Residue theorem V

Series involving hyperbolic cotangent

We show the proof following result about series involving hyperbolic functions posted by @infseriesbot: \[ \pi \sum_{n=1}^{\infty} \frac{\coth \pi n}{n^{4m-1}} = \zeta(4m) + \sum_{n=1}^{2m-1}(-1)^{k-1}\zeta(2k)\zeta(4m-2k)\] The proof will rely on the residue theorem.

Proof

In order to calculate the residue, we will need the expansion in zeta numbers of $\displaystyle \coth(\pi x)$ and $\cot(x)$ \[ \coth(\pi x) = \frac{1}{\pi x} - \frac{2}{\pi } \sum_{n=1}^{\infty} (-1)^n\zeta(2n)x^{2n-1}\] \[ \cot(\pi x) = \frac{1}{\pi x} - \frac{2}{\pi } \sum_{n=1}^{\infty}\zeta(2n)x^{2n-1}\]

1. To obtain these expansions we need the Fourier series expansion of $\cos(ax)$

Since $\cos(ax)$ is defined for all $x$, and represents a continuous, piece-wise smooth, odd function is everywhere equal to its Fourier series, which is absolutely and uniformly convergent.

Therefore,
\[ \cos ax = \sum_{n=-\infty}^{\infty} c_{n}e^{inx} \quad a\neq n\] where \[ c_{n} = \frac{1}{2\pi} \int_{-\pi}^{\pi} \cos(ax)e^{-inx}dx \] Hence \begin{align*} c_{n} = \frac{1}{2\pi} \int_{-\pi}^{\pi} \cos(ax)e^{-inx}dx =& \frac{1}{2\pi} \int_{-\pi}^{\pi} \left[\frac{e^{iax}+ e^{-iax}}{2}\right]e^{-inx}dx \\ =& \frac{1}{4\pi} \int_{-\pi}^{\pi} e^{ix(a-n)}dx + \frac{1}{4\pi} \int_{-\pi}^{\pi} e^{ix(-a-n)}dx\\ =& \frac{1}{4\pi}\left[\frac{e^{ i(a-n)} -e^{- i(a-n)}}{i(a-n)}\right] + \frac{1}{4\pi}\left[\frac{e^{ i(a+n)} -e^{- i(a+n)}}{i(a+n)}\right]\\ =& \frac{1}{4\pi}\left[\frac{e^{ a} e^{-\pi in} -e^{-\pi i a}e^{- \pi i n}}{i(a-n)}\right] + \frac{1}{4\pi}\left[\frac{e^{\pi i a}e^{\pi in)} -e^{-\pi i a}e^{i\pi n}}{i(a+n)}\right]\\ =&\frac{e^{i\pi n} (e^{i \pi a}-e^{-i \pi a} )}{4\pi i}\left[\frac{1}{a-n}+\frac{1}{a+n}\right]\\ =& (-1)^n \frac{\sin(a\pi)}{2\pi}\left[ \frac{2a}{a^2-n^2}\right]\\ =& (-1)^n \frac{a\sin(a\pi)}{\pi(a^2-n^2)}\end{align*} Therefore \begin{align*} \cos(ax) = \sum_{n=-\infty}^{\infty} (-1)^n \frac{a\sin(a\pi)}{\pi(a^2-n^2)}e^{inx} =&\frac{\sin(\pi a)}{\pi a}+ \sum_{j=-\infty}^{j=-1} (-1)^j \frac{a\sin(a\pi)}{\pi(a^2-j^2)}e^{ijx} + \sum_{n=1}^{\infty} (-1)^n \frac{a\sin(a\pi)}{\pi(a^2-n^2)}e^{inx} \\ =&\frac{\sin(\pi a)}{\pi a}+ \sum_{n=1}^{\infty} (-1)^n \frac{a\sin(a\pi)}{\pi(a^2-n^2)}e^{-inx} + \sum_{n=1}^{\infty} (-1)^n \frac{a\sin(a\pi)}{\pi(a^2-n^2)}e^{inx} \quad (n= -j)\\ =&\frac{\sin(\pi a)}{\pi a}+ \sum_{n=1}^{\infty} (-1)^n \frac{a\sin(a\pi)}{\pi(a^2-n^2)}(e^{-inx} +e^{inx})\\ =&\frac{\sin(\pi a)}{\pi a}+ \frac{2}{\pi}\sum_{n=1}^{\infty} (-1)^n \frac{a\sin(a\pi)\cos(nx)}{\pi(a^2-n^2)}\\ =&\frac{2}{\pi}\sin(\pi a)\left[\frac{1}{2 a}+ \sum_{n=1}^{\infty} (-1)^n \frac{a\cos(nx)}{(a^2-n^2)}\right] \end{align*} Hence \[ \boxed{\cos(ax) = \frac{2}{\pi}\sin(\pi a)\left[\frac{1}{2 a}+ \sum_{n=1}^{\infty} (-1)^n \frac{a\cos(nx)}{(a^2-n^2)}\right] } \tag{1}\]
2. With this result we can obtain the expansion in partial fraction of $\coth(x)$ and $\cot(x)$:

If we put $\displaystyle x=\pi \Longrightarrow \cos(n\pi) = (-1)^n$ in the Fourier series expansion of $\cot(ax)$ then \[ \cos(a \pi) = \frac{2}{\pi}\sin(\pi a)\left[\frac{1}{2 a}+ \sum_{n=1}^{\infty} \frac{a}{(a^2-n^2)}\right]\] If we divide by $\sin(a\pi) $ \[ \cot(a \pi) = \frac{2}{\pi}\left[\frac{1}{2 a}+ \sum_{n=1}^{\infty} \frac{a}{(a^2-n^2)}\right]\] Now, if we allow $a$ to take complex values: let $a= iv$ \[ \cot(-iv \pi) = \frac{2}{\pi}\left[-\frac{i}{2 v} - \sum_{n=1}^{\infty} \frac{iv}{(v^2+n^2)}\right]\] Multiplying by $i$ \[ \coth(v \pi) = \frac{2}{\pi}\left[\frac{1}{2 v} + \sum_{n=1}^{\infty} \frac{v}{(v^2+n^2)}\right]\] We have found the expansion in partial fractions of $\cot$ and $\coth$ \[\boxed{ \cot(x \pi) = \frac{2}{\pi}\left[\frac{1}{2 x}+ \sum_{n=1}^{\infty} \frac{x}{(x^2-n^2)}\right] \quad x\neq \pm 1, \pm 2, \pm 3,..} \tag{2}\] \[ \boