Series over Bessel Functions as Series in Terms of Riemann’s Zeta Function

Abstract

Relying on the Hurwitz formula, we find closed-form formulas for the series over sine and cosine functions through the Hurwitz zeta functions, and using them and another summation formula for trigonometric series, we obtain a finite sum for some series over the Riemann zeta functions. We apply these results to the series over Bessel functions, expressing them first as series over the Riemann zeta functions.

1. Preliminaries

Apostol [1] (Theorem 12.6, p. 257) derived Hurwitz’s formula

$$\zeta (1-s,a)={\displaystyle \frac{\Gamma \left(s\right)}{{\left(2\pi \right)}^s}}\left(e^{-i\pi s/2}\sum _{n=1}^{\infty }{\displaystyle \frac{e^{2i\pi na}}{n^s}}+e^{i\pi s/2}\sum _{n=1}^{\infty }{\displaystyle \frac{e^{-2i\pi na}}{n^s}}\right),$$

(1)

where $\zeta (1-s,a)$ is the Hurwitz zeta function with $\mathrm{Re}s>1$ and $0<a⩽1$. We can write it as

$$\zeta (1-s,a)={\displaystyle \frac{2\Gamma \left(s\right)}{{\left(2\pi \right)}^s}}\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^s}}cos(\frac{\pi }{2}s-2n\pi a).$$

(2)

Theorem 1.

For $\mathrm{Re}s>1$ and $0<x<2\pi$, there holds the following formulas:

$$\begin{array}{cc}\hfill & \sum _{n=1}^{\infty }{\displaystyle \frac{sin(nx+y)}{n^s}}={\displaystyle \frac{{\left(2\pi \right)}^s}{2\Gamma \left(s\right)sin\pi s}}(cos\left(y-{\displaystyle \frac{\pi s}{2}}\right)\zeta \left(1-s,{\displaystyle \frac{x}{2\pi }}\right)\hfill \\ \hfill & -cos\left(y+{\displaystyle \frac{\pi s}{2}}\right)\zeta \left(1-s,1-{\displaystyle \frac{x}{2\pi }}\right)),\hfill \\ \hfill & \sum _{n=1}^{\infty }{\displaystyle \frac{cos(nx+y)}{n^s}}={\displaystyle \frac{{\left(2\pi \right)}^s}{2\Gamma \left(s\right)sin\pi s}}(sin\left(y+{\displaystyle \frac{\pi s}{2}}\right)\zeta \left(1-s,1-{\displaystyle \frac{x}{2\pi }}\right)\hfill \\ \hfill & -sin\left(y-{\displaystyle \frac{\pi s}{2}}\right)\zeta \left(1-s,{\displaystyle \frac{x}{2\pi }}\right)),\hfill \end{array}$$

(3)

where $y\in \mathbb{R}$.

Proof. 

We replace a in Equation (1) first with $1-x/2\pi$, and then with $x/2\pi$. Further, we multiply the first equality by $e^{i(y+\pi s/2)}$, and the latter by $e^{i(y-\pi s/2)}$. After subtracting, we obtain

$$\begin{array}{cc}\hfill & e^{i(y+\pi s/2)}\zeta \left(1-s,1-{\displaystyle \frac{x}{2\pi }}\right)-e^{i(y-\pi s/2)}\zeta \left(1-s,{\displaystyle \frac{x}{2\pi }}\right)\hfill \\ \hfill & ={\displaystyle \frac{\Gamma \left(s\right)}{{\left(2\pi \right)}^s}}\left(e^{i\pi s}-e^{-i\pi s}\right)\sum _{n=1}^{\infty }{\displaystyle \frac{e^{i(nx+y)}}{n^s}}\hfill \\ \hfill & ={\displaystyle \frac{2\Gamma \left(s\right)sin\pi s}{{\left(2\pi \right)}^s}}\left(i\sum _{n=1}^{\infty }{\displaystyle \frac{cos(nx+y)}{n^s}}-\sum _{n=1}^{\infty }{\displaystyle \frac{sin(nx+y)}{n^s}}\right).\hfill \end{array}$$

Hence, comparing the corresponding real and imaginary parts, we simultaneously arrive at both formulas in Equation (3). □

For $y=0$ in the first formula of Equation (3), we have

$$\begin{array}{c}\hfill \sum _{n=1}^{\infty }{\displaystyle \frac{sinnx}{n^s}}={\displaystyle \frac{{\left(2\pi \right)}^s}{4\Gamma \left(s\right)}}{\displaystyle \frac{\zeta \left(1-s,\frac{x}{2\pi }\right)-\zeta \left(1-s,1-\frac{x}{2\pi }\right)}{sin\frac{\pi s}{2}}}.\end{array}$$

Since $\underset{s\to 2m}{lim}sin{\displaystyle \frac{\pi s}{2}}=0$, and based on Equation (2) there holds

$$\underset{s\to 2m}{lim}\left(\zeta (1-s,a)-\zeta (1-s,1-a)\right)=\underset{s\to 2m}{lim}\left({\displaystyle \frac{4\Gamma \left(s\right)}{{\left(2\pi \right)}^s}}\sum _{n=1}^{\infty }{\displaystyle \frac{sin\frac{\pi s}{2}sin2n\pi a}{n^s}}\right)=0,$$

applying the L’Hôpital rule yields

$$\begin{array}{cc}\hfill \sum _{n=1}^{\infty }{\displaystyle \frac{sinnx}{n^{2m}}}& ={\displaystyle \frac{{\left(2\pi \right)}^{2m}}{4(2m-1)!}}\underset{s\to 2m}{lim}{\displaystyle \frac{\zeta \left(1-s,\frac{x}{2\pi }\right)-\zeta \left(1-s,1-\frac{x}{2\pi }\right)}{sin\frac{\pi s}{2}}}\hfill \\ \hfill & ={\displaystyle \frac{{(-1)}^m{\left(2\pi \right)}^{2m-1}}{(2m-1)!}}\left({\zeta }^{\prime }\left(1-2m,1-{\displaystyle \frac{x}{2\pi }}\right)-{\zeta }^{\prime }\left(1-2m,{\displaystyle \frac{x}{2\pi }}\right)\right).\hfill \end{array}$$

(4)

The left-hand series is called the Clausen functions, denoted by ${\mathrm{Cl}}_{2m}\left(x\right)$.

Similarly, setting $y=0$ in the second formula of Equation (3), we have

$$\sum _{n=1}^{\infty }{\displaystyle \frac{cosnx}{n^s}}={\displaystyle \frac{{\left(2\pi \right)}^s}{4\Gamma \left(s\right)}}{\displaystyle \frac{\zeta \left(1-s,1-\frac{x}{2\pi }\right)+\zeta \left(1-s,\frac{x}{2\pi }\right)}{cos\frac{\pi s}{2}}},$$

and after repeating the preceding procedure, we obtain

$$\begin{array}{cc}\hfill \sum _{n=1}^{\infty }{\displaystyle \frac{cosnx}{n^{2m-1}}}& ={\displaystyle \frac{{\left(2\pi \right)}^{2m-1}}{4(2m-2)!}}\underset{s\to 2m-1}{lim}{\displaystyle \frac{\zeta \left(1-s,1-\frac{x}{2\pi }\right)+\zeta \left(1-s,\frac{x}{2\pi }\right)}{cos\frac{\pi s}{2}}}\hfill \\ \hfill & ={\displaystyle \frac{{(-1)}^{m-1}{\left(2\pi \right)}^{2m-2}}{(2m-2)!}}\left({\zeta }^{\prime }\left(2-2m,1-{\displaystyle \frac{x}{2\pi }}\right)+{\zeta }^{\prime }\left(2-2m,{\displaystyle \frac{x}{2\pi }}\right)\right),\hfill \end{array}$$

(5)

The left-hand series is also called the Clausen functions, but denoted by ${\mathrm{Cl}}_{2m-1}\left(x\right)$.

