can be achieved by applying (1.1) and (1.5):

$\lim_{t\to 1/2}\frac{E_{2k}(t)-\cfrac{E_{2k}}{2^{2k}}\sin(\pi t)}{2\cos(\pi
t)}\ =\ 0.$

Therefore, $g(t)$ is differentiable with continuous derivative on $[0,1/2]$.

We now consider the integral on the right-hand side of the last equation of
(3). Writing $(2N+2)\pi=R$ and integrating by parts give

$\int_{0}^{1/2}g(t)\sin(Rt)\,dt\ =\
-\cfrac{\cos(R/2)}{R}g(1/2)+\frac{1}{R}g(0)+\int_{0}^{1/2}g^{\prime}(t)\cfrac{\cos(Rt)}{R}\,dt.\\\
$

The boundedness of $g(0)$, $g(1/2)$ and $g^{\prime}(t)$ shows that each term
in the above sum approaches zero as $R$ approaches infinity, and therefore

$\lim_{N\to\infty}\sum_{m=0}^{N}(-1)^{m}I^{*}(k,m)=0.$

Thus, Equation (3.5) is simplified as

$\sum_{m=0}^{\infty}(-1)^{m}I(k,m)\ =\
-\cfrac{B_{2k+1,\chi}}{(2k+1)2^{2k+1}}.$

This, together with Equation (3.3), completes the proof. ∎

##  4\. An Integral Representation of $\beta(2k)$

This section is devoted to obtaining the integral representation of
$\beta(2k)$ as stated in Theorem 3. In this case, we split the proof into two
steps.

###### Proof of Theorem 3.

We consider a slightly different auxiliary integrals

$J(k,m)\ =\ \int_{0}^{1/2}E_{2k+1}(t)\cos((2m+1)\pi t)\,dt$ (4.1)

with integers $k,m\geq 0$.

1) Summing auxiliary functions. Similar to the case of $I(0,m)$, applying
(1.5) and integrating by parts yield

$J(0,m)\ =\
\frac{E_{1}\left(\frac{1}{2}\right)\sin\left(\frac{(2m+1)}{2}\pi\right)-E_{1}(0)\sin(0)}{(2m+1)\pi}-\int_{0}^{1/2}E_{0}(t)\frac{\sin((2m+1)\pi
t)}{(2m+1)\pi}\,dt.$

Then, using (1.1), together with the facts that $E_{0}(t)=1$ and $\sin(0)=0$,
we are left with

$J(0,m)\ =\ -\int_{0}^{1/2}\frac{\sin((2m+1)\pi t)}{(2m+1)\pi}\,dt\ =\
-\frac{1}{(2m+1)^{2}\pi^{2}}.$ (4.2)

Now we consider Equation (4.1) when $k\geq 1$. Integrating by parts twice,
along with (1.1), (1.4), and (1.5), gives us

$J(k,m)=-\frac{(2k+1)(2k)}{(2m+1)^{2}\pi^{2}}J(k-1,m).$ (4.3)

Putting Equations (4.2) and (4.3) together provides the closed form of
$J(k,m)$ as

$J(k,m)\ =\ \frac{(-1)^{k+1}(2k+1)!}{(2m+1)^{2k+2}\pi^{2k+2}}.$

Therefore, we can relate $J(k,m)$ to $\beta(2k)$ as

$\sum_{m=0}^{\infty}(-1)^{m}J(k-1,m)\ =\
\sum_{m=0}^{\infty}(-1)^{m}\frac{(-1)^{k}(2k-1)!}{(2m+1)^{2k}\pi^{2k}}\ =\
\frac{(-1)^{k}(2k-1)!}{\pi^{2k}}\beta(2k),$

for any $k\geq 1$, or equivalently,

$\beta(2k)\ =\
\frac{(-1)^{k}\pi^{2k}}{(2k-1)!}\sum_{m=0}^{\infty}(-1)^{m}J(k-1,m).$ (4.4)

2) Computing telescoping series. We now simplify the right-hand side of
Equation (4.4) by exploiting a telescoping series and the trigonometric
identity

$\cos((2m+1)\pi t)\ =\ \frac{\cos(2m\pi t)+\cos((2m+2)\pi t)}{2\cos(\pi t)}.$
(4.5)

Applying Equation (4.5) in Equation (4.1), we obtain that

$J(k-1,m)\ =\ \int_{0}^{1/2}E_{2k-1}(t)\frac{\cos(2m\pi t)+\cos((2m+2)\pi
t)}{2\cos(\pi t)}\ dt.$

Multiplying each $J(k-1,m)$ by $(-1)^{m}$ and summing up over nonnegative
integers $m$ give

$\begin{split}&\sum_{m=0}^{\infty}(-1)^{m}J(k-1,m)\\\ &\hskip
14.45377pt=\lim_{N\to\infty}\int_{0}^{1/2}\left(E_{2k-1}(t)\frac{\cos(0)+\cos(2\pi
t)}{2\cos(\pi t)}-\cdots\right.\\\ &\hskip
108.405pt\left.+(-1)^{N}E_{2k-1}(t)\frac{\cos(2N\pi t)+\cos((2N+2)\pi
t)}{2\cos(\pi t)}\right)\ dt.\end{split}$

Then we extend the function

$h(t)\ :=\ \frac{E_{2k-1}(t)}{2\cos(\pi t)},\hskip 21.68121ptt\in[0,1/2),$

in each summand, to $t=1/2$ by continuity. By applying L'Hôpital's rule, we
obtain

$\lim_{t\to 1/2}\frac{E_{2k-1}(t)}{2\cos(\pi t)}\ =\
\frac{(2k-1)E_{2k-2}(1/2)}{-2\pi},$

which is some constant. Thus, $h(t)$ is differentiable on $[0,1/2]$ with a
continuous derivative, and, hence, most of the terms in the above integral now
get cancelled, leaving us

$\displaystyle\sum_{m=0}^{\infty}(-1)^{m}J(k-1,m)$
$\displaystyle=\lim_{N\to\infty}\int_{0}^{1/2}\left(E_{2k-1}(t)\frac{\cos(0)}{2\cos(\pi
t)}+(-1)^{N}E_{2k-1}(t)\frac{\cos((2N+2)\pi t)}{2\cos(\pi t)}\right)\,dt$
$\displaystyle=\int_{0}^{1/2}\frac{E_{2k-1}(t)\sec(\pi
t)}{2}\,dt+\lim_{N\to\infty}(-1)^{N}\int_{0}^{1/2}E_{2k-1}(t)\frac{\cos((2N+2)\pi
t)}{2\cos(\pi t)}\,dt.$ (4.6)

