On Closed Forms of Some Trigonometric Series

Abstract

We have derived alternative closed-form formulas for the trigonometric series over sine or cosine functions when the immediate replacement of the parameter appearing in the denominator with a positive integer gives rise to a singularity. By applying the Choi–Srivastava theorem, we reduce these trigonometric series to expressions over Hurwitz’s zeta function derivative.

1. Preliminaries

All particular cases of the general summation formula for the trigonometric series

$$T_{\alpha }^f={\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{{\left(s\right)}^{n-1}f\left((an-b)x\right)}{{(an-b)}^{\alpha }}},$$

(1)

with parameters $s=1\mathrm{or}s=-1,f=sin\mathrm{or}f=cos,a=\left\{{\displaystyle \genfrac{}{}{0pt}{}{1}{2}}\right\}b=\left\{{\displaystyle \genfrac{}{}{0pt}{}{0}{1}}\right\},\alpha >0$, we derived in [1] and expressed as a single formula via the power series

$$T_{\alpha }^f={\displaystyle \frac{c\pi x^{\alpha -1}}{2\Gamma \left(\alpha \right)f\left(\frac{\pi \alpha }{2}\right)}}+{\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{(-1)}^{k}F(\alpha -2k-\delta )}{(2k+\delta )!}}{x}^{2k+\delta },$$

(2)

and one can obtain each of them by looking up in

Table 1. Particular and closed-form cases of (2).

a	b	s	c	F	f	$\mathit{\delta }$	p	Convergence Region
1	0	1	1	ζ	sin	1	0	0 < x < 2π
					cos	0	1
		−1	0	η	sin	1	0	−π < x < π
					cos	0	1
2	1	1	${\displaystyle \frac{1}{2}}$	λ	sin	1	0	0 < x < π
					cos	0	1
		−1	0	β	sin	1	1	$-{\displaystyle \frac{\pi }{2}}<x<{\displaystyle \frac{\pi }{2}}$
					cos	0	0

and taking the corresponding parameters, including $s,a,b,f,c$ and $\delta$, with F standing for Riemann’s $\zeta$ function, initially defined by the series [2]

$$\zeta \left(z\right)={\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n^z}},\mathrm{Re}z>1,$$

(3)

or Dirichlet $\eta ,\lambda ,\beta$ functions defined as follows:

$$\eta \left(z\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^{n-1}}{n^z}},\lambda \left(z\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{1}{{(2n-1)}^z}},\beta \left(z\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^{n-1}}{{(2n-1)}^z}}.$$

(4)

Riemann’s $\zeta$ function satisfies the functional equation [3]

$$\zeta (1-z)={\displaystyle \frac{2\zeta \left(z\right)}{{\left(2\pi \right)}^z}}\Gamma \left(z\right)cos{\displaystyle \frac{\pi z}{2}},$$

(5)

where $\Gamma \left(s\right)$ is the gamma function [4]

$$\Gamma \left(z\right)={\int }_0^{\infty }{x}^{z-1}{e}^{-x}\mathrm{d}x,\mathrm{Re}z>0,$$

(6)

introduced by Euler. On the other hand, integration by parts of the integral (6) gives rise to the basic relation

$$\Gamma (z+1)=z\Gamma \left(z\right),$$

(7)

which can be extended for arbitrary $n\in \mathbb{N}$:

$$\Gamma (z+n)=(z+n-1)(z+n-2)\cdots z\Gamma \left(z\right),$$

and this formula one can use to express the Pochhammer symbol:

$${\left(z\right)}_n=z(z+1)(z+2)\cdots (z+n-1)={\displaystyle \frac{\Gamma (z+n)}{\Gamma \left(z\right)}}.$$

(8)

Rewriting integral (6) as

$$\begin{array}{cc}\hfill \Gamma \left(z\right)={\int }_0^{\infty }{x}^{z-1}{e}^{-x}\mathrm{d}x& ={\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{{(-1)}^n}{n!}}{\int }_0^1{x}^{z+n-1}\mathrm{d}x+{\int }_1^{\infty }{x}^{s-1}{e}^{-x}\mathrm{d}x\hfill \\ \hfill & ={\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{{(-1)}^n}{n!}}{\displaystyle \frac{1}{z+n}}+{\int }_1^{\infty }{x}^{z-1}{e}^{-x}\mathrm{d}x\hfill \end{array}$$

provides the analytic continuation of the gamma function for all complex numbers, except for integers less than or equal to zero.

For $z\notin \mathbb{Z}$, Euler’s reflection formula [4] (p. 35) holds:

$$\Gamma (1-z)\Gamma \left(z\right)={\displaystyle \frac{\pi }{sin\pi z}}.$$

(9)

The first logarithmic derivative of the gamma function [5] is the function $\psi$ (also known as the digamma), i.e.,

$$\psi \left(s\right)={\displaystyle \frac{\mathrm{d}}{\mathrm{d}s}}log\Gamma \left(s\right)={\displaystyle \frac{{\Gamma }^{\prime }\left(s\right)}{\Gamma \left(s\right)}},$$

(10)

whence we find $\psi \left(1\right)={\Gamma }^{\prime }\left(1\right)$. It is connected to the nth harmonic number $H_n$, given as the sum of reciprocal values of the first n positive integers by the equality

$$H_{n-1}=\psi \left(n\right)+\gamma ,n\in \mathbb{N},H_0=0,$$

(11)

where $\gamma$ is the Euler–Mascheroni constant, obtained as the limiting value of the sequence $H_n-logn$ as $n\to \infty$.

As an extension of the Riemann zeta function the Hurwitz zeta function appears, initially defined by the series [2]

$$\zeta (z,a)=\sum _{k=0}^{\infty }{\displaystyle \frac{1}{{(k+a)}^z}},\mathrm{Re}z>1,,a\ne 0,-1,-2,\cdots .$$

(12)

Obviously, $\zeta (z,1)=\zeta \left(z\right)$. A straightforward consequence of (12) is

$$\zeta (z,a)={\displaystyle \frac{1}{a^z}}+{\displaystyle \frac{1}{{(1+a)}^z}}+{\displaystyle \frac{1}{{(2+a)}^z}}+\cdots =a^{-z}+\zeta (z,a+1),$$

(13)

and because of that, for most purposes, it suffices to restrict a to $0<\mathrm{Re}a⩽1$. In addition, most references treat a as a real number with $0<a⩽1$. Considering a function of a, with $z\ne 1$ fixed, $\zeta (z,a)$ is analytic in the half-plane $\mathrm{Re}a>0$.