Remark 1.

The results (4) and (5) are new, seeing as we have expressed the Clausen functions through the Hurwitz functions, i.e., their first derivative. Because of the similarity regarding the form with some trigonometric series over the sine and cosine functions, there is confusion that these results already exist in some books of tables and integrals. For instance, in [2], (p. 726, Section 5.4.2, entries 5. and 6.) the denominator is $k^{2n-1}$ with no entries, where the denominator in the sine series is $k^{2n}$. Also, in [2], (Section 5.4.2, entries 7. and 8.) the denominator is $k^{2n}$ with no entries, where the denominator in the cosine series is $k^{2n-1}$. In [2], (Section 5.4.2, entries 1. and 2.), the authors state general formulas for trigonometric series over the sine and cosine functions. Nevertheless, they do not thence derive the cases $s=2n$ for the sine series nor $s=2n-1$ for the cosine series. In addition, we have derived more general formulas comprising theirs.

2. Results Related to Series over Zeta Function

Based on the results in the initial section, we shall obtain a finite sum of the series over the zeta functions.

Theorem 2.

For $m\in \mathbb{N}$, there holds

$$\begin{array}{cc}\hfill & \sum _{k=1}^{\infty }{\displaystyle \frac{\zeta \left(2k\right)}{{\left(2k\right)}_{2m}}}{\left({\displaystyle \frac{x}{2\pi }}\right)}^{2k}={\displaystyle \frac{logx-H_{2m-1}}{2(2m-1)!}}-{\displaystyle \frac{{\left(\frac{2\pi }{x}\right)}^{2m-1}}{2(2m-1)!}}({\zeta }^{\prime }\left(1-2m,1-{\displaystyle \frac{x}{2\pi }}\right)\hfill \\ \hfill & -{\zeta }^{\prime }\left(1-2m,{\displaystyle \frac{x}{2\pi }}\right))+\sum _{k=0}^{m-2}{\displaystyle \frac{{(-1)}^{m+k}{x}^{2k-2m+2}\zeta (2m-2k-1)}{2(2k+1)!}},\hfill \end{array}$$

(6)

where $H_n$ stands for the nth harmonic number.

Proof. 

For $\mathrm{Re}s>1$, we consider the first formula ([3], p. 445)

$$\sum _{n=1}^{\infty }{\displaystyle \frac{sinnx}{n^s}}={\displaystyle \frac{\pi x^{s-1}}{2\Gamma \left(s\right)sin\frac{\pi s}{2}}}+\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k\zeta (s-2k-1)}{(2k+1)!}}{x}^{2k+1},0<x<2\pi .$$

Since we encounter singularities if we set $s=2m$ in the first term and the series member for $k=m-1$, we must take the limit $s\to 2m$, $m=1,2,\cdots$, i.e.,

$$\begin{array}{cc}\hfill & \sum _{n=1}^{\infty }{\displaystyle \frac{sinnx}{n^{2m}}}=\underset{s\to 2m}{lim}\left({\displaystyle \frac{\pi x^{s-1}}{2\Gamma \left(s\right)sin\frac{\pi s}{2}}}+{\displaystyle \frac{{(-1)}^{m-1}\zeta (s-2m+1)}{(2m-1)!}}{x}^{2m-1}\right)\hfill \\ \hfill & +\sum _{k=0}^{m-2}{\displaystyle \frac{{(-1)}^k\zeta (2m-2k-1)}{(2k+1)!}}{x}^{2k+1}+\sum _{k=m}^{\infty }{\displaystyle \frac{{(-1)}^k\zeta (2m-2k-1)}{(2k+1)!}}{x}^{2k+1}.\hfill \end{array}$$

(7)

Relying on the L’Hôpital rule, we obtain

$$\begin{array}{cc}\hfill & \sum _{n=1}^{\infty }{\displaystyle \frac{sinnx}{n^{2m}}}={\displaystyle \frac{{(-1)}^m{x}^{2m-1}}{(2m-1)!}}\left(logx-H_{2m-1}\right)+\sum _{k=0}^{m-2}{\displaystyle \frac{{(-1)}^k\zeta (2m-2k-1)}{(2k+1)!}}{x}^{2k+1}\hfill \\ \hfill & -2{(-1)}^m{x}^{2m-1}\sum _{j=1}^{\infty }{\displaystyle \frac{\zeta \left(2j\right)}{{\left(2j\right)}_{2m}}}{\left({\displaystyle \frac{x}{2\pi }}\right)}^{2j}.\hfill \end{array}$$

We substitute in the right-hand-side series Equation (7) the running index j for k by $k=m+j-1$, and then apply the relation

$$\zeta (1-2j)={(-1)}^j{\displaystyle \frac{2(2j-1)!\zeta \left(2j\right)}{{\left(2\pi \right)}^{2j}}}={(-1)}^j{\displaystyle \frac{2\Gamma \left(2j\right)\zeta \left(2j\right)}{{\left(2\pi \right)}^{2j}}}.$$

(8)

Finally, we replace j with k and leverage Equation (4) to arrive at Equation (6). □

Theorem 3.

For $m\in \mathbb{N}$, there holds

$$\begin{array}{cc}\hfill & \sum _{k=1}^{\infty }{\displaystyle \frac{\zeta \left(2k\right)}{{\left(2k\right)}_{2m-1}}}{\left({\displaystyle \frac{x}{2\pi }}\right)}^{2k}={\displaystyle \frac{logx-H_{2m-2}}{2(2m-2)!}}+{\displaystyle \frac{{\left(\frac{2\pi }{x}\right)}^{2m-2}}{2(2m-2)!}}({\zeta }^{\prime }\left(2-2m,1-{\displaystyle \frac{x}{2\pi }}\right)\hfill \\ \hfill & +{\zeta }^{\prime }\left(2-2m,{\displaystyle \frac{x}{2\pi }}\right))+\sum _{k=0}^{m-2}{\displaystyle \frac{{(-1)}^{m+k}{x}^{2k-2m+2}\zeta (2m-2k-1)}{2\left(2k\right)!}}.\hfill \end{array}$$

(9)

Proof. 

For $\mathrm{Re}s>1$, taking account of the second formula ([3], p. 445)

$$\sum _{n=1}^{\infty }{\displaystyle \frac{cosnx}{n^s}}={\displaystyle \frac{\pi x^{s-1}}{2\Gamma \left(s\right)cos\frac{\pi s}{2}}}+\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k\zeta (s-2k)}{\left(2k\right)!}}{x}^{2k},0<x<2\pi ,$$

after letting $s\to 2m-1$, $m=1,2,\cdots$, it follows

$$\begin{array}{cc}\hfill & \sum _{n=1}^{\infty }{\displaystyle \frac{cosnx}{n^{2m-1}}}=\sum _{k=0}^{m-2}{\displaystyle \frac{{(-1)}^k\zeta (2m-2k-2)}{\left(2k\right)!}}{x}^{2k}+\underset{s\to 2m-1}{lim}({\displaystyle \frac{\pi x^{s-1}}{2\Gamma \left(s\right)cos\frac{\pi s}{2}}}\hfill \\ \hfill & +{\displaystyle \frac{{(-1)}^{m-1}\zeta (s-2m+2)}{(2m-2)!}}{x}^{2m-2})+\sum _{k=m}^{\infty }{\displaystyle \frac{{(-1)}^k\zeta (2m-2k-1)}{\left(2k\right)!}}{x}^{2k}.\hfill \end{array}$$

(10)

By bringing the fractions in brackets to the same denominator and applying L’Hôpital’s rule, we determine the above limiting value

$$\begin{array}{c}\underset{s\to 2m-1}{lim}\left({\displaystyle \frac{\pi x^{s-1}}{2\Gamma \left(s\right)cos\frac{\pi s}{2}}}+{\displaystyle \frac{{(-1)}^{m-1}\zeta (s-2m+2)}{(2m-2)!}}{x}^{2m-2}\right)\hfill \\ \hfill ={\displaystyle \frac{{(-1)}^m{x}^{2m-2}}{(2m-2)!}}\left(logx-H_{2m-2}\right).\end{array}$$

As for the remainder in Equation (10), we introduce a new running index by $k=m+j-1$; then, reverting to k and recalling Equation (8), we come to the formula

$$\begin{array}{cc}\hfill & \sum _{n=1}^{\infty }{\displaystyle \frac{cosnx}{n^{2m-1}}}={\displaystyle \frac{{(-1)}^m{x}^{2m-2}}{(2m-2)!}}\left(logx-H_{2m-2}\right)+\sum _{k=0}^{m-2}{\displaystyle \frac{{(-1)}^k{x}^{2k}\zeta (2m-2k-1)}{\left(2k\right)!}}\hfill \\ \hfill & -2{(-1)}^m{x}^{2m-2}\sum _{k=1}^{\infty }{\displaystyle \frac{\zeta \left(2k\right)}{{\left(2k\right)}_{2m-1}}}{\left({\displaystyle \frac{x}{2\pi }}\right)}^{2k},m=1,2,\cdots .\hfill \end{array}$$

Using Equation (5), we obtain Equation (9). □

We are dealing now with a more general series over the Riemann zeta function, reducing it to the series Equations (6) and (9).

Theorem 4.

For $m,p\in \mathbb{N}$ and $0<x<2\pi$, there holds

$$\begin{array}{cc}\hfill \sum _{n=1}^{\infty }{\displaystyle \frac{{(m+n)}_p}{{\left(2n\right)}_{2p+2m}}}& \zeta \left(2n\right){\left({\displaystyle \frac{x}{2\pi }}\right)}^{2n}=\sum _{k=1}^p\left({\displaystyle \genfrac{}{}{0pt}{}{p-1}{k-1}}\right)\times \hfill \\ \hfill & \sum _{j=0}^{2k-1}{\displaystyle \frac{{(-1)}^{j+k-1}}{2^{2p-1}(p-1)!}}\prod _{i=1}^j(2k-i)\sum _{n=1}^{\infty }{\displaystyle \frac{\zeta \left(2n\right)}{{\left(2n\right)}_{2m+j+1}}}{\left({\displaystyle \frac{x}{2\pi }}\right)}^{2n}.\hfill \end{array}$$

(11)

Proof. 

We make a rational function decomposition of the left-hand side in Equation (11)

$$\begin{array}{cc}\hfill {\displaystyle \frac{{(m+n)}_p}{{\left(2n\right)}_{2p+2m}}}& ={\displaystyle \frac{(m+n)\cdots (m+n+p-1)}{{\left(2n\right)}_{2m}(2n+2m)\cdots (2n+2m+2p-1)}}={\displaystyle \frac{1}{2^p{\left(2n\right)}_{2m}Q\left(z\right)}},\hfill \end{array}$$

with $Q\left(z\right)=(2z+2m+1)(2z+2m+3)\cdots (2z+2m+2p-1)$. Further, we have

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{Q\left(z\right)}}& ={\displaystyle \frac{1}{(2z+2m+1)(2z+2m+3)\cdots (2z+2m+2p-1)}}\hfill \\ \hfill & ={\displaystyle \frac{C_1}{2z+2m+1}}+{\displaystyle \frac{C_2}{2z+2m+3}}+\cdots +{\displaystyle \frac{C_p}{2z+2m+2p-1}},\hfill \end{array}$$

and determine the constants $C_1,C_2,\cdots ,C_p$ by applying Heaviside’s method. Knowing that for $a_k=-(m+k-\frac{1}{2}),Q\left(a_k\right)=0,k=1,\cdots ,p$, it yields

$$C_k={\displaystyle \frac{2}{Q^{\prime }\left(a_k\right)}},k=1,\cdots ,p,{\displaystyle \frac{1}{Q\left(z\right)}}={\displaystyle \frac{2^{1-p}}{(p-1)!}}\sum _{k=1}^p{\displaystyle \frac{{(-1)}^{k-1}\left(\genfrac{}{}{0pt}{}{p-1}{k-1}\right)}{2z+2m+2k-1}}.$$

(12)

Thus, we can express the left-hand-side series in Equation (11) as follows

$$\sum _{n=1}^{\infty }{\displaystyle \frac{{(m+n)}_p}{{\left(2n\right)}_{2p+2m}}}\zeta \left(2n\right){\left({\displaystyle \frac{x}{2\pi }}\right)}^{2n}=\sum _{k=1}^p\sum _{n=1}^{\infty }{\displaystyle \frac{2^{1-2p}{(-1)}^{k-1}\left(\genfrac{}{}{0pt}{}{p-1}{k-1}\right)\zeta \left(2n\right){\left(\frac{x}{2\pi }\right)}^{2n}}{(p-1)!{\left(2n\right)}_{2m}(2n+2m+2k-1)}}.$$

(13)

In the next step, we shall calculate constants $c_1,\cdots ,c_{2k}$ in the representation

$${\displaystyle \frac{1}{2n+2m+2k-1}}={\displaystyle \frac{c_1}{2n+2m}}+{\displaystyle \frac{c_2}{{(2n+2m)}_2}}+\cdots +{\displaystyle \frac{c_{2k}}{{(2n+2m)}_{2k}}},$$

where $k=1,\cdots ,p$. So, we first multiply the numerator and denominator of the left-hand-side fraction by the missing factors between $(2n+2m+2k-1)$ and $(2n+2m)$, including the latter. Then, we bring the right-hand side fractions to the same denominator, obtaining this way the equality of the numerators

$$\begin{array}{c}(2n+2m)\cdots (2n+2m+2k-2)=c_1(2n+2m+1)\cdots (2n+2m+2k-1)\hfill \\ \hfill +c_2(2n+2m+2)\cdots (2n+2m+2k-1)+\cdots +c_{2k},k=1,\cdots ,p.\end{array}$$