To further simplify the above expression, we will show that, in fact,

$\lim_{N\to\infty}(-1)^{N}\int_{0}^{1/2}E_{2k-1}(t)\frac{\cos((2N+2)\pi
t)}{2\cos(\pi t)}dt$

is null. Let $R$ denote $(2N+2)\pi$. The integral above then equals

$\begin{split}&\lim_{N\to\infty}(-1)^{N}\int_{0}^{1/2}h(t)\cos(Rt)dt\\\
&=\lim_{N\to\infty}(-1)^{N}\left(h(1/2)\frac{\sin(R/2)}{R}-h(0)\frac{\sin(0)}{R}-\int_{0}^{1/2}h^{\prime}(t)\frac{\sin(Rt)}{R}dt\right).\end{split}$

Since $h(1/2),h(0)$, and $h^{\prime}(t)$ are bounded, each summand approaches
$0$ as $R\to\infty$, and therefore this limit is indeed 0. Thus, the summation
(4) is

$\sum_{m=0}^{\infty}(-1)^{m}J(k-1,m)\ =\
\int_{0}^{1/2}\frac{E_{2k-1}(t)\sec(\pi t)}{2}dt.$

Substituting this back into Equation (4.4) yields

$\beta(2k)\ =\
\frac{(-1)^{k-1}\pi^{2k}}{2(2k-1)!}\int_{0}^{1/2}E_{2k-1}(t)\sec({\pi t})dt,$

for all positive integers $k$, as desired. ∎

Acknowledgement. We are grateful for the support from the Kibbe Science
Fellowship from Bowdoin College.

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#  Channel Reuse for Backhaul in UAV Mobile Networks with User QoS Guarantee

Mohammadsaleh Nikooroo1, Zdenek Becvar1, Omid Esrafilian2, David Gesbert2 This
work was supported by the project No. LTT 20004 funded by Ministry of
Education, Youth and Sports, Czech Republic, and by the grant of Czech
Technical University in Prague No. SGS20/169/OHK3/3T/13, and partially by the
HUAWEI France supported Chair on Future Wireless Networks at EURECOM. 1
Faculty of Electrical EngineeringCzech Technical University in Prague, Prague,
Czech Republic
2Communication Systems Department, EURECOM, Sophia Antipolis, France
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

In mobile networks, unmanned aerial vehicles (UAVs) acting as flying base
stations (FlyBSs) can effectively improve performance. Nevertheless, such
potential improvement requires an efficient positioning of the FlyBS. In this
paper, we study the problem of sum downlink capacity maximization in FlyBS-
assisted networks with mobile users and with a consideration of wireless
backhaul with channel reuse while a minimum required capacity to every user is
guaranteed. The problem is formulated under constraints on the FlyBS’s flying
speed, propulsion power consumption, and transmission power for both of flying
and ground base stations. None of the existing solutions maximizing the sum
capacity can be applied due to the combination of these practical constraints.
This paper pioneers in an inclusion of all these constraints together with
backhaul to derive the optimal 3D positions of the FlyBS and to optimize the
transmission power allocation for the channels at both backhaul and access
links as the users move over time. The proposed solution is geometrical based,
and it shows via simulations a significant increase in the sum capacity (up by
19%-47%) compared with baseline schemes where one or more of the aspects of
backhaul communication, transmission power allocation, and FlyBS’s positioning
are not taken into account.

###### Index Terms:

Flying base station, UAV, Backhaul, Relaying, Transmission power, Sum
capacity, Mobile networks, 6G.

##  I Introduction

Unmanned aerial vehicles (UAVs) have received an extensive attention in
wireless communications in the recent years. Due to a high flexibility and
adaptability to the environment, UAVs can be regarded as flying base stations
(FlyBSs) that potentially bring a significant enhancement in the performance
of mobile networks [1]. Such potential enhancements, however, are essentially
subject to an effective management of several aspects, including propulsion
power consumption, transmission power consumption/allocation, FlyBS’s
positioning , etc. As another crucial aspect, a backhaul communication of the
FlyBSs with the ground base station (GBS) or access point (AP) must be ensured
in order to integrate FlyBSs into mobile networks.

Many recent work investigate the performance in FlyBS-assisted networks with
inclusion of backhaul. In [2], they address the FlyBS’s positioning and
bandwidth allocation to optimize the total profit gained from the users in a
network. Furthermore, the authors in [3] investigate an optimization of the
FlyBS’s position, user association, and resource allocation, to maximize the
utility in software-defined cellular networks. In [4], they maximize energy
efficiency in a relaying network with static BSs via optimization of
transmission power allocation to the BSs. Then, the problem of joint 2D
trajectory design and resource allocation is investigated in [5] to minimize
the network latency in a space-air-ground network with millimeter wave
(mmWave) backhaul. In [6], the authors study a joint placement, resource
allocation, and user association of FlyBSs to maximize the network’s utility.
Then, in [7], they maximize the minimum rate of the delay-tolerant users via a
joint resource allocation and FlyBS’s positioning. Then, in [8], the authors
consider a scenario where a set of relaying FlyBSs establish a communication
between multiple sources and multiple destinations. The goal is to maximize
the minimum average rate among the relays via transmission power allocation
and FlyBSs’ positioning. Furthermore, in [9], the operation cost in a mobile
edge computing network is minimized via FlyBS’s positioning and resource
allocation. The solutions provided in [2]-[9] do not assume constraints on the
user’s instantaneous capacity and, hence, they cannot be applied in scenarios
with delay-sensitive users where a minimum capacity is demanded by the users.

Several works also consider the individual user’s quality of service in terms
of instantaneous capacity. In [10] and [11], the problem of FlyBS positioning
and resource allocation is investigated to minimize the transmission power of
the FlyBS. Then, the minimum capacity of the users is maximized via the
FlyBS’s positioning and the transmission power allocation in [12].
Furthermore, the problem of transmission power allocation is investigated in
[13] for FlyBS networks to maximize the energy efficiency, i.e., the ratio of
the sum capacity to the total transmission power consumption. The authors in
[14] minimize the number of FlyBSs in a network while ensuring both coverage
to all ground users. Then, in [15], the problem of resource allocation and
circular-trajectory design for fixed-wing FlyBSs is investigated to minimize
the power consumption of the FlyBS. Furthermore, the minimum capacity of the
users is maximized in [16] via resource allocation and positioning. Also, the
minimum downlink throughput is maximized in [17] by optimizing the FlyBSs’
positioning, bandwidth, and power allocation.

In our prior work [18], the problem of FlyBS’s positioning and user
association is investigated in mobile networks assisted by relaying FlyBSs.
Also, in [19] and [20], a positioning of the FlyBS and transmission power
allocation is proposed at the access link to maximize the sum capacity and to
minimize the FlyBS’s total power consumption, respectively, where a minimum
required capacity for the users is guaranteed.

To our best knowledge, there is no work targeting the sum capacity
maximization in a practical scenario with moving users and with the minimum
capacity guaranteed to the individual users where a backhaul link is also
provided. All related works either target scenario where no minimum capacity
is guaranteed to the users and/or a backhaul connection (together with related
backhaul constraints [10], [21]) is missing. It is also noted that, existing
solutions maximizing the minimum capacity among users cannot be applied in
many scenarios where users require different instantaneous capacities. To this
end, we target the case with both backhaul and user’s required capacity and we
propose an analytical solution based on an alternating optimization of the
FlyBS’s positioning and the transmission power allocation at the backhaul and
at the access links. Due to a non-convex nature of the problem, a heuristic
solution is proposed with respect to the feasibility region that is determined
via constraints in the problem, i.e., 1) user’s required capacity at all time,
2) FlyBS’s maximum speed, 3) maximum propulsion power consumption of FlyBS,
and 4) flow conservation constraint regarding the backhaul and access links.

##  II System model and problem formulation

In this section, we first define the system model. Then, we formulate the
constrained problem of sum capacity maximization with inclusion of backhaul
communication.

In our system model, the FlyBS serves $N$ mobile users in an area as shown in
Fig. 1. The FlyBS connects to the GBS located at
$\bm{l_{G}}=[X_{G},Y_{G},H_{G}]$ via backhaul. Let ,
$\bm{l_{F}}[k]=\big{[}X[k],Y[k],H[k]\big{]}^{T}$ and
$\bm{u_{n}}[k]=\big{[}x_{n}[k],\ y_{n}[k],z_{n}[k]\big{]}^{T}$ denote the
location of the FlyBS and the user $n$ at the time step $k$, respectively.
Also, let $d_{n,F}[k]$ and $d_{n,G}[k]$ denote the Euclidean distance of the
user $u_{n}$ to the FlyBS and to the GBS at the time step $k$, respectively.

Suppose the whole available radio band is divided into a set of $S$ channels
$\textbf{J}=\\{J_{1},…,J_{S}\\}$, where channel $J_{s}$ has a bandwidth of
$B_{s}$ $(1\leq s\leq S)$. At the FlyBSs, we adopt orthogonal downlink channel
allocation to all users. Furthermore, all the $S$ channels are reused at the
backhaul link to alleviate the scarcity of radio resources. Let
$g_{n}\in[1,S]$ be the index of the channel allocated to user $n$. Note that,
we do not target an optimization of channel allocation in this paper, and we
leave that for future work. Nevertheless, our model works with any channel
allocation.

Let $p_{n,F}^{R}$ be the received power at the user $n$ from the FlyBS.
Furthermore, $p_{F,G,s}^{R}$ denotes the received power at the FlyBS from the
GBS over the channel $s$. Then, the channel capacity of the user $n$ is:

$\displaystyle
C_{n}[k]=B_{g_{n}}\log_{2}{\left(1+\frac{p_{n,F}^{R}[k]}{\sigma^{2}_{n}+p_{n,G}^{R}[k]}\right)},$
(1)

where $p_{n,G}^{R}$ is the interference power received at user $n$ from the
GBS, $\sigma^{2}_{n}$ is noise’s power. Similarly, the link’s capacity between
the GBS and the FlyBS is:

$\displaystyle
C_{G,F}[k]=\sum_{s=1}^{S}B_{s}\log_{2}{\left(1+\frac{p_{F,G,s}^{R}[k]}{\sigma^{2}_{F,s}}\right)},$
(2)

where $\sigma^{2}_{F,s}$ is the noise power over the channel $s$.

Let $\bm{p_{F}}^{T}=[p_{F,1}^{T},...,p_{F,N}^{T}]$ denote the FlyBS’s
transmission power vector to all the users. Also, for the GBS-to-FlyBS
communication, let $\bm{p_{G}}^{T}=[p_{G,1}^{T},...,p_{G,S}^{T}]$ be the GBS’s
transmission power vector over the $S$ channels. According to the Friis’
transmission equation, we have

$\displaystyle
p_{n,F}^{R}=Q_{n,F}p_{F,n}^{T}{d_{n,F}}^{-\alpha_{n,F}},n\in[1,N],$ (3)

where the coefficient $Q_{n,F}$ is the parameter depending on the
communication frequency and the gain of antennas, and $\alpha_{n,F}$ is the
pathloss exponent of the channel between the FlyBS and the user $n$. Similar
relation can be derived between the GBS’s transmission power and the received
power at the user $n$ and at the FlyBS as

$\displaystyle
p_{n,G}^{R}=Q_{n,G}p_{G,g_{n}}^{T}{d_{n,G}}^{-\alpha_{n,G}},n\in[1,N],$ (4)
$\displaystyle
p_{F,G}^{R}=Q_{F,G}p_{G,s}^{T}{d_{F,G}}^{-\alpha_{F,G}},n\in[1,N],s\in[1,S].$

For the propulsion power consumption, we refer to the model provided in [22]
for rotary-wing UAVs, where the propulsion power is expressed as:

$\displaystyle
P_{pr}[k]=L_{0}\big{(}1+\frac{3V_{F}^{2}[k]}{U_{\text{tip}}^{2}}\big{)}+\frac{\eta_{0}\rho
s_{r}AV_{F}^{3}[k]}{2}+$ $\displaystyle
L_{i}\big{(}\sqrt{1+\frac{V_{F}^{4}[k]}{4v^{4}_{0,h}}}-\frac{V_{F}^{2}[k]}{2v^{2}_{0,h}}\big{)}^{\frac{1}{2}},$
(5)

where $V_{F}[k]$ is the FlyBS’s speed at the time step $k$, $L_{0}$ and
$L_{i}$ are the blade profile and induced powers in hovering status,
respectively, $U_{\text{tip}}$ is the tip speed of the rotor blade, $v_{0,h}$
is the mean rotor induced velocity during hovering, ${\eta_{0}}$ is the
fuselage drag ratio, $\rho$ is the air density, $s_{r}$ is the rotor solidity,
and $A$ is the rotor disc area.

Figure 1: System model with mobile users placed within the coverage area of
the FlyBS. The channels at the access link are reused for GBS-to-FlyBS
communication

Our goal is to find the optimized position of the FlyBS and to determine the
transmission power allocation over each channel both at the backhaul and at
the access link to maximize the sum capacity at every time step $k$ under
practical constraints as follows:

$\displaystyle\operatorname*{max}_{\bm{p_{G}}^{T}[k],\bm{p_{F}}^{T}[k],\bm{l_{F}}[k]}{\;}\sum_{n=1}^{N}C_{n}[k],\quad\quad\quad\quad\quad\forall
k,$ (6) $\displaystyle\text{s.t.}\quad C_{n}[k]\geq
C_{n,min}[k],n\in[1,N],\quad(\text{\ref{eqn:problem_formulation_original}a})$
$\displaystyle H_{min}[k]\leq H[k]\leq
H_{max}[k],\quad\quad\quad\quad(\text{\ref{eqn:problem_formulation_original}b})$
$\displaystyle\big{|}\big{|}\bm{l}[k]-\bm{l}[{k-1}]\big{|}\big{|}\leq
V_{F,max}\delta_{k},\quad\quad\quad{\;}(\text{\ref{eqn:problem_formulation_original}c})$
$\displaystyle P_{pr}[k]\leq
P_{pr,th}[k],\quad\quad\quad\quad\quad\quad\quad\quad{\;}(\text{\ref{eqn:problem_formulation_original}d})$
$\displaystyle\sum\nolimits_{n=1}^{N}C_{n}[k]\leq
C_{G,F}[k],\quad\quad\quad\quad\quad{\;}{\;}(\text{\ref{eqn:problem_formulation_original}e})$
$\displaystyle\sum\nolimits_{n=1}^{N}p_{G,n}^{T}\leq p_{G,max}^{T},\quad
p_{G,n}^{T}\geq
0\quad{\;}{\;}(\text{\ref{eqn:problem_formulation_original}f})$
$\displaystyle\sum\nolimits_{s=1}^{S}p_{F,s}^{T}\leq p_{F,max}^{T},\quad
p_{F,s}^{T}\geq
0\quad{\;}{\;}{\;}{\;}(\text{\ref{eqn:problem_formulation_original}g})$

where $\delta_{k}$ is the duration between the time steps $k-1$ and $k$, and
$||.||$ is the $\mathcal{L}_{2}$ norm. The constraint (6a) ensures that every
user always receives their minimum required capacity $C_{n,min}[k]$. The
constraint (6b) restricts the FlyBS’s altitude within $[H_{min},H_{max}]$
where $H_{min}$ and $H_{max}$ are the minimum and maximum allowed flying
altitude, respectively, and are set according to the environment and also the
flying regulations. The constraint (6c) ensures the FlyBS’s speed would not
exceed the maximum supported speed $V_{F,max}$, and (6d) assures that the
FlyBS’s movement would not incur the propulsion power larger than a threshold
$P_{pr,th}$. In practice, the value of $P_{pr,th}$ can be set arbitrarily at
every time step and according to available remaining energy in the FlyBS’s
battery to prolong the FlyBS’s operation. Furthermore, (6f) and (6g) limit the
total transmission powers of the GBS and the FlyBS to the maximum values of
$p_{G,max}^{T}$ and $p_{F,max}^{T}$, respectively.

In the next section, we elaborate on our proposed solution to the formulated
problem in (6).

##  III FlyBS positioning and transmission power allocation on access and
backhaul links

In this section, we present our proposed solution to (6). We provide a high
level overview of the optimization of the transmission power allocation on
both access and backhaul links as well as the FlyBS’s positioning. Then, we
describe in details individual steps of the optimization in following
subsections.

###  III.A Overview of the proposed solution

Solving (6) in general is challenging, since the objective (i.e., sum
capacity) is only convex with respect to $\bm{p_{G}}^{T}[k]$, as it is concave
with respect to $\bm{p_{F}}^{T}[k]$ and also not convex (nor concave) with
respect to $\bm{l_{F}}$. In addition, the constraints (6a), (6d), and (6e) are
also not convex (nor concave) with respect to $\bm{l_{F}}$. Therefore, we
propose a solution based on an alternating optimization of the power
allocation and the FlyBS’s positioning. In particular, the optimization in (6)
is done via iterating the following three steps: 1) optimize $p_{F}^{T}$ at a
given position of the FlyBS $q_{F}$ and for a fixed power allocation
$p_{G}^{T}$, 2) optimize $p_{G}^{T}$ at the same given position of the FlyBS
in step 1 and for the updated $p_{F}^{T}$ from step 1, 3) optimize the FlyBS’s
position $\bm{l_{F}}$ for the updated power allocation derived from steps 1
and 2. Furthermore, to tackle the non-convexity of the objective, we propose
an approximation form of the objective that intuits us to what direction for
the FlyBS’s movement incurs an increase in the sum capacity. The idea of step-
wise solving of (6) facilitates to deal with the non-convexity of the
constraints. Each step is solved with respect to the related constraints in
(6). In the next section, we elaborate our proposed solution.