The Hurwitz zeta function satisfies an identity that generalizes the functional equation of the Riemann zeta function [3] (p. 257, Theorem 12.6)

$$\zeta (1-z,a)={\displaystyle \frac{\Gamma \left(z\right)}{{\left(2\pi \right)}^z}}\left(e^{-i\pi z/2}\sum _{k=1}^{\infty }{\displaystyle \frac{e^{2k\pi ia}}{k^z}}+e^{i\pi z/2}\sum _{k=0}^{\infty }{\displaystyle \frac{e^{-2k\pi ia}}{k^z}}\right)$$

but we rewrite it in a more suitable form:

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

(14)

For $z=1$ in (14), by relying on Euler’s famous formula $e^{i\phi }=cos\phi +isin\phi$ and elementary properties of the complex logarithm, we evaluate

$$\zeta (0,a)={\displaystyle \frac{1}{\pi }}\sum _{n=1}^{\infty }{\displaystyle \frac{sin2n\pi a}{n}}={\displaystyle \frac{i}{2\pi }}log{\displaystyle \frac{1-e^{2i\pi a}}{1-e^{-2i\pi a}}}={\displaystyle \frac{1}{2}}-a.$$

(15)

Multiplying $\zeta (z,a)$ by $\Gamma \left(z\right)$, taking account of (12) and (6), we have

$$\zeta (z,a)\Gamma \left(z\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{1}{{(k+a)}^z}}{\int }_0^{\infty }{x}^{z-1}{e}^{-x}\mathrm{d}x.$$

Introducing the substitution $x=(k+a)t$, then interchanging the sum and integral, we obtain the integral representation of the Hurwitz zeta function:

$$\zeta (z,a)={\displaystyle \frac{1}{\Gamma \left(z\right)}}{\int }_0^{\infty }{t}^{z-1}{e}^{-at}\sum _{k=0}^{\infty }{e}^{-kt}\mathrm{d}t={\displaystyle \frac{1}{\Gamma \left(z\right)}}{\int }_0^{\infty }{\displaystyle \frac{t^{z-1}{e}^{-at}}{1-e^{-t}}}\mathrm{d}t.$$

(16)

Given (4), one easily derives the relations of the $\zeta$ function to $\eta$ and $\lambda$:

$$\eta \left(z\right)=(1-2^{1-z})\zeta \left(z\right),\lambda \left(z\right)=(1-2^{-z})\zeta \left(z\right).$$

(17)

We conclude that the $\eta$ function extends analytically to the whole complex plane, including $z=1$, since

$$\eta \left(1\right)={\displaystyle \underset{z\to 1}{lim}}\eta \left(z\right)={\displaystyle \underset{z\to 1}{lim}}(1-2^{1-z})\zeta \left(z\right)={\displaystyle \underset{z\to 1}{lim}}{\displaystyle \frac{1-2^{1-z}}{z-1}}(z-1)\zeta \left(z\right).$$

Further, using (5), then setting $a=1$ in (15), we find

$$\eta \left(1\right)=log2{\displaystyle \underset{z\to 1}{lim}}{\displaystyle \frac{\frac{\pi }{2}(z-1)}{sin\frac{\pi }{2}(1-z)}}{\displaystyle \frac{{\left(2\pi \right)}^z\zeta (1-z)}{\pi \Gamma \left(z\right)}}=-2\zeta \left(0\right)log2=log2.$$

However, (17) tells us that, as for the $\zeta$ function, $\lambda$ is analytical for all complex numbers except $z=1$, where it has a non-removable singularity. In addition, (17) immediately gives rise to $\eta (-2n)=\lambda (-2n)=0$, $n\in \mathbb{N}$, because by setting $s=2n+1,n\in \mathbb{N}$, in the functional equation for the Riemann zeta function (5), we find

$$\zeta (-2n)={\displaystyle \frac{2\left(2n\right)!}{{\left(2\pi \right)}^z}}\zeta (2n+1)cos{\displaystyle \frac{\pi }{2}}(2n+1)=0.$$

Because of

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{{(-1)}^n}{{(2n+1)}^s}}& =1-{\displaystyle \frac{1}{3^s}}+{\displaystyle \frac{1}{5^s}}-{\displaystyle \frac{1}{7^s}}+\cdots ={\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{1}{{(4n+1)}^z}}-{\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{1}{{(4n+3)}^z}}\hfill \\ \hfill & ={\displaystyle \frac{1}{4^z}}\left({\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{1}{{(n+\frac{1}{4})}^z}}-{\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{1}{{(n+\frac{3}{4})}^z}}\right)={\displaystyle \frac{1}{4^z}}\zeta \left(s,\frac{1}{4}\right)-\zeta \left(s,\frac{3}{4}\right),\hfill \end{array}$$

we express the $\beta$ function through a difference of two Hurwitz functions:

$$\beta \left(z\right)=4^{-s}\left(\zeta \left(s,\frac{1}{4}\right)-\zeta \left(s,\frac{3}{4}\right)\right).$$

(18)

Using (16), we find its integral, represented as

$$\begin{array}{cc}\hfill \beta \left(z\right)& ={\displaystyle \frac{1}{4^z}}\left(\zeta \left(s,\frac{1}{4}\right)-\zeta \left(s,\frac{3}{4}\right)\right)={\displaystyle \frac{1}{\Gamma \left(z\right)}}{\int }_0^{\infty }{\displaystyle \frac{t^{z-1}{e}^{-\frac{t}{4}}}{1+e^{-\frac{t}{2}}}}\mathrm{d}t\hfill \\ \hfill & ={\displaystyle \frac{1}{\Gamma \left(s\right)}}{\int }_0^{\infty }{\displaystyle \frac{x^{s-1}{e}^{-x}}{1+e^{-2x}}}\mathrm{d}x,\hfill \end{array}$$

(19)

which defines $\beta$ as an analytical function for $\mathrm{Re}s>0$, but employing the functional equations conjectured by Euler in 1749 and proved by the Swedish mathematician Malmsten in 1842,

$$\beta (1-z)={\left({\displaystyle \frac{2}{\pi }}\right)}^{z}sin{\displaystyle \frac{\pi s}{2}}\Gamma \left(z\right)\beta \left(z\right).$$

(20)

$\beta$ extends to the whole complex plane. By virtue of (9), Equation (20) can be rewritten as follows:

$$\beta \left(z\right)={\left({\displaystyle \frac{\pi }{2}}\right)}^{z-1}\beta (1-z)\Gamma (1-z)cos{\displaystyle \frac{\pi z}{2}},$$

(21)

whence we find $\beta (-2n+1)=0$, $n\in \mathbb{N}$.