The highest power on the left-hand side and the product at $c_1$ is ${\left(2n\right)}^{2k-1}$. There follows $c_1=1$. After rearrangements, we have

$$\begin{array}{cc}\hfill & -(2k-1)(2n+2m+1)\cdots (2n+2m+2k-2)\hfill \\ \hfill & =c_2(2n+2m+2)\cdots (2n+2m+2k-1)+c_3(2n+2m+3)\cdots \times \hfill \\ \hfill & \cdots (2n+2m+2k-1)+\cdots +c_{2k-1}(2n+2m+2k-1)+c_{2k}.\hfill \end{array}$$

(14)

Regarding both sides as polynomial functions in n of the degree $2k-2$, we take the $(2k-2)$th derivative. As a result, we have

$$-(2k-1)2^{2k-2}(2k-2)!=c_2{2}^{2k-2}(2k-2)!.$$

Hence, we find $c_2=-(2k-1)$. Replacing this value in Equation (14), we can determine $c_3$ similarly

$$\begin{array}{cc}\hfill & (2k-1)(2k-2)(2n+2m+2)\cdots (2n+2m+2k-2)\hfill \\ \hfill & =c_3(2n+2m+3)\cdots (2n+2m+2k-1)+\hfill \\ \hfill & c_4(2n+2m+4)\cdots (2n+2m+2k-1)+\cdots +c_{2k-1}(2n+2m+2k-1)+c_{2k}.\hfill \end{array}$$

Now, the polynomials on both sides are of the degree $2k-3$, and we take the $(2k-3)$th derivative and obtain

$$(2k-1)(2k-2)2^{2k-3}(2k-3)!=c_3{2}^{2k-3}(2k-3)!,$$

which implies $c_3=(2k-1)(2k-2)$. By repeating this procedure, we obtain

$${\displaystyle \frac{1}{(2n+2m+2k-1)}}=\sum _{j=0}^{2k-1}{\displaystyle \frac{{(-1)}^j{\prod }_{i=1}^j(2k-i)}{{(2n+2m)}_{j+1}}}.$$

Since

$$\begin{array}{c}{\left(2n\right)}_{2m}{(2n+2m)}_{j+1}=2n(2n+1)\cdots (2n+2m-1)\times \hfill \\ \hfill (2n+2m)(2n+2m+1)\cdots (2n+2m+j)={\left(2n\right)}_{2m+j+1},\end{array}$$

we arrive at the relation

$${\displaystyle \frac{1}{{\left(2n\right)}_{2m}(2n+2m+2k-1)}}=\sum _{j=0}^{2k-1}{\displaystyle \frac{{(-1)}^j{\prod }_{i=1}^j(2k-i)}{{\left(2n\right)}_{2m+j+1}}}.$$

Placing this in Equation (13) gives rise to

$$\begin{array}{c}\sum _{k=1}^p\sum _{n=1}^{\infty }{\displaystyle \frac{2^{1-2p}{(-1)}^{k-1}\left(\genfrac{}{}{0pt}{}{p-1}{k-1}\right)\zeta \left(2n\right){\left(\frac{x}{2\pi }\right)}^{2n}}{(p-1)!{\left(2n\right)}_{2m}(2n+2m+2k-1)}}\hfill \\ \hfill =\sum _{k=1}^p\left({\displaystyle \genfrac{}{}{0pt}{}{p-1}{k-1}}\right)\sum _{j=0}^{2k-1}{\displaystyle \frac{{(-1)}^{j+k-1}}{2^{2p-1}(p-1)!}}\prod _{i=1}^j(2k-i)\sum _{n=1}^{\infty }{\displaystyle \frac{\zeta \left(2n\right)}{{\left(2n\right)}_{2m+j+1}}}{\left({\displaystyle \frac{x}{2\pi }}\right)}^{2n},\end{array}$$

(15)

which proves Equation (11). □

Thus, the procedure to evaluate the left-hand-side sum of Equation (11) reduces to a multiple application of Equation (9).

3. Applications to Some Series over Bessel Functions

Bessel functions $J_{\nu }\left(x\right)$, $x>0$, defined by the Swiss mathematician Daniel Bernoulli, then generalized and developed by Friedrich Bessel while studying the dynamics of gravitational systems in the second decade of the 19th century, are canonical solutions of Bessel’s homogenous differential equation

$$x^2{\displaystyle \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}}+x{\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}}+(x^2-{\nu }^2)y=0$$

(16)

for an arbitrary complex number $\nu$. The particular solution

$$J_{\nu }\left(x\right)=\sum _{m=0}^{\infty }{\displaystyle \frac{{(-1)}^m{\left(\frac{x}{2}\right)}^{2m+\nu }}{m!\Gamma (m+\nu +1)}},x>0,$$

(17)

is called Bessel’s function of the first kind order $\nu$ [4]. Another way to represent it is by Poisson’s integral

$$J_{\nu }\left(z\right)=\frac{2{\left(\frac{z}{2}\right)}^{\nu }}{\Gamma \left(\frac{1}{2}\right)\Gamma \left(\nu +\frac{1}{2}\right)}{\int }_0^{\pi /2}{sin}^{2\nu }\theta cos(zcos\theta )d\theta ,\mathrm{Re}\nu >-\frac{1}{2},$$

(18)

proved by Poisson [5] and Lommel [6] to be a solution of Bessel’s homogenous differential equation for $2\nu \in {\mathbb{N}}_0$; relying on the summation of trigonometric series, in [7], we derived summation formulas for the series

$$\sum _{n=1}^{\infty }{\displaystyle \frac{J_{\nu }\left(nx\right)}{n^{\alpha }}}={\displaystyle \frac{\pi {\left(\frac{x}{2}\right)}^{\alpha -1}sec\frac{\pi (\alpha -\nu )}{2}}{2\Gamma \left(\frac{\alpha -\nu +1}{2}\right)\Gamma \left(\frac{\alpha +\nu +1}{2}\right)}}+\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k{\left(\frac{x}{2}\right)}^{\nu +2k}\zeta (\alpha -\nu -2k)}{k!\Gamma (\nu +k+1)}},$$

(19)

where $\alpha >0,\alpha >\nu >-{\displaystyle \frac{1}{2}},0<x<2\pi$.

Providing $\alpha -\nu =2m$, $m\in \mathbb{N}$, the right-hand-side series in Equation (19) truncates because Riemann’s zeta function equals zero if the argument is a negative even integer. So, setting $\alpha =\nu +2m$ in (19) brings the series in closed form

$$\sum _{n=1}^{\infty }{\displaystyle \frac{J_{\nu }\left(nx\right)}{n^{\nu +2m}}}={\displaystyle \frac{m!{(-1)}^m{x}^{\nu +2m-1}\sqrt{\pi }}{2^{\nu }\left(2m\right)!\Gamma (\nu +m+\frac{1}{2})}}+\sum _{k=0}^m{\displaystyle \frac{{(-1)}^k\zeta (2m-2k){\left(\frac{x}{2}\right)}^{\nu +2k}}{k!\Gamma (\nu +k+1)}},$$

(20)

where $0<x<2\pi ,\mathrm{Re}\nu >-2m-{\displaystyle \frac{1}{2}}$.