###  III.B Transmission power allocation for access link

At a fixed position of the FlyBS and for a given setting of transmission power
at the backhaul link ($\bm{p_{G}}^{T}$), the problem of transmission power
optimization at the access link to maximize the sum capacity is formulated as
follows:

$\displaystyle\operatorname*{max}_{\bm{p_{F}}^{T}[k]}{\;}\sum_{n=1}^{N}C_{n}[k],\quad\quad\quad\quad\quad\forall
k,$ (7)
$\displaystyle\text{s.t.}\quad(\text{\ref{eqn:problem_formulation_original}a}),(\text{\ref{eqn:problem_formulation_original}e}),(\text{\ref{eqn:problem_formulation_original}g}).$

Note that, only constraints from (6) that directly relate to the optimization
variable $\bm{p_{F}}^{T}$ are included in (7). According to (1) and (3), the
objective in (7) is concave and the constraint (6g) is convex with respect to
$\bm{p_{F}}^{T}[k]$. Furthermore, the constraint (6a) is rewritten as

$\displaystyle
B_{g_{n}}\log_{2}{\left(1+\frac{Q_{n,F}p_{F,n}^{T}}{{d_{n,F}}^{\alpha_{n,F}}(\sigma^{2}_{n}+p_{n,G}^{R}[k])}\right)}\geq
C_{n,min}[k],$ (8)

or equivalently

$\displaystyle p_{F,n}^{T}\geq
Q^{-1}_{n,F}({2^{\frac{C_{n,min}[k]}{B_{g_{n}}}}-1})(\sigma^{2}_{n}+p_{n,G}^{R}[k]){d_{n,F}}^{\alpha_{n,F}},$
(9)

which is linear with respect to $p_{F,n}^{T}$. Next, we rewrite (6e) by the
means of (1) and (3) as:

$\displaystyle\sum_{n=1}^{N}B_{g_{n}}\log_{2}{\left(1+\frac{Q_{n,F}p_{F,n}^{T}}{d_{n,F}^{\alpha_{n,F}}(\sigma^{2}_{n}+p_{n,G}^{R}[k])}\right)}\leq
C_{G,F}[k],$ (10)

which is non-convex with respect to $p_{F,n}^{T}$. To tackle this issue, we
consider the inequality
$\text{log}_{2}(1+a\mathcal{X})\leq\text{log}_{2}(a)+\frac{1}{\text{ln}(2)}(\text{ln}(\frac{1+s\tau}{a})+\frac{a\mathcal{X}-s\tau}{1+s\tau})$
for arbitrary values $a$ and $\mathcal{X}$, where
$s=\lfloor\frac{a\mathcal{X}}{\tau}\rfloor$, and $\tau$ is an approximation
parameter in the Taylor series and choosing a smaller $\tau$ leads to a
smaller gap between the two sides of the mentioned inequality. By adopting
$s_{n}=\lfloor\frac{p_{F,n}^{T}}{\tau}\rfloor$ and by applying the inequality
to the left-hand side in (10), we get:

$\displaystyle\sum_{n=1}^{N}B_{g_{n}}\log_{2}{\left(1+\frac{Q_{n,F}p_{F,n}^{T}}{d_{n,F}^{\alpha_{n,F}}(\sigma^{2}_{n}+p_{n,G}^{R}[k])}\right)}\leq$
(11) $\displaystyle
C_{\text{tot}}^{\text{ub}}=\sum_{n=1}^{N}B_{g_{n}}({Q^{-1}_{n,F}d_{n,F}^{\alpha_{n,F}}(\sigma^{2}_{n}+p_{n,G}^{R}[k])})+$
$\displaystyle\sum_{n=1}^{N}\frac{B_{g_{n}}}{\text{ln}(2)}(\text{ln}({Q^{-1}_{n,F}d_{n,F}^{\alpha_{n,F}}(\sigma^{2}_{n}+p_{n,G}^{R}[k])}(1+s_{n}\tau))-$
$\displaystyle\frac{s_{n}\tau_{n}}{1+s_{n}\tau}+\frac{p_{F,n}^{T}}{(1+s_{n}\tau)(\sigma^{2}_{n}+p_{n,G}^{R}[k])}).$

$C_{\text{tot}}^{\text{ub}}$ substitutes the upper bound for sum capacity in
the right-hand side in (11). Using (11), we replace (6e) by the linear (with
respect to $p_{F,n}^{T}$) inequality $C_{\text{tot}}^{\text{up}}\leq
C_{G,F}[k]$. Then, the problem in (7) is solved using CVX.

###  III.C Transmission power allocation for backhaul link

Once the power allocation $\bm{p_{G}}^{T}[k]$ over the access link channels is
optimized, we optimize the power allocation over the backhaul. To this end, we
derive the subproblem of the transmission power optimization at the access
link from (6) as

$\displaystyle\operatorname*{max}_{\bm{p_{G}}^{T}[k]}{\;}\sum_{n=1}^{N}C_{n}[k],\quad\quad\quad\quad\quad\forall
k,$ (12)
$\displaystyle\text{s.t.}\quad(\text{\ref{eqn:problem_formulation_original}a}),(\text{\ref{eqn:problem_formulation_original}e}),(\text{\ref{eqn:problem_formulation_original}f}).$

From (1) and (4), we observe that the objective as well as (6f) are convex
with respect to $\bm{p_{G}}^{T}$. Furthermore, the constraint (6a) is
rewritten similarly as for (8) and (9) as

$\displaystyle
p_{G,g_{n}}^{T}\leq(\frac{p_{n,F}^{R}[k]}{2^{\frac{C_{n,min}[k]}{B_{g_{n}}}}-1}-\sigma^{2}_{n})Q^{-1}_{n,G}{d_{n,G}}^{\alpha_{n,G}},$
(13)

which is linear with respect to $p_{G,g_{n}}^{T}$. Next, we rewrite (6e) by
the means of (1), (3), and (4) as:

$\displaystyle\sum\nolimits_{n=1}^{N}B_{g_{n}}\log_{2}{\left(1+\frac{Q_{n,F}p_{F,n}^{T}}{d_{n,F}^{\alpha_{n,F}}(\sigma^{2}_{n}+\frac{Q_{n,G}p_{G,g_{n}}^{T}}{d_{n,G}^{\alpha_{n,G}}})}\right)}-$
$\displaystyle\sum_{s=1}^{S}B_{s}\log_{2}{\left(1+\frac{Q_{F,G}p_{G,s}^{T}[k]}{\sigma^{2}_{F,s}d_{F,G}^{\alpha_{F,G}}}\right)}\leq
0,$ (14)

which is convex with respect to $p_{G,g_{n}}^{T}$ for $p_{G,g_{n}}^{T}\geq 0$.
Hence, the problem in (12) is a concave programming problem. Similar to the
convex optimization, efficient solutions are developed in the literature for
such class of problems in case that constraints define a convex compact set
(like in (12)).