Thus, owing to

$$\zeta (-2n)=\eta (-2n)=\lambda (-2n)=\beta (-2n+1)=0,n\in \mathbb{N}$$

the right-hand side series in (2) truncates. Thus, we obtain [1], one type of closed-form formula. So, for $\alpha =2m+p-1$, where $m\in \mathbb{N}$ and $p=0$ or $p=1$, we have all these cases comprised by the general formula [1]

$$\begin{array}{cc}\hfill T_{2m+p-1}^f={\displaystyle \sum _{n=1}^{\infty }}& {\displaystyle \frac{{\left(s\right)}^{n-1}f\left((an-b)x\right)}{{(an-b)}^{2m+p-1}}}={\displaystyle \frac{c\pi x^{2m+p-2}}{2(2m+p-2)!f(m\pi +\frac{\pi }{2}(p-1))}}\hfill \\ \hfill & +{\displaystyle \sum _{k=0}^{m+p-1}}{\displaystyle \frac{{(-1)}^{k}F(2m+p-1-2k-\delta )}{(2k+\delta )!}}{x}^{2k+\delta }.\hfill \end{array}$$

We can obtain each from Table 1 by choosing the corresponding parameters.

2. Alternative Closed-Form Formulas

For $a=1,b=0,s=1$ in (1), and putting $f=sin$, then $f=cos$, we have

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{sinnx}{n^{\alpha }}}& ={\displaystyle \frac{\pi x^{\alpha -1}}{2\Gamma \left(\alpha \right)sin\frac{\pi }{2}\alpha }}+{\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{(-1)}^k\zeta (\alpha -2k-1)}{(2k+1)!}}{x}^{2k+1},\hfill \\ \hfill {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{cosnx}{n^{\alpha }}}& ={\displaystyle \frac{\pi x^{\alpha -1}}{2\Gamma \left(\alpha \right)cos\frac{\pi }{2}\alpha }}+{\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{(-1)}^k\zeta (\alpha -2k)}{\left(2k\right)!}}{x}^{2k},0<x<2\pi ,\hfill \end{array}$$

(22)

where $\zeta$ presents Riemann’s zeta function.

However, by taking limits $\alpha \to 2m$ and $\alpha \to 2m-1$ in the first and the second formula of (22), respectively, one encounters singularities of the $\zeta$ function, so we have to act differently.

Theorem 1. 

Letting $\alpha \to 2m$ in the first formula of (22), one brings the sine series in closed form:

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

(23)

where $0<x<2\pi$. This series defines the Clausen function ${\mathrm{Cl}}_{2m}$ [1] (p. 452).

Proof. 

We are not allowed to replace $\alpha$ with $2m$ in the first formula of (22) since $sin{\displaystyle \frac{\pi }{2}}\alpha =0$ and in the term for $k=m-1$ in the series, and we encounter singularities. We recall the Hurwitz formula connecting a trigonometric series and the Hurwitz zeta function.

In the first derivative of the Formula (14) for $z=2m$ and $a={\displaystyle \frac{x}{2\pi }}$, we have

$$\begin{array}{c}-{\zeta }^{\prime }\left(1-2m,{\displaystyle \frac{x}{2\pi }}\right)={\displaystyle \frac{2{(-1)}^m\Gamma \left(2m\right)}{{\left(2\pi \right)}^{2m}}}\left(\psi \left(2m\right)-log2\pi \right){\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{cosnx}{n^{2m}}}\hfill \\ \hfill -{\displaystyle \frac{2{(-1)}^m\Gamma \left(2m\right)}{{\left(2\pi \right)}^{2m}}}{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{cosnx}{n^{2m}}}logn-{\displaystyle \frac{\pi {(-1)}^{m-1}\Gamma \left(2m\right)}{{\left(2\pi \right)}^{2m}}}{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{sinnx}{n^{2m}}}.\end{array}$$

For $a=1-{\displaystyle \frac{x}{2\pi }}$, we have

$$\begin{array}{c}-{\zeta }^{\prime }\left(1-2m,1-{\displaystyle \frac{x}{2\pi }}\right)={\displaystyle \frac{2{(-1)}^m\Gamma \left(2m\right)}{{\left(2\pi \right)}^{2m}}}\left(\psi \left(2m\right)-log2\pi \right){\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{cosnx}{n^{2m}}}\hfill \\ \hfill -{\displaystyle \frac{2{(-1)}^m\Gamma \left(2m\right)}{{\left(2\pi \right)}^{2m}}}{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{cosnx}{n^{2m}}}logn-{\displaystyle \frac{\pi {(-1)}^m\Gamma \left(2m\right)}{{\left(2\pi \right)}^{2m}}}{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{sinnx}{n^{2m}}}.\end{array}$$

By subtracting these equalities, we obtain

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

Hence, we obtain the sum of the sine series. □

Theorem 2. 

Letting $\alpha \to 2m-1$ in the second formula of (22), one brings the cosine series in closed form:

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

(24)

where $0<x<2\pi$. This series defines the Clausen function ${\mathrm{Cl}}_{2m-1}$ [1] (p. 452).

Proof. 