In the case $\alpha -\nu =2m-1,m\in \mathbb{N}$, we cannot immediately place this on the right-hand side of the relation Equation (19) because one encounters singularities, i.e., in $sec{\displaystyle \frac{\pi (\alpha -\nu )}{2}}$ within the numerator of the first term and in the member of the right-hand series for the index $k=m-1$.

Theorem 5.

For $\alpha =\nu +2m-1,\nu >-{\displaystyle \frac{1}{2}},m\in \mathbb{N}$, there holds

$$\begin{array}{cc}\hfill \sum _{n=1}^{\infty }& {\displaystyle \frac{J_{\nu }\left(nx\right)}{n^{\nu +2m-1}}}={\displaystyle \frac{{(-1)}^{m-1}{\left(\frac{x}{2}\right)}^{\nu +2m-2}}{2\Gamma \left(m\right)\Gamma (\nu +m)}}\left(H_{m-1}+H_{\nu +m-1}-2log\frac{x}{2}\right)\hfill \\ \hfill & +\sum _{k=0}^{m-2}{\displaystyle \frac{{(-1)}^k\zeta (2m-2k-1){\left(\frac{x}{2}\right)}^{\nu +2k}}{k!\Gamma (\nu +k+1)}}+\sum _{k=1}^{\infty }{\displaystyle \frac{\Gamma \left(2k\right)\zeta \left(2k\right){\left(\frac{x}{4\pi }\right)}^{2k}}{\Gamma (m+k)\Gamma (\nu +m+k)}},\hfill \end{array}$$

(21)

with harmonic numbers $H_{m-1}$ and $H_{\nu +m-1}$.

Proof. 

Because of what is said above, it is necessary to take the limiting value in Equation (19) when $\alpha \to \nu +2m-1$

$$\underset{\alpha \to \nu +2m-1}{lim}({\displaystyle \frac{\frac{\pi }{2}{\left(\frac{x}{2}\right)}^{\alpha -1}sec\frac{\pi (\alpha -\nu )}{2}}{\Gamma \left(\frac{\alpha -\nu +1}{2}\right)\Gamma \left(\frac{\alpha +\nu +1}{2}\right)}}+\sum _{k=0}^{m-1}{\displaystyle \frac{{(-1)}^k\zeta (\alpha -\nu -2k){\left(\frac{x}{2}\right)}^{\nu +2k}}{k!\Gamma (\nu +k+1)}}.$$

For $k=0,1,\cdots ,m-2$, all the terms have no singularities if $\alpha =\nu +2m-1$, so it suffices to deal only with the term for $k=m-1$; then, we find

$$\begin{array}{c}\hfill \underset{\alpha \to \nu +2m-1}{lim}\left({\displaystyle \frac{\frac{\pi }{2}{\left(\frac{x}{2}\right)}^{\alpha -1}sec\frac{\pi (\alpha -\nu )}{2}}{\Gamma \left(\frac{\alpha -\nu +1}{2}\right)\Gamma \left(\frac{\alpha +\nu +1}{2}\right)}}+{\displaystyle \frac{{(-1)}^{m-1}{\left(\frac{x}{2}\right)}^{\nu +2m-2}\zeta (\alpha -\nu -2m+2)}{(m-1)!\Gamma (\nu +m)}}\right)\\ \hfill ={\displaystyle \frac{{(-1)}^{m-1}{\left(\frac{x}{2}\right)}^{\nu +2m-2}}{2\Gamma \left(m\right)\Gamma (\nu +m)}}\left(H_{m-1}+H_{\nu +m-1}-2log\frac{x}{2}\right),\end{array}$$

where $\gamma$ is Euler–Mascheroni’s constant. Thus, we have

$$\begin{array}{cc}\hfill \sum _{n=1}^{\infty }& {\displaystyle \frac{J_{\nu }\left(nx\right)}{n^{\nu +2m-1}}}={\displaystyle \frac{{(-1)}^{m-1}{\left(\frac{x}{2}\right)}^{\nu +2m-2}}{2\Gamma \left(m\right)\Gamma (\nu +m)}}\left(H_{m-1}+H_{\nu +m-1}-2log\frac{x}{2}\right)\hfill \\ \hfill & +\sum _{k=0}^{m-2}{\displaystyle \frac{{(-1)}^k\zeta (2m-2k-1){\left(\frac{x}{2}\right)}^{\nu +2k}}{k!\Gamma (\nu +k+1)}}+\sum _{k=m}^{\infty }{\displaystyle \frac{{(-1)}^k\zeta (2m-2k-1){\left(\frac{x}{2}\right)}^{\nu +2k}}{k!\Gamma (\nu +k+1)}}.\hfill \end{array}$$

As for the remainder, we transform it by shifting the index, so we introduce the substitution $k=m+j-1$ and then revert to k instead of using j, i.e.,

$$\sum _{k=m}^{\infty }{\displaystyle \frac{{(-1)}^k\zeta (2m-2k-1){\left(\frac{x}{2}\right)}^{\nu +2k}}{k!\Gamma (\nu +k+1)}}={(-1)}^{m-1}{\left(\frac{x}{2}\right)}^{\nu +2m-2}\sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^k\zeta (1-2k){\left(\frac{x}{2}\right)}^{2k}}{\Gamma (m+k)\Gamma (\nu +m+k)}}.$$

Using Equation (8), substituting k for n, the last series becomes

$$2{(-1)}^{m-1}{\left(\frac{x}{2}\right)}^{\nu +2m-2}\sum _{k=1}^{\infty }{\displaystyle \frac{\Gamma \left(2k\right)\zeta \left(2k\right){\left(\frac{x}{4\pi }\right)}^{2k}}{\Gamma (m+k)\Gamma (\nu +m+k)}},$$

whereby we arrive at Equation (21). □

In a special case, setting $\nu ={\displaystyle \frac{1}{2}}$ in Equation (21), we have

$$\begin{array}{cc}\hfill & \sum _{n=1}^{\infty }{\displaystyle \frac{J_{1/2}\left(nx\right)}{n^{2m-1/2}}}={\displaystyle \frac{{(-1)}^{m-1}{\left(\frac{x}{2}\right)}^{2m-3/2}}{2\Gamma \left(m\right)\Gamma (\frac{1}{2}+m)}}\left(H_{m-1}+H_{m-{\displaystyle \frac{1}{2}}}-2log\frac{x}{2}\right)\hfill \\ \hfill & +\sqrt{{\displaystyle \frac{x}{2}}}\sum _{k=0}^{m-2}{\displaystyle \frac{{(-1)}^k\zeta (2m-2k-1){\left(\frac{x}{2}\right)}^{2k}}{k!\Gamma (\frac{1}{2}+k+1)}}+\sum _{k=1}^{\infty }{\displaystyle \frac{\Gamma \left(2k\right)\zeta \left(2k\right){\left(\frac{x}{4\pi }\right)}^{2k}}{\Gamma (m+k)\Gamma (\frac{1}{2}+m+k)}}.\hfill \end{array}$$

(22)