We develop a solution based on an iterative construction of level sets for the
objective function and derivation of local solutions with respect to the level
sets using linear programming (LP), see [23].

###  III.D FlyBS positioning

After optimizing the power allocation over the access and backhaul channels,
we propose a solution to the FlyBS’s positioning. To this end, we formulate
the problem as:

$\displaystyle\operatorname*{max}_{\bm{l_{F}}[k]}{\;}\sum_{n=1}^{N}C_{n}[k],\quad\quad\quad\quad\quad\forall
k,$ (15)
$\displaystyle\text{s.t.}\quad(\text{\ref{eqn:problem_formulation_original}a}),(\text{\ref{eqn:problem_formulation_original}b}),(\text{\ref{eqn:problem_formulation_original}c}),(\text{\ref{eqn:problem_formulation_original}d}),(\text{\ref{eqn:problem_formulation_original}e}).$

The objective and the constraints (6c) and (6e) are not convex with respect to
$\bm{l_{F}}[k]$. Before dealing with the mentioned non-convexity, let us first
discuss the constraints (6a), (6c), and (6d).

C.1) Interpretation of constraints

The constraint (6a) is rewritten as

$\displaystyle{d_{n,F}}\leq(\frac{Q_{n,F}p_{F,n}^{T}}{(2^{\frac{C_{n,min}[k]}{B_{g_{n}}}}-1)(\sigma^{2}_{n}+p_{n,G}^{R}[k])})^{\frac{1}{\alpha_{n,F}}},\quad\forall
n$ (16)

which defines as the FlyBS’s next possible position as the border and inside
of a sphere with a center at $\bm{u_{n}}[k]$ and with a radius of the right-
hand side in 16.

According to Fig. 3 the constraint (6d) is equivalent to $V_{F}$ being upper
bounded by a threshold $V_{F,th}$, i.e., $V_{F}\leq V_{F,th}$. By combining
this inequality with (6c) we get

$\displaystyle||\bm{l}[k]-\bm{l}[k-1]||\leq(\text{min}\\{V_{F,max},V_{F,th}\\})\delta_{k},$
(17)

Equation (17) defines the FlyBS’s next possible position as the border or
inside of the region enclosed by two spheres centered at $\bm{l}[k-1]$ (i.e.,
the FlyBS’s position at the previous time step), one with a radius of
$V_{F,th}\delta_{k}$ and the other one with
($\text{min}\\{V_{F,max},V_{F,th}\\})\delta_{k}$.

Next, to deal with the non-convexity in (6e), let us first derive an upper
bound for the left-hand side in (6e). To this end, we use the fact the FlyBS’s
next position is bounded due to the limit on the FlyBS’s speed as well as
altitude. More specifically, from (17) we find a lower bound to the FlyBS’s
distance from user $n$ ($n\in[1,N]$) at time step $k$ in terms of the FlyBS’s
position at time step $k-1$ as

$\displaystyle d_{n,F}[k]\geq d_{n,F,min}[k]=\text{max}\\{H_{min}[k],$ (18)
$\displaystyle||\bm{l_{F}}[k-1]-\bm{u_{n}}[k]||+(\text{min}\\{V_{F,max},,V_{F,th}\\})\delta_{k}\\},$

Then, by using (18), we get the following upper bound for the left-hand side
in (6e):

$\displaystyle\sum_{n=1}^{N}C_{n}[k]\leq$ (19)
$\displaystyle\sum_{n=1}^{N}B_{g_{n}}\log_{2}{\left(1+\frac{Q_{n,F}p_{F,n}^{T}}{{d_{n,F,min}}^{\alpha_{n,F}}(\sigma^{2}_{n}+p_{n,G}^{R}[k])}\right)},$

Thus, we replace (6e) with the following constraint:

$\displaystyle\sum_{n=1}^{N}B_{g_{n}}\log_{2}{\left(1+\frac{Q_{n,F}p_{F,n}^{T}}{{d_{n,F,min}}^{\alpha_{n,F}}(\sigma^{2}_{n}+p_{n,G}^{R}[k])}\right)}\leq$
$\displaystyle
C_{G,F}[k]=\sum_{s=1}^{S}B_{s}\log_{2}{\left(1+\frac{Q_{F,G}p_{G,s}^{T}[k]}{\sigma^{2}_{F,s}d_{F,G}^{\alpha_{F,G}}}\right)}.$
(20)

Figure 2: two-dimensional depiction of feasibility region (hatched in yellow)
with respect to the constraints in (22).

Once (20) is fulfilled, the constraint (6e) is automatically fulfilled as
well. The left-hand side in (20) is a constant. Furthermore, the right-hand
side in (20) is strictly decreasing with respect to $d_{F,G}[k]$. Hence, we
use bisection method to find the upper bound $d_{F,G,max}[k]$ such that the
inequality

$\displaystyle d_{F,G}[k]\leq d_{F,G,max}[k],$ (21)

is equivalent to (20). The derived upper bound in (21) defines the next
allowed position of the FlyBS as a sphere centered at the GBS’s transmitter
and with a radius of $d_{F,G,max}[k]$.

With the above-provided analysis of the constraints in (15), we now target the
following substitute optimization problem

$\displaystyle\operatorname*{max}_{\bm{l_{F}}[k]}{\;}\sum_{n=1}^{N}C_{n}[k],\quad\quad\quad\quad\quad\forall
k,$ (22) $\displaystyle\text{s.t.}\quad(\ref{eqn:
5a_in_positioning_subproblem}),(\text{\ref{eqn:problem_formulation_original}b}),(\ref{eqn:4c-4d_combined_1}),(\ref{eqn:5e_lesser}).$

Note that, in later discussions, we refer to the combination of constraints in
(22) as the feasibility region at the time step $k$ and we denote it as
$\mathcal{R}_{f}$, i.e., $\mathcal{R}_{f}=\\{\bm{l_{F}}|(\ref{eqn:
5a_in_positioning_subproblem}),(\text{\ref{eqn:problem_formulation_original}b}),(\ref{eqn:4c-4d_combined_1}),(\ref{eqn:5e_lesser})\\}$).
Fig. 2 shows a 2D instance of $\mathcal{R}_{f}$.