Similarly, one cannot immediately replace $\alpha$ with $2m-1$, so it is necessary to perform a different method. Taking the first derivative of Formula (14) with respect to z for $a={\displaystyle \frac{x}{2\pi }}$, we have

$$\begin{array}{c}-{\zeta }^{\prime }\left(1-s,{\displaystyle \frac{x}{2\pi }}\right)={\displaystyle \frac{2\Gamma \left(s\right)}{{\left(2\pi \right)}^s}}\left(\psi \left(s\right)-log2\pi \right){\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{cos\left(\frac{\pi s}{2}-nx\right)}{n^s}}\hfill \\ \hfill -{\displaystyle \frac{2\Gamma \left(s\right)}{{\left(2\pi \right)}^s}}{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{cos\left(\frac{\pi s}{2}-nx\right)}{n^s}}logn-{\displaystyle \frac{\pi \Gamma \left(s\right)}{{\left(2\pi \right)}^s}}{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{sin\left(\frac{\pi s}{2}-nx\right)}{n^s}}.\end{array}$$

Setting here $z=2m-1$, we find

$$\begin{array}{c}-{\zeta }^{\prime }\left(2-2m,{\displaystyle \frac{x}{2\pi }}\right)={\displaystyle \frac{2{(-1)}^{m-1}\Gamma (2m-1)}{{\left(2\pi \right)}^{2m-1}}}\left(\psi (2m-1)-log2\pi \right){\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{sinnx}{n^{2m-1}}}\hfill \\ \hfill -{\displaystyle \frac{2{(-1)}^{m-1}\Gamma (2m-1)}{{\left(2\pi \right)}^{2m-1}}}{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{sinnx}{n^{2m-1}}}logn-{\displaystyle \frac{\pi {(-1)}^{m-1}\Gamma (2m-1)}{{\left(2\pi \right)}^{2m-1}}}{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{cosnx}{n^{2m-1}}}.\end{array}$$

For $a=1-{\displaystyle \frac{x}{2\pi }}$, we have

$$\begin{array}{c}-{\zeta }^{\prime }\left(2-2m,1-{\displaystyle \frac{x}{2\pi }}\right)={\displaystyle \frac{2{(-1)}^m\Gamma (2m-1)}{{\left(2\pi \right)}^{2m-1}}}\left(\psi (2m-1)-log2\pi \right){\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{sinnx}{n^{2m-1}}}\hfill \\ \hfill -{\displaystyle \frac{2{(-1)}^m\Gamma (2m-1)}{{\left(2\pi \right)}^{2m-1}}}{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{sinnx}{n^{2m-1}}}logn-{\displaystyle \frac{\pi {(-1)}^{m-1}\Gamma (2m-1)}{{\left(2\pi \right)}^{2m-1}}}{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{cosnx}{n^{2m-1}}}.\end{array}$$

By adding these equalities, we obtain

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

Thence, we obtain the sum of the cosine series. □

For $a=1,b=0,s=-1$ and putting $f=sin$ in (1), based on (22) and using its relation to the zeta function, (17), we obtain the alternating series over the sines and cosines expressed in terms of the Dirichlet eta function:

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{{(-1)}^{n-1}sinnx}{n^{\alpha }}}& ={\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{{(-1)}^{n-1}sinnx}{n^{\alpha }}}-{\displaystyle \frac{1}{2^{\alpha -1}}}{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{sin2nx}{n^{\alpha }}}\hfill \\ \hfill & ={\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{(-1)}^k\eta (\alpha -2k-1)}{(2k+1)!}}{x}^{2k+1},-\pi <x<\pi .\hfill \end{array}$$

(25)

Similarly, for $a=1,b=0,s=-1$ and putting $f=cos$ in (1), we find

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

(26)

Theorem 3. 

For $\alpha =2m$, the series (25) takes the closed form

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

(27)

Proof. 

These replacements are legitimate since we encounter no singularities in (25).

Further, from (25), we have

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{{(-1)}^{n-1}sinnx}{n^{2m}}}=\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k\eta (2m-2k-1)}{(2k+1)!}}{x}^{2k+1},$$

and split the right-hand side series as follows:

$$\begin{array}{c}{\displaystyle \sum _{k=0}^{m-2}}{\displaystyle \frac{{(-1)}^k\eta (2m-2k-1)}{(2k+1)!}}{x}^{2k+1}+{\displaystyle \frac{{(-1)}^{m-1}{x}^{2m-1}}{(2m-1)!}}log2\hfill \\ \hfill +{\displaystyle \sum _{k=m}^{\infty }}{\displaystyle \frac{{(-1)}^k\eta (2m-2k-1)}{(2k+1)!}}{x}^{2k+1}.\end{array}$$

(28)

Taking account of $\eta (2m-2k-1)=(1-2^{1-(2m-2k-1)})\zeta (2m-2k-1)$, we can express the last right-hand side series in terms of the zeta function as a difference of two series:

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

(29)

If we set $z=2n$ in (5), we can determine values of Riemann’s zeta function at negative odd integers:

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

(30)

By shifting indices in both series of (29) and applying (30) and (8), the difference (29) becomes

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

(31)

where ${\left(2k\right)}_{2m}$ denotes Pochhammer’s symbol, given by (8). We refer now to the theorem in [6] (p. 419), which we state here in a slightly modified form.

Theorem 1. 

For every non-negative integer n there holds

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{k=2}^{\infty }}{\displaystyle \frac{\zeta (k,a)}{{\left(k\right)}_{n+1}}}{t}^{n+k}={\displaystyle \frac{{(-1)}^n}{n!}}({\zeta }^{\prime }(-n,a-t)-{\zeta }^{\prime }(-n,a)+{\displaystyle \sum _{k=1}^n}{(-t)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\times \hfill \\ \hfill & \left(\zeta (k-n,a)(H_n-H_{n-k})-{\zeta }^{\prime }(k-n,a)\right))+\left(H_n+\psi \left(a\right)\right){\displaystyle \frac{t^{n+1}}{(n+1)!}},\hfill \end{array}$$

(32)

with $\left|t\right|<\left|a\right|,n\in {\mathbb{N}}_0$.