Relying on Legendre’s duplication formula for the gamma function ([8], p. 35)

$$\sqrt{\pi }\Gamma \left(2z\right)=2^{2z-1}\Gamma \left(z\right)\Gamma \left(z+\frac{1}{2}\right).$$

(23)

where we set $z=m+n$, we modify the right-hand side series in Equation (22)

$$\sum _{k=1}^{\infty }{\displaystyle \frac{\Gamma \left(2k\right)\zeta \left(2k\right){\left(\frac{x}{4\pi }\right)}^{2k}}{\Gamma (m+k)\Gamma (\frac{1}{2}+m+k)}}={\displaystyle \frac{{(-1)}^{m-1}4x^{2m-3/2}}{\sqrt{2\pi }}}\sum _{k=1}^{\infty }{\displaystyle \frac{\zeta \left(2k\right)}{{\left(2k\right)}_{2m}}}{\left({\displaystyle \frac{x}{2\pi }}\right)}^{2k}.$$

and referring to Equation (9), we have

$$\begin{array}{cc}\hfill & {\displaystyle \frac{{(-1)}^{m-1}4x^{2m-3/2}}{\sqrt{2\pi }}}\sum _{k=1}^{\infty }{\displaystyle \frac{\zeta \left(2k\right)}{{\left(2k\right)}_{2m}}}{\left({\displaystyle \frac{x}{2\pi }}\right)}^{2k}\hfill \\ \hfill & ={\displaystyle \frac{{(-1)}^{m}2x^{2m-3/2}}{(2m-1)!\sqrt{2\pi }}}\left(H_{2m-1}-log2\pi \right)-\sqrt{{\displaystyle \frac{2x}{\pi }}}\sum _{k=0}^{m-2}{\displaystyle \frac{{(-1)}^k\zeta (2m-2k-1)x^{2k+1}}{(2k+1)!}}\hfill \\ \hfill & +{\displaystyle \frac{{(-1)}^{m}2{\left(2\pi \right)}^{2m-3/2}}{(2m-1)!\sqrt{x}}}\left({\zeta }^{\prime }\left(1-2m,1-{\displaystyle \frac{x}{2\pi }}\right)-{\zeta }^{\prime }\left(1-2m,1+{\displaystyle \frac{x}{2\pi }}\right)\right).\hfill \end{array}$$

(24)

By differentiating the basic property of Hurwitz’s function

$$\zeta (s,a)=a^{-s}+\zeta (s,a+1),\mathrm{Re}s>1,0<a⩽1,$$

with respect to s and then putting $a={\displaystyle \frac{x}{2\pi }}$ ($0<x⩽2\pi$) and $s=1-2m$ ($m\in \mathbb{N}$), for the last term of Equation (24), we find

$$-{\zeta }^{\prime }\left(1-2m,1+{\displaystyle \frac{x}{2\pi }}\right)=-{\left({\displaystyle \frac{x}{2\pi }}\right)}^{2m-1}log{\displaystyle \frac{x}{2\pi }}-{\zeta }^{\prime }\left(1-2m,{\displaystyle \frac{x}{2\pi }}\right).$$

Taking this substitution to the formula Equation (24), we have

$$\begin{array}{c}\sum _{k=1}^{\infty }{\displaystyle \frac{\Gamma \left(2k\right)\zeta \left(2k\right){\left(\frac{x}{4\pi }\right)}^{2k}}{\Gamma (m+k)\Gamma (\frac{1}{2}+m+k)}}={\displaystyle \frac{{(-1)}^{m-1}4x^{2m-3/2}}{\sqrt{2\pi }}}\sum _{k=1}^{\infty }{\displaystyle \frac{\zeta \left(2k\right)}{{\left(2k\right)}_{2m}}}{\left({\displaystyle \frac{x}{2\pi }}\right)}^{2k}\hfill \\ ={\displaystyle \frac{{(-1)}^{m}2x^{2m-3/2}}{(2m-1)!\sqrt{2\pi }}}(H_{2m-1}-logx)-\sqrt{{\displaystyle \frac{2x}{\pi }}}\sum _{k=0}^{m-2}{\displaystyle \frac{{(-1)}^k\zeta (2m-2k-1)x^{2k+1}}{(2k+1)!}}\\ \hfill +{\displaystyle \frac{{(-1)}^{m}2{\left(2\pi \right)}^{2m-3/2}}{(2m-1)!\sqrt{x}}}\left({\zeta }^{\prime }\left(1-2m,1-{\displaystyle \frac{x}{2\pi }}\right)-{\zeta }^{\prime }\left(1-2m,{\displaystyle \frac{x}{2\pi }}\right)\right).\end{array}$$

(25)

To calculate the harmonic numbers for half-integer indexes in the formula Equation (22), we use the formula

$$H_{m-{\displaystyle \frac{1}{2}}}=2H_{2m-1}-H_{m-1}-2log2,m\in \mathbb{N},$$

(26)

derived by taking the logarithmic derivative of Legendre’s duplication formula Equation (23), making use of the identity $\psi \left(z\right)={\displaystyle \frac{{\Gamma }^{\prime }\left(z\right)}{\Gamma \left(z\right)}}$, and the relation $\psi \left(z\right)=H_{z-1}-\gamma$.

We now express the left-hand side series Equation (22) as a finite sum, taking account of the formula Equation (25). So, we obtain

$$\sum _{n=1}^{\infty }{\displaystyle \frac{J_{\frac{1}{2}}\left(nx\right)}{n^{2m-\frac{1}{2}}}}={\displaystyle \frac{{(-1)}^m{\left(2\pi \right)}^{2m-\frac{1}{2}}}{(2m-1)!\pi \sqrt{x}}}\left({\zeta }^{\prime }\left(1-2m,1-{\displaystyle \frac{x}{2\pi }}\right)-{\zeta }^{\prime }\left(1-2m,{\displaystyle \frac{x}{2\pi }}\right)\right).$$

(27)

4. Series over Spherical Bessel Functions

We can generalize Equation (19) and (27) by considering spherical Bessel functions

$$j_p\left(x\right)=\sqrt{{\displaystyle \frac{\pi }{2x}}}{J}_{p+{\displaystyle \frac{1}{2}}}\left(x\right),p\in {\mathbb{N}}_0,x>0.$$

So, multiplying both sides of Equation (19) by $\sqrt{{\displaystyle \frac{\pi }{2x}}}$ and setting $\nu =p+{\displaystyle \frac{1}{2}},p\in {\mathbb{N}}_0$, for $0<x<2\pi$, we have

$$\sum _{n=1}^{\infty }{\displaystyle \frac{j_p\left(nx\right)}{n^{\alpha }}}={\displaystyle \frac{\pi \sqrt{\pi }{x}^{\alpha -\frac{3}{2}}sec(\frac{\pi (\alpha -p)}{2}-\frac{\pi }{4})}{2^{\alpha +\frac{1}{2}}\Gamma (\frac{\alpha -p}{2}+\frac{1}{4})\Gamma (\frac{\alpha +p}{2}+\frac{3}{4})}}+{\displaystyle \frac{\sqrt{\pi }}{2}}\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k\zeta (\alpha -p-\frac{1}{2}-2k){\left(\frac{x}{2}\right)}^{p+2k}}{k!\Gamma (p+\frac{1}{2}+k+1)}}.$$

(28)

If we place $\alpha =p+2m+{\displaystyle \frac{1}{2}},m\in \mathbb{N}$ in Equation (28), we obtain a finite sum on the right-hand side

$$\sum _{n=1}^{\infty }{\displaystyle \frac{j_p\left(nx\right)}{n^{p+2m+\frac{1}{2}}}}={\displaystyle \frac{\sqrt{\pi }}{2}}\left({\displaystyle \frac{{(-1)}^m\pi {\left(\frac{x}{2}\right)}^{p+2m-1}}{(p+m)!\Gamma (m+\frac{1}{2})}}+\sum _{k=0}^m{\displaystyle \frac{{(-1)}^k\zeta (2m-2k){\left(\frac{x}{2}\right)}^{p+2k}}{k!\Gamma (p+\frac{1}{2}+k+1)}}\right),$$

because the series over $\zeta$ values truncate for $k>m$ since $\zeta (-2n)=0,\in \mathbb{N}$.