C.2) Radial approximation of sum capacity

Now, to tackle the non-convexity of the objective, we propose a radial-basis
approximation for the sum capacity. Such approach helps to express the sum
capacity as a union of level surfaces determining the direction of FlyBS’s
movement towards the optimum position. In the following, we explain the steps
towards the derivation of radial approximation. Firstly, using (3), the log(.)
term in (1) is rewritten as

$\displaystyle\log_{2}{\left(1+\frac{p_{n,F}^{R}[k]}{\sigma^{2}_{n}+p_{n,G}^{R}[k]}\right)}=\log_{2}{\left(1+\frac{{Q_{n,F}p_{F,n}^{T}d_{n,F}^{-\alpha_{n,F}}}}{(\sigma^{2}_{n}+{p_{n,G}^{R}[k]})}\right)}$
(23)

Next, the linear approximation $\log_{2}(1+X)\approx\frac{X}{\text{ln}(2)}$ is
applied to the right-hand side in (23) to derive a linear expression with
respect to $d_{n,F}^{-\alpha_{n,F}}$. Then, we further derive a linear
approximation of $d_{n,F}^{-\alpha_{n,F}}$ with respect to ${d^{2}_{n,F}}$. In
particular, we use the Taylor approximation $(a+X)^{k}\approx(a+\delta
a\xi)^{k}+k(a+\delta a\xi)^{k-1}(X-\delta a\xi)$ where
$\delta=\lfloor\frac{X}{a\xi}\rfloor$, and the parameter $\xi$ determines the
accuracy in approximation (smaller $\xi$ leads to smaller error). Using the
mentioned approximation for $n\in[1,N]$, we get a sum of quadratic terms in
the form of ${d^{2}_{n,F}}=(X-x_{n})^{2}+(Y-y_{n})^{2}+(Z-z_{n})^{2}$. Since a
sum of quadratic terms is also a quadratic expression, the sum capacity
$\sum_{n=1}^{N}C_{n}$ is rewritten as

Figure 3: Propulsion power model vs. speed for rotary-wing FlyBS.

$\displaystyle\sum_{n=1}^{N}C_{n}[k]\approx
W(\bm{p^{T}},k)-\zeta(\bm{p^{T}},k){\big{|}\big{|}\bm{l_{F}}[k]-\bm{l_{F,o}}[k]\big{|}\big{|}}^{2},$
(24)

where the substitutions $W(\bm{p^{T}},k)$ and $\zeta(\bm{p^{T}},k)$ are
constants with respect to $\bm{l_{F}}[k]$. The details about the derivation of
(24) are not entirely shown to avoid distraction from the main discussion in
this section. Nevertheless, interested readers can refer to [19] where more
steps of a similar derivation is presented (Appendix A in [19] in particular).

In order to make the approximation in (24) efficient, we derive the
approximation at a position "close" to the actual optimal position, since the
objective (sum capacity) is a continuous function. Let $\bm{l_{F,c}}$ denote
such a position. We propose to choose $\bm{l_{F,c}}$ by solving an
optimization problem derived from (6) as explained in the following Remark 1.

Remark 1: By using the inequality
$\log_{2}(1+\frac{1}{ax})\geq\frac{1}{\ln(2)}(\frac{-x}{ax_{0}^{2}}+\frac{1}{ax_{0}}+\ln(1+\frac{1}{ax_{0}}))$
for arbitrary $a$ and $x$ at any point $x_{0}$, a lower bound
$C_{\text{tot}}^{\text{lb}}$ for the sum capacity is obtained as

$\displaystyle\sum_{n=1}^{N}C_{n}[k]=\sum_{n=1}^{N}B_{g_{n}}\log_{2}{\left(1+\frac{Q_{n,F}p_{F,n}^{T}}{d_{n,F}^{\alpha_{n,F}}(\sigma^{2}_{n}+p_{n,G}^{R}[k])}\right)}\geq
C_{\text{tot}}^{\text{lb}}=$
$\displaystyle\sum_{n=1}^{N}\frac{B_{g_{n}}}{\ln(2)}\big{(}\frac{-Q_{n,F}{d_{n,F}}^{\alpha_{n,F}}}{{(\sigma^{2}_{n}p_{n,G}^{R}[k])d_{n,F}^{2\alpha_{n,F}}[k-1]}}+\frac{Q_{n,F}}{{{(\sigma^{2}_{n}+p_{n,G}^{R}[k])d_{n,F}^{\alpha_{n,F}}[k-1]}}}\big{)}$
(25)

The right-hand side in (25) is a concave function with respect to
$\bm{l_{F}}[k]$. Hence,

$\displaystyle\bm{l_{F,c}}=\operatorname*{argmax}_{\bm{l_{F}}[k]}{\;}C_{\text{tot}}^{\text{lb}},\quad\forall
k,$ (26) $\displaystyle\text{s.t.}\quad(\ref{eqn:
5a_in_positioning_subproblem}),(\text{\ref{eqn:problem_formulation_original}b}),(\ref{eqn:4c-4d_combined_1}),(\ref{eqn:5e_lesser}).$

The convex problem in (26) is solved using CVX. Next, we set $\bm{l_{F,c}}[k]$
as the reference point for the approximation in (24) as it is a "close" point
to the optimal position.

C.3) Solution to FlyBS positioning

Now, we elaborate the solution to the FlyBS’s positioning. According to (24),
the sum capacity increases with a decrease in the distance between
$\bm{l_{F,o}}$ and $\bm{l_{F}}$. Thus, the maximum value of sum capacity is
achieved at the closest point to $\bm{l_{F,o}}$ that fulfills all the
constraints in (22). According to the discussion in this subsection, each of
the constraints (16), (17), and (21) in (22) limits the FlyBS’s position to
the border and interior of a sphere and, hence, are convex. Combined with
(6b), the feasibility region $\mathcal{R}_{f}$ for the FlyBS’s position is
convex.