For the first series of (31), we set in (32) $a=1$, $n=2m-1$, $t={\displaystyle \frac{x}{2\pi }}$, then again, $a=1$, $n=2m-1$, $t=-{\displaystyle \frac{x}{2\pi }}$, and subtract the second equality from the first. In this way, we obtain

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

Knowing that $\zeta (2k-2m)=0$ for $k=1,\cdots ,m-1$ and $\zeta \left(0\right)=-1/2$, $H_0=0$, the right-hand side sum becomes

$${\left({\displaystyle \frac{x}{2\pi }}\right)}^{2m-1}{H}_{2m-1}+2{\displaystyle \sum _{k=1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{2m-1}{2k-1}}\right){\left({\displaystyle \frac{x}{2\pi }}\right)}^{2k-1}{\zeta }^{\prime }(2k-2m).$$

Taking the derivative with respect to s on both sides of (13), then putting there $a={\displaystyle \frac{x}{2\pi }}$ and $s=1-2m$, we find

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

We obtain the same structure for the second series, with the sole difference that instead of ${\displaystyle \frac{x}{2\pi }}$, there appears ${\displaystyle \frac{x}{\pi }}$. The subtraction yields the sum of (31), which is the sum of the right-hand side series in (28). So, for (31) we find

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

(33)

By differentiating (5) we can evaluate ${\zeta }^{\prime }(-2n)$ for positive integers n. So, the left-hand side is $-{\zeta }^{\prime }(1-s)$, but on the right-hand side, we obtain a sum of four terms; and notice that in three of them, $cos{\displaystyle \frac{\pi s}{2}}$ appears while only one contains $sin{\displaystyle \frac{\pi s}{2}}$, i.e.,

$$-{\displaystyle \frac{\pi \zeta \left(s\right)}{{\left(2\pi \right)}^s}}\Gamma \left(s\right)sin{\displaystyle \frac{\pi s}{2}}.$$

With $s=2n+1$, all the terms except for the latter become zero, and we find

$${\zeta }^{\prime }(-2n)={\displaystyle \frac{{(-1)}^n\left(2n\right)!\zeta (2n+1)}{2{\left(2\pi \right)}^{2n}}},n\in \mathbb{N}.$$

(34)

Because of (34) and $\eta \left(s\right)=(1-2^{1-s})\zeta \left(s\right)$, for the last row of (33), we have

$$-{(-1)}^m{\displaystyle \frac{(2m-1)!}{{\pi }^{2m-1}}}\sum _{k=0}^{m-2}{\displaystyle \frac{{(-1)}^k\eta (2m-2k-1)}{(2k+1)!}}{x}^{2k+1}.$$

Adding this modified Formula (33) to (28), after a rearrangement, we arrive at (27). □

Theorem 5. 

For $\alpha =2m-1$, the series (26) takes the closed form

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

(35)

Proof. 

Following a similar procedure as in the proof of the preceding theorem, we replace $\alpha$ with $2m-1$ in (26) and have

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{{(-1)}^{n-1}cosnx}{n^{2m-1}}}={\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{(-1)}^k\eta (2m-2k-1)}{\left(2k\right)!}}{x}^{2k}.$$

The function $\eta \left(s\right)$ is analytic in the whole complex plane, so to bring the latter series into closed form, we express it first as a sum comprising the value of $\eta \left(1\right)$, obtained for $k=m-1$, and its remainder, i.e.,

$$\begin{array}{c}{\displaystyle \sum _{k=0}^{m-2}}{\displaystyle \frac{{(-1)}^k\eta (2m-2k-1)}{\left(2k\right)!}}{x}^{2k}+{\displaystyle \frac{{(-1)}^{m-1}{x}^{2m-2}}{(2m-2)!}}log2\hfill \\ \hfill +{\displaystyle \sum _{k=m}^{\infty }}{\displaystyle \frac{{(-1)}^k\eta (2m-2k-1)}{\left(2k\right)!}}{x}^{2k}.\end{array}$$

(36)

Further, we act in the same manner as in the previous proof. Relying on the relation $\eta \left(s\right)=(1-2^{1-s})\zeta \left(s\right)$ and (30), we determine the remainder in (36) as a difference of the two series

$$\begin{array}{c}{\displaystyle \sum _{k=m}^{\infty }}{\displaystyle \frac{{(-1)}^k\zeta (2m-2k-1)}{\left(2k\right)!}}{x}^{2k}-{\displaystyle \sum _{k=m}^{\infty }}{\displaystyle \frac{{(-1)}^k\zeta (2m-2k-1)}{2^{2m-2k-2}\left(2k\right)!}}{x}^{2k}\hfill \\ \hfill =2{(-1)}^{m-1}{x}^{2m-2}\left({\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{\zeta \left(2k\right)}{{\left(2k\right)}_{2m-1}}}{\left({\displaystyle \frac{x}{2\pi }}\right)}^{2k}-{\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{\zeta \left(2k\right)}{{\left(2k\right)}_{2m-1}}}{\left({\displaystyle \frac{x}{\pi }}\right)}^{2k}\right).\end{array}$$

Applying again Theorem 4, where we take $n=2m-2$ and $a=1$, then in succession set $t={\displaystyle \frac{x}{2\pi }}$ and $t={\displaystyle \frac{x}{\pi }}$, we obtain two formulas of the same structure. Subtracting them gives rise to the sum of the right-hand side series in (36), which is

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

(37)

Adding (37) to (36), we obtain (27). □

By using the identity $\zeta \left(s\right)+\eta \left(s\right)=2\lambda \left(s\right)$, and summing up the first and second equations of (22) with (25) and (26), respectively, we obtain

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{sin(2n-1)x}{{(2n-1)}^{\alpha }}}& ={\displaystyle \frac{\pi x^{\alpha -1}}{4\Gamma \left(\alpha \right)sin\frac{\pi }{2}\alpha }}+{\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{(-1)}^k\lambda (\alpha -2k-1)}{(2k+1)!}}{x}^{2k+1},\hfill \\ \hfill {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{cos(2n-1)x}{{(2n-1)}^{\alpha }}}& ={\displaystyle \frac{\pi x^{\alpha -1}}{4\Gamma \left(\alpha \right)cos\frac{\pi }{2}\alpha }}+{\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{(-1)}^k\lambda (\alpha -2k)}{\left(2k\right)!}}{x}^{2k},0<x<\pi ,\hfill \end{array}$$

(38)

Theorem 6. 

If $\alpha \to 2m$ in the first formula of (38), the following holds:

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

(39)

Proof. 