However, we cannot set right away $\alpha =2m+p-{\displaystyle \frac{1}{2}},m\in \mathbb{N}$ in the right-hand side of Equation (28), since in the numerator of the first term is then $secm\pi =0$. There appears $\zeta \left(1\right)$ as well for the running index $k=m-1$ of the right-hand side series, which implies that we encounter in both cases singularities, so we must take a limit. We choose in the right-hand series only term for $k=m-1$ to evaluate

$$\underset{\alpha \to p+2m-{\displaystyle \frac{1}{2}}}{lim}{\displaystyle \frac{\sqrt{\pi }}{2}}\left({\displaystyle \frac{\pi x^{\alpha -\frac{3}{2}}sec(\frac{\pi (\alpha -p)}{2}-\frac{\pi }{4})}{2^{\alpha -\frac{1}{2}}\Gamma (\frac{\alpha -p}{2}+\frac{1}{4})\Gamma (\frac{\alpha +p}{2}+\frac{3}{4})}}-{\displaystyle \frac{{(-1)}^m\zeta (\alpha -p-2m+\frac{3}{2}){\left(\frac{x}{2}\right)}^{p+2m-2}}{(m-1)!\Gamma (p+\frac{1}{2}+m)}}\right)$$

by bringing to the same denominator and applying the L’Hôpital rule. Afterward, we use Legendre’s duplication formula Equation (23) to rearrange the right-hand side series of (28). Thus, we obtain

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{j_p\left(nx\right)}{n^{p+2m-1/2}}}={\left(2x\right)}^p(\sum _{k=0}^{m-2}{\displaystyle \frac{{(-x^2)}^k(p+k)!\zeta (2m-1-2k)}{k!(2p+2k+1)!}}\hfill \\ +{(-x^2)}^{m-1}({\displaystyle \frac{(p+m)!\left(H_{m-1}+H_{p+m-\frac{1}{2}}-2log\frac{x}{2}\right)}{(m-1)!(2p+2m)!}}\\ \hfill +2\sum _{n=1}^{\infty }{\displaystyle \frac{{(m+n)}_p}{{\left(2n\right)}_{2p+2m}}}\zeta \left(2n\right){\left({\displaystyle \frac{x}{2\pi }}\right)}^{2n}\left)\right).\end{array}$$

(29)

Referring to Equation (11) and Equation (9), we evaluate the sum of the right-hand side series. To calculate the harmonic numbers for half-integer indexes, we use the formula Equation (26) with $n=p+m$, i.e.

$$H_{p+m-{\displaystyle \frac{1}{2}}}=2H_{2p+2m-1}-H_{p+m-1}-2log2.$$

Example 1.

Setting $m=3$ and $p=2$ in Equation (29), taking account of Equation (11), we have

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{j_2\left(nx\right)}{n^{15/2}}}={\displaystyle \frac{8x^2\zeta \left(5\right)}{5!}}-{\displaystyle \frac{24x^4\zeta \left(3\right)}{7!}}+{\displaystyle \frac{5!\left(H_2+H_{\frac{9}{2}}-2log\frac{x}{2}\right)4x^6}{2!·10!}}\hfill \\ \hfill +2x^6\sum _{n=1}^{\infty }{\displaystyle \frac{\zeta \left(2n\right)}{{\left(2n\right)}_8}}{\left({\displaystyle \frac{x}{2\pi }}\right)}^{2n}-6x^6\sum _{n=1}^{\infty }{\displaystyle \frac{\zeta \left(2n\right)}{{\left(2n\right)}_9}}{\left({\displaystyle \frac{x}{2\pi }}\right)}^{2n}+6x^6\sum _{n=1}^{\infty }{\displaystyle \frac{\zeta \left(2n\right)}{{\left(2n\right)}_{10}}}{\left({\displaystyle \frac{x}{2\pi }}\right)}^{2n}.\end{array}$$

Applying now Equation (26), Equation (6) and Equation (9), we find

$$\begin{array}{c}\sum _{n=1}^{\infty }{\displaystyle \frac{j_2\left(nx\right)}{n^{15/2}}}={\displaystyle \frac{8{\pi }^7}{315x}}\left({\zeta }^{\prime }\left(-7,{\displaystyle \frac{x}{2\pi }}\right)-{\zeta }^{\prime }\left(-7,1-{\displaystyle \frac{x}{2\pi }}\right)\right)-{\displaystyle \frac{2{\pi }^8}{105x^2}}{\zeta }^{\prime }\left(-8,{\displaystyle \frac{x}{2\pi }}\right)\hfill \\ \hfill +{\zeta }^{\prime }\left(-8,1-{\displaystyle \frac{x}{2\pi }}\right))+{\displaystyle \frac{4{\pi }^9}{945x^3}}\left({\zeta }^{\prime }\left(-9,{\displaystyle \frac{x}{2\pi }}\right)-{\zeta }^{\prime }\left(-9,1-{\displaystyle \frac{x}{2\pi }}\right)\right).\end{array}$$

We can obtain the summation formula for the alternating series over spherical Bessel functions. First, considering that there holds

$$\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^{n-1}{J}_{\nu }\left(nx\right)}{n^{\alpha }}}=\sum _{n=1}^{\infty }{\displaystyle \frac{J_{\nu }\left(nx\right)}{n^{\alpha }}}-{\displaystyle \frac{1}{2^{\alpha -1}}}\sum _{n=1}^{\infty }{\displaystyle \frac{J_{\nu }\left(2nx\right)}{n^{\alpha }}},$$

applying Equation (19), we easily find

$$\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^{n-1}{J}_{\nu }\left(nx\right)}{n^{\alpha }}}=\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k{\left(\frac{x}{2}\right)}^{\nu +2k}\eta (\alpha -\nu -2k)}{k!\Gamma (\nu +k+1)}},$$

where we used the relation $\eta (\alpha -\nu -2k)=(1-2^{1-(\alpha -\nu -2k)})\zeta (\alpha -\nu -2k)$, and $\eta$ is Dirichlet’s eta function. For $\nu =p+{\displaystyle \frac{1}{2}}$, after multiplying this equality by ${\displaystyle \sqrt{\frac{\pi }{2x}}}$, we have

$$\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^{n-1}{j}_p\left(nx\right)}{n^{\alpha }}}={\displaystyle \frac{\sqrt{\pi }}{2}}\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k{\left(\frac{x}{2}\right)}^{p+2k}\eta (\alpha -p-\frac{1}{2}-2k)}{k!\Gamma (p+\frac{3}{2}+k)}},$$