Then, the problem of FlyBS’s positioning is transformed to

$\displaystyle\operatorname*{min}_{\Lambda\in\mathcal{R}_{f}}{\;}||\Lambda-\bm{l_{F,o}}[k]||^{2},\quad\forall
k.$ (27)

The objective and the domain in (27) is convex and, hence, it is solved using
CVX.

Once the FlyBS’s position $\bm{l_{F}}$ is updated to the solution derived from
(27), the power allocation $\bm{p^{T}}$ is again optimized at the updated
position of the FlyBS. Consequently, the updated $\bm{p^{T}}$ changes the
spheres corresponding to (16), (17), (21) in (22). Thus, an updated solution
to (27) is derived. This optimization of $\bm{p^{T}}$ and $\bm{l_{F}}$ is
repeated until the FlyBS’s movement at some iteration falls below a given
threshold $\epsilon$ or until the maximum number of iterations is reached.

##  IV Simulations and results

This section provides the details for our adopted simulation scenario followed
by the results and discussions to show superiority of the proposed solution
over state-of-the-art.

###  IV.A Simulation scenario and models

We assume a 500 m $\times$ 500 m square area with 100 to 600 users initially
distributed randomly. The GBS is located at a distance of 1500 m from the
center of the area. We adopt the user’s mobility model from [24] where a half
of the users move at a speed of 1 m/s according to random-walk model and, the
other half are randomly divided into six clusters of crowds. A simulation
duration of 1200 seconds is assumed.

A total bandwidth of 100 MHz is divided equally among the users at the access
link. The background interference and the noise’s spectral density are set to
–90 dBm and –174 dBm/Hz, respectively. Pathloss exponents of
$\alpha_{n,F}=2.3$, $\alpha_{n,G}=2.8$, and $\alpha_{F,G}=2.1$ for FlyBS-user,
GBS-user, and GBS-FlyBS channels are assumed, respectively [18]. An altitude
range of [100, 300] m and a maximum transmission power limit of
$P_{F,max}^{T}=30$ dBm is considered for the FlyBS. Also, an altitude of 30 m
and a maximum transmission power of $P_{G,max}^{T}=36$ dBm (5 W) is assumed
for GBS. The results are averaged out over 100 simulation drops.

We benchmark our proposed solution to backhaul-aware sum capacity maximization
against the following state-of-the-art schemes: $i$) maximization of sum
capacity, referred to as MSC, via FlyBS’s positioning and transmission power
allocation at the access link, published in [19], $ii$) minimum capacity
maximization, referred to as mCM, via optimization of FlyBS’s positioning and
transmission power allocation to the users at the access link, published in
[12], $iii$) maximization of energy efficiency, referred to as EEM, via
transmission power allocation at the access link, as introduced in [13]. Note
that the original solution in [13] does not provide a positioning of the
FlyBS, thus, the benchmark scheme EEM is an enhanced version of the solution
[13] and the FlyBS’s positioning is solved using K-means.

Figure 4: Sum capacity vs. number of users for $C_{min}=$1 Mbps.

###  IV.B Simulation results

In this subsection, we present the simulation results and we discuss the
performance of different schemes.

Fig. 4 shows the sum capacity versus number of users ($N$) for different
schemes. A minimum required capacity of $C_{min}=1$ Mbps is assumed for all
users. According to Fig. 4, the sum capacity decreases if more users are
served by the FlyBS. This is due to two main reasons 1) the bandwidth
allocated to each user becomes smaller, and 2) the FlyBS’s total transmission
power is divided among more users. Nevertheless, the proposed solution
outperforms other schemes in the achieved sum capacity. More specifically, the
sum capacity is increased by up to 21%, 28%, and 47% with respect to MSC, mCM,
and EEM, respectively.

Figure 5: Sum capacity vs. $C_{n,min}$ for $N=$ 300. Figure 6: Sum capacity
vs. $C_{n,min}$ for $N=$ 600.

Next, Figs. 5 and 6 show the impact of minimum user’s capacity $C_{min}$ on
the sum capacity for $N=$ 300 and $N=$ 600, respectively. The maximum depicted
$C_{min}$ represents the largest $C_{min}$ for which a feasible solution
exists. However, the value of $C_{min}$ in mCM is not set manually and
beforehand, as it is directly derived by the scheme itself (which maximizes
the minimum capacity). Hence, the sum capacity is constant in Figs. 5 and 6.
However, for the proposed solution, MSC, and EEM, increasing $C_{min}$ reduces
the sum capacity. This is because increasing $C_{min}$ leads to a tighter
feasibility region according to (16) and, hence, it limits the FlyBS’s
movement to maximize the sum capacity. The proposed solution increases the sum
capacity with respect to MSC, mCM, and EEM by 24%, 25%, and 49%, respectively,
for $N=$ 300, and by 19%, 33%, and 49%, respectively, for $N=$ 600.

Next, we also demonstrate the fast convergence of our proposed iterative
algorithm in Figs. 7 and 8 by showing an evolution of the sum capacity over
iterations the alternating optimization of transmission power allocation and
FlyBS’s positioning. Note that, the benchmark schemes mCM and EEM are not
iterative and, hence, their sum capacity is constant and they are shown in the
Figs. 7 and 8 only to show their performance. The proposed solution converges
very fast and in only few iterations. This confirms that the iterative manner
of the proposed solution does not limit its feasibility and practical
application. Note that, although the mCM scheme outperforms our proposal in
the first iteration in Fig. 8, only the converged results should be subject to
comparison as the performance at early iterations can be greatly impacted by
the initialization of FlyBS’s position and power allocation.

Figure 7: Convergence of the proposed scheme for $N=$300. Figure 8:
Convergence of the proposed scheme for $N=$600.

##  V Conclusions

In this paper, we have provided an analytical approach to maximize the sum
capacity via a positioning of the FlyBS, allocation of transmission power to
the backhaul channels, and an allocation of the transmission power to the
users at the access channel. The problem is constrained by the minimum
required instantaneous capacity to each user and practical real world
limitations of the FlyBSs. We have shown that the proposed solution enhances
the sum capacity by tens of percent compared to state-of-the-art works. In the
future work, a scenario with multiple FlyBSs should be studied along with
related aspects, such as a management of interference among FlyBSs and an
association of users to FlyBSs.

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