Since $sin{\displaystyle \frac{2m\pi }{2}}=0$, in the first formula of (38) for $\alpha =2m$ singularity is encountered, and for $k=m-1$ there is singularity of the lambda function at 1, so we are not permitted to immediately replace $\alpha$ with $2m$ but have to take the limit

$$L_{2m}=\underset{\alpha \to 2m}{lim}\left({\displaystyle \frac{\pi x^{\alpha -1}}{4\Gamma \left(\alpha \right)sin\frac{\pi }{2}\alpha }}+{\displaystyle \frac{{(-1)}^{m-1}\lambda (\alpha -2m+1)}{(2m-1)!}}{x}^{2m-1}\right).$$

(40)

Further, by bringing the fractions to the same denominator, we have

$$\underset{\alpha \to 2m}{lim}{\displaystyle \frac{\pi x^{\alpha -1}(2m-1)!+4{(-1)}^{m-1}{x}^{2m-1}\lambda (\alpha -2m+1)\Gamma \left(\alpha \right)sin\frac{\pi }{2}\alpha }{4(2m-1)!\Gamma \left(\alpha \right)sin\frac{\pi }{2}\alpha }}.$$

The denominator tends to zero as $\alpha$ tends to $2m$.

As for the numerator, knowing that $\lambda \left(s\right)=(1-2^{-s})\zeta \left(s\right)$, we have

$$\begin{array}{cc}\hfill & {\displaystyle \underset{\alpha \to 2m}{lim}}\left(\pi x^{\alpha -1}(2m-1)!+4x^{2m-1}\Gamma \left(\alpha \right)\lambda (\alpha -2m+1){(-1)}^{m-1}sin{\displaystyle \frac{\pi }{2}}\alpha \right)\hfill \\ \hfill & =\pi x^{2m-1}(2m-1)!\left(1-2{\displaystyle \underset{\alpha \to 2m}{lim}}(\alpha -2m)\lambda (\alpha -2m+1){\displaystyle \frac{sin\frac{\pi }{2}(\alpha -2m)}{\frac{\pi }{2}(\alpha -2m)}}\right)=0,\hfill \end{array}$$

where, relying on the functional equation for the Riemann zeta function (5), we make use of the limiting value

$$\begin{array}{cc}\hfill {\displaystyle \underset{z\to 0}{lim}}z\lambda (z+1)& ={\displaystyle \underset{z\to 1}{lim}}(z-1)\lambda \left(z\right)={\displaystyle \underset{z\to 1}{lim}}{\displaystyle \frac{{\left(2\pi \right)}^z(z-1)(1-2^{-z})\zeta (1-z)}{2\Gamma \left(z\right)cos\frac{z\pi }{2}}}\hfill \\ \hfill & =-{\displaystyle \underset{z\to 1}{lim}}{\displaystyle \frac{{\pi }^{z-1}\frac{\pi }{2}(1-z)\zeta (1-z)}{\Gamma \left(z\right)sin\frac{\pi }{2}(1-z)}}=-\zeta \left(0\right)=\frac{1}{2}.\hfill \end{array}$$

So, we can apply L’Hopital’s rule here. For the limiting value of the first derivative of the denominator, we find

$$4(2m-1)!{\displaystyle \underset{\alpha \to 2m}{lim}}\left({\displaystyle \frac{\pi }{2}}cos{\displaystyle \frac{\pi \alpha }{2}}\Gamma \left(\alpha \right)+sin{\displaystyle \frac{\pi \alpha }{2}}\Gamma \left(\alpha \right)\psi \left(\alpha \right)\right)={(-1)}^{m}2\pi (2m-1){!}^2,$$

Now, we take the limit of the first derivative of the numerator

$$\begin{array}{cc}\hfill & \pi (2m-1)!x^{2m-1}logx+{(-1)}^{m-1}{x}^{2m-1}{\displaystyle \underset{\alpha \to 2m}{lim}}(4{\lambda }^{\prime }(\alpha -2m+1)\Gamma \left(\alpha \right)sin{\displaystyle \frac{\pi \alpha }{2}}\hfill \\ \hfill & +4\lambda (\alpha -2m+1)\Gamma \left(\alpha \right)\psi \left(\alpha \right)sin{\displaystyle \frac{\pi \alpha }{2}}+2\pi \lambda (\alpha -2m+1)\Gamma \left(\alpha \right)cos{\displaystyle \frac{\pi \alpha }{2}}).\hfill \end{array}$$

(41)

First of all, using the relation $\lambda \left(z\right)=(1-2^{-z})\zeta \left(z\right)$, we find

$${\displaystyle \frac{{\lambda }^{\prime }\left(z\right)}{\lambda \left(z\right)}}={\displaystyle \frac{{\zeta }^{\prime }\left(z\right)}{\zeta \left(z\right)}}+{\displaystyle \frac{log2}{2^z-1}}.$$

It is easy to see that in the neighborhood of $z=1$, there holds

$${\displaystyle \frac{log2}{2^z-1}}=log2+O\left(\right|z-1\left|\right),$$

and looking up in [7] (p. 23), we read

$${\displaystyle \frac{{\zeta }^{\prime }\left(s\right)}{\zeta \left(z\right)}}=-{\displaystyle \frac{1}{z-1}}+\gamma +O\left(\right|z-1\left|\right).$$

Thus, there follows

$${\lambda }^{\prime }(\alpha -2m+1)=-{\displaystyle \frac{\lambda (\alpha -2m+1)}{\alpha -2m}}+(\gamma +log2)\lambda (\alpha -2m+1)+O\left(\right|\alpha -2m\left|\right).$$

By replacing this in (41), omitting $O\left(\right|\alpha -2m\left|\right)$, it remains to calculate

$$\begin{array}{c}{\displaystyle \underset{\alpha \to 2m}{lim}}\Gamma \left(\alpha \right)\lambda (\alpha -2m+1)(-2\pi {\displaystyle \frac{{(-1)}^{m}sin\frac{\pi }{2}(\alpha -2m)}{\frac{\pi }{2}(\alpha -2m)}}\hfill \\ \hfill +4sin{\displaystyle \frac{\pi \alpha }{2}}(\psi \left(\alpha \right)+\gamma +log2)+2\pi cos{\displaystyle \frac{\pi \alpha }{2}}).\end{array}$$