Further, using the relation

$$\sum _{n=1}^{\infty }{\displaystyle \frac{J_{\nu }\left((2n-1)x\right)}{{(2n-1)}^{\alpha }}}=\sum _{n=1}^{\infty }{\displaystyle \frac{J_{\nu }\left(nx\right)}{n^{\alpha }}}-{\displaystyle \frac{1}{2^{\alpha }}}\sum _{n=1}^{\infty }{\displaystyle \frac{J_{\nu }\left(2nx\right)}{n^{\alpha }}}$$

and the equality $(1-2^{-(\alpha -\nu -2k)})\zeta (\alpha -\nu -2k)=\lambda (\alpha -\nu -2k)$, where $\lambda$ is Dirichlet’s lambda function, we come to the summation formula for the series over a Bessel function containing odd arguments

$$\sum _{n=1}^{\infty }{\displaystyle \frac{J_{\nu }\left((2n-1)x\right)}{{(2n-1)}^{\alpha }}}={\displaystyle \frac{\pi {\left(\frac{x}{2}\right)}^{\alpha -1}}{4\Gamma \left(\frac{\alpha -\nu +1}{2}\right)\Gamma \left(\frac{\alpha +\nu +1}{2}\right)cos\frac{\pi (\alpha -\nu )}{2}}}+\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k{\left(\frac{x}{2}\right)}^{\nu +2k}\lambda (\alpha -\nu -2k)}{k!\Gamma (\nu +k+1)}}.$$

(30)

Setting $\nu =p+{\displaystyle \frac{1}{2}}$ and multiplying this equality by ${\displaystyle \sqrt{\frac{\pi }{2x}}}$ yields

$$\sum _{n=1}^{\infty }{\displaystyle \frac{j_p\left((2n-1)x\right)}{{(2n-1)}^{\alpha }}}={\displaystyle \frac{{\left(\frac{\pi }{2x}\right)}^{\frac{3}{2}}{\left(\frac{x}{2}\right)}^{\alpha }sec\frac{\pi (\alpha -p-\frac{1}{2})}{2}}{\Gamma \left(\frac{\alpha -p+\frac{1}{2}}{2}\right)\Gamma \left(\frac{\alpha +p+\frac{3}{2}}{2}\right)}}+{\displaystyle \frac{\sqrt{\pi }}{2}}\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k{\left(\frac{x}{2}\right)}^{p+2k}\lambda (\alpha -p-\frac{1}{2}-2k)}{k!\Gamma (p+\frac{3}{2}+k)}}.$$

Finally, replacing Poisson’s integral Equation (18) in the alternating series of Equation (30), we have

$$\begin{array}{cc}\hfill \sum _{n=1}^{\infty }& {\displaystyle \frac{{(-1)}^{n-1}{J}_{\nu }\left((2n-1)x\right)}{{(2n-1)}^{\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{2{\left(\frac{x}{2}\right)}^{\nu }}{\sqrt{\pi }\Gamma (\nu +\frac{1}{2})}}\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^{n-1}}{{(2n-1)}^{\alpha -\nu }}}{\int }_0^{\pi /2}{sin}^{2\nu }\theta cos((2n-1)xcos\theta )d\theta .\hfill \end{array}$$

(31)

Interchanging summation and integration and referring to the formula ([3], p. 446)

$$\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^{n-1}cos(2n-1)x}{{(2n-1)}^{\alpha }}}=\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k\beta (\alpha -2k)}{\left(2k\right)!}}{x}^{2k},-\frac{\pi }{2}<x<\frac{\pi }{2},$$

where $\beta$ is Dirichlet’t beta function, for the right-hand side of Equation (31), we obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{2{\left(\frac{x}{2}\right)}^{\nu }}{\sqrt{\pi }\Gamma (\nu +\frac{1}{2})}}{\int }_0^{\pi /2}{sin}^{2\nu }\theta \left(\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k\beta (\alpha -\nu -2k)x^{2k}}{\left(2k\right)!}}{cos}^{2k}\theta \right)d\theta ,\end{array}$$

and then again, interchanging summation and integration, the identity Equation (31) becomes

$$\begin{array}{cc}\hfill \sum _{n=1}^{\infty }& {\displaystyle \frac{{(-1)}^{n-1}{J}_{\nu }\left((2n-1)x\right)}{{(2n-1)}^{\alpha }}}\hfill \\ \hfill & ={\displaystyle \frac{{\left(\frac{x}{2}\right)}^{\nu }}{\sqrt{\pi }\Gamma (\nu +\frac{1}{2})}}\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k\beta (\alpha -\nu -2k)x^{2k}}{\left(2k\right)!}}2{\int }_0^{\pi /2}{sin}^{2\nu }\theta {cos}^{2k}\theta d\theta \hfill \end{array}$$

The last integral is Euler’s beta function

$$\mathrm{B}(2\nu +1,2k+1)=2{\int }_0^{\pi /2}{sin}^{2(\nu +{\displaystyle \frac{1}{2}})-1}\theta {cos}^{2(k+{\displaystyle \frac{1}{2}})-1}\theta d\theta ={\displaystyle \frac{\Gamma (\nu +\frac{1}{2})\Gamma (k+\frac{1}{2})}{\Gamma (\nu +k+1)}}.$$

Thus, applying Equation (23), we come to the summation formula of an alternating series over odd arguments Bessel functions:

$$\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^{n-1}{J}_{\nu }\left((2n-1)x\right)}{{(2n-1)}^{\alpha }}}=\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k\beta (\alpha -\nu -2k){\left(\frac{x}{2}\right)}^{\nu +2k}}{k!\Gamma (k+\nu +1)}}.$$

The beta function vanishes at negative odd integers ([3], p. 447). That is why, for $\alpha =\nu +2m-1,m\in \mathbb{N}$, the series over Bessel functions in last formula has a finite sum

$$\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^{n-1}{J}_{\nu }\left((2n-1)x\right)}{{(2n-1)}^{\nu +2m-1}}}=\sum _{k=0}^{m-1}{\displaystyle \frac{{(-1)}^k\beta (2m-1-2k){\left(\frac{x}{2}\right)}^{\nu +2k}}{k!\Gamma (k+\nu +1)}}.$$

Putting $\nu =p+{\displaystyle \frac{1}{2}}$ and multiplying both sides by ${\displaystyle \sqrt{\frac{\pi }{2x}}}$, we arrive at the summation formula for the alternating series over spherical Bessel functions

$$\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^{n-1}{j}_p\left((2n-1)x\right)}{{(2n-1)}^{\alpha }}}={\displaystyle \frac{\sqrt{\pi }}{2}}\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k\beta (\alpha -p-\frac{1}{2}-2k){\left(\frac{x}{2}\right)}^{p+2k}}{k!\Gamma \left(k+p+\frac{3}{2}\right)}}.$$