Because of ${\displaystyle \underset{\alpha \to 2m}{lim}}cos{\displaystyle \frac{\pi \alpha }{2}}=cos\pi m={(-1)}^m$, we conclude that the limiting value of the sum of only the first and third terms is zero, and finding the limit reduces to

$$\begin{array}{c}{\displaystyle \underset{\alpha \to 2m}{lim}}{\displaystyle \frac{2{(-1)}^m\pi \Gamma \left(\alpha \right)sin\frac{\pi }{2}(\alpha -2m)}{\frac{\pi }{2}(\alpha -2m)}}(\alpha -2m)\lambda (\alpha -2m+1)(\psi \left(\alpha \right)+\gamma +log2)\hfill \\ \hfill ={(-1)}^m\pi (2m-1)!(\psi \left(2m\right)+\gamma +log2)={(-1)}^m\pi (2m-1)!(H_{2m-1}+log2).\end{array}$$

We have applied here the relation (11).

Thus, taking account of all of this, the value of (40) is

$$L_{2m}={\displaystyle \frac{{(-1)}^m{x}^{2m-1}(logx-H_{2m-1}-log2)}{2(2m-1)!}},$$

and the trigonometric series (39) can now be expressed as

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

(42)

and after shifting the summation index in the last series, making use of the relation $\lambda (2m-2k-1)=(1-2^{-(2m-2k-1)})\zeta (2m-2k-1)$ and applying (30), one can represent it as a difference of two series in terms of the zeta function:

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

We have already dealt with these series in Formula (31), but here we express this difference as follows:

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

Adding the right-hand side to (42), after a rearrangement, and using (13), we obtain (39). □

Theorem 7. 

If $\alpha \to 2m-1$ in the second formula of (38), it holds that

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

(43)

Proof. 

After taking the limit $\alpha \to 2m-1$ in the second formula of (38), acting in the same manner as in the proof of the previous theorem, we first find

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

(44)

We deal again with the remainder and rewrite it as follows:

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

Employing a similar procedure as in the case of the preceding theorem, we come to the closed-form Formula (43). □

For alternating series related to (38) we can express them as a power series involving Dirichlet’s beta function

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{{(-1)}^{n-1}sin(2n-1)x}{{(2n-1)}^{\alpha }}}& ={\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{(-1)}^k\beta (\alpha -2k-1)}{(2k+1)!}}{x}^{2k+1},\hfill \\ \hfill {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{{(-1)}^{n-1}cos(2n-1)x}{{(2n-1)}^{\alpha }}}& ={\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{(-1)}^k\beta (\alpha -2k)}{\left(2k\right)!}}{x}^{2k},-\frac{\pi }{2}<x<\frac{\pi }{2}.\hfill \end{array}$$

(45)

Theorem 8. 

If α is replaced with $2m-1$ in the first formula of (45), we obtain the following closed form:

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{{(-1)}^{n-1}sin(2n-1)x}{{(2n-1)}^{2m-1}}}={\displaystyle \frac{{(-1)}^{m-1}{\left(2\pi \right)}^{2m-2}}{2(2m-2)!}}({\zeta }^{\prime }\left(2-2m,{\displaystyle \frac{1}{4}}-{\displaystyle \frac{x}{2\pi }}\right)\hfill \\ \hfill & -{\zeta }^{\prime }\left(2-2m,{\displaystyle \frac{3}{4}}-{\displaystyle \frac{x}{2\pi }}\right)-{\zeta }^{\prime }\left(2-2m,{\displaystyle \frac{1}{4}}+{\displaystyle \frac{x}{2\pi }}\right)+{\zeta }^{\prime }\left(2-2m,{\displaystyle \frac{3}{4}}+{\displaystyle \frac{x}{2\pi }}\right)),\hfill \end{array}$$

(46)

Proof. 

These replacements are legitimate because we do not encounter singularities since the function $\beta$ is analytic in the whole complex plane; knowing that $\beta \left(1\right)=\mathrm{arctg}1={\displaystyle \frac{\pi }{4}}$ and by using (20), we easily calculate

$$\beta \left(0\right)={\displaystyle \frac{2}{\pi }}sin{\displaystyle \frac{\pi }{2}}\Gamma \left(1\right)\beta \left(1\right)={\displaystyle \frac{2}{\pi }}·{\displaystyle \frac{\pi }{4}}={\displaystyle \frac{1}{2}}.$$

Further, from (45), we have

$$\begin{array}{c}\hfill {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{{(-1)}^{n-1}sin(2n-1)x}{{(2n-1)}^{2m-1}}}={\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{(-1)}^k\beta (2m-2k-2)}{(2k+1)!}}{x}^{2k+1},\end{array}$$

splitting the right-hand side series in three, but writing it for brevity now as

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

By shifting the summation index and making use of (21), the last series becomes

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

(47)

Taking account of (18), we further change (47) to

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

(48)

That means we are dealing with two series over the Hurwitz zeta functions and set $n=2m-2$ and $t={\displaystyle \frac{x}{2\pi }}$ in (32), considering, apart from t, the same formula for $-t$ as well.

Replacing a in succession with ${\displaystyle \frac{1}{4}}$ and ${\displaystyle \frac{3}{4}}$ in (32), then subtracting these equalities, we obtain (48) on the left-hand side. The right-hand side consists of a sum of four terms, the first one being

$$\begin{array}{cc}\hfill & {\displaystyle \frac{{(-1)}^{m-1}{\left(2\pi \right)}^{2m-2}}{2(2m-2)!}}({\zeta }^{\prime }\left(-2m+2,{\displaystyle \frac{1}{4}}-{\displaystyle \frac{x}{2\pi }}\right)-{\zeta }^{\prime }\left(-2m+2,{\displaystyle \frac{3}{4}}-{\displaystyle \frac{x}{2\pi }}\right)\hfill \\ \hfill & -{\zeta }^{\prime }\left(-2m+2,{\displaystyle \frac{1}{4}}+{\displaystyle \frac{x}{2\pi }}\right)+{\zeta }^{\prime }\left(-2m+2,{\displaystyle \frac{3}{4}}+{\displaystyle \frac{x}{2\pi }}\right))\hfill \end{array}$$

then of the two sums

$$\begin{array}{cc}\hfill & {(-1)}^m{\left({\displaystyle \frac{\pi }{2}}\right)}^{2m-2}{\displaystyle \sum _{k=1}^{m-1}}{\left({\displaystyle \frac{2x}{\pi }}\right)}^{2k-1}{\displaystyle \frac{H_{2m-2}-H_{2m-2k-1}}{(2m-2k-1)!(2k-1)!}}\beta (2k-2m+1)\hfill \\ \hfill & +{\displaystyle \frac{{(-1)}^m{\left(2\pi \right)}^{2m-2}}{(2m-2)!}}{\displaystyle \sum _{k=1}^{m-1}}{\left({\displaystyle \frac{x}{2\pi }}\right)}^{2k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{2m-2}{2k-1}}\right)4^{2k-2m+1}{\beta }^{\prime }(2k-2m+1)\hfill \end{array}$$

(49)

and finally of

$$\left(\psi \left(\frac{1}{4}\right)-\psi \left(\frac{3}{4}\right)\right){\displaystyle \frac{{(-1)}^{m-1}{\left(2\pi \right)}^{2m-2}}{(2m-1)!}}{\left({\displaystyle \frac{x}{2\pi }}\right)}^{2m-1}.$$

(50)

For $k=1,\cdots ,m-1$, the expressions $2k-2m+1$ present negative odd integers, which means $\beta (2k-2m+1)=0$, so the first sum in (49) equals zero.

Differentiating the relation (21) at $z=2k-2m+1$, for $1⩽k⩽m-1$, yields

$${\beta }^{\prime }(2k-2m+1)=-{\left({\displaystyle \frac{\pi }{2}}\right)}^{2k-2m+1}{(-1)}^{k-m}\Gamma (2m-2k)\beta (2m-2k),$$

whereby the second sum in (49) becomes

$$-{\displaystyle \sum _{k=1}^{m-1}}{\displaystyle \frac{{(-1)}^{k-1}\beta (2m-2k)}{(2k-1)!}}=-{\displaystyle \sum _{k=0}^{m-2}}{\displaystyle \frac{{(-1)}^k\beta (2m-2k-2)}{(2k+1)!}}.$$

Expression (50) is transformed into

$$-4\beta \left(1\right){\displaystyle \frac{{(-1)}^{m-1}{\left(2\pi \right)}^{2m-2}}{(2m-1)!}}{\left({\displaystyle \frac{x}{2\pi }}\right)}^{2m-1}=4·{\displaystyle \frac{\pi }{4}}{\displaystyle \frac{{(-1)}^m{x}^{2m-1}}{2\pi (2m-1)!}}={\displaystyle \frac{{(-1)}^m{x}^{2m-1}}{2(2m-1)!}},$$

and this is obtained by setting $n=1$ in the relation

$$\beta \left(n\right)={\displaystyle \frac{{(-1)}^n}{2^{2n}(2n-1)!}}\left({\psi }^{(n-1)}\left(\frac{1}{4}\right)-{\psi }^{(n-1)}\left(\frac{3}{4}\right)\right),n\in \mathbb{N},$$

which in turn we obtain from (18) by relying on the identity [5]

$${\psi }^{\left(n\right)}\left(a\right)={(-1)}^{n-1}n!\zeta (n+1,a).$$

Collecting all the cases, we arrive at the closed-form Formula (46). □

Theorem 9. 

If α is replaced with $2m$ in the second formula of (45), we obtain the following closed forms:

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

(51)

Proof. 

Acting in the same manner as in deducing Formula (46), we come to Formula (51). □

We can organize all these formulas by writing them as a general closed-form formula comprising the Formulas (23), (24), (27), (35), (39), (43), (46), and (51), i.e.,

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=1}^{\infty }}& {\displaystyle \frac{{\left(s\right)}^{n-1}f\left((an-b)x\right)}{{(an-b)}^{2m+p-1}}}={\displaystyle \frac{{(-1)}^{m+p-1}{\pi }^{2m+p-2}{2}^{2m+r-2}}{(2m+p-2)!}}\times \hfill \\ \hfill & (2^{2k-2}{\zeta }^{\prime }\left(2-p-2m,q-{\displaystyle \frac{x}{2\pi }}\right)+2^{2k-2}\delta {\zeta }^{\prime }\left(2-p-2m,1-q+{\displaystyle \frac{x}{2\pi }}\right)\hfill \\ \hfill & -j\left({\zeta }^{\prime }\left(2-p-2m,1-q-{\displaystyle \frac{x}{2c\pi }}\right)+\delta {\zeta }^{\prime }\left(2-p-2m,q+{\displaystyle \frac{x}{2c\pi }}\right)\right)),\hfill \end{array}$$

and one can obtain all its particular cases from

Table 2. Closed-form cases, $\alpha =2m+p-1$.

a	b	s	f	p	r	c	$\mathit{\delta }$	q	k	j	Convergence Region
1	0	1	sin	1	1		−1	1	1	0	$0<x<2\pi$
			cos	0	0		1	1	1	0
		−1	sin	1	$2-2m$	$-{\displaystyle \frac{1}{2}}$	−1	1	$m+{\displaystyle \frac{1}{2}}$	−1	$-\pi <x<\pi$
			cos	0	$2-2m$	$-{\displaystyle \frac{1}{2}}$	1	1	m	1
2	1	1	sin	1	$1-2m$	$-{\displaystyle \frac{1}{2}}$	−1	1	$m+1$	−1	$0<x<\pi$
			cos	0	$1-2m$	$-{\displaystyle \frac{1}{2}}$	1	1	$m+{\displaystyle \frac{1}{2}}$	1
		−1	sin	0	−1	1	1	${\displaystyle \frac{1}{4}}$	1	1	$-{\displaystyle \frac{\pi }{2}}<x<{\displaystyle \frac{\pi }{2}}$
			cos	1	0	1	−1	${\displaystyle \frac{1}{4}}$	1	−1

by choosing the corresponding parameters.

Thus, we have two types of closed-form formulas for each case of trigonometric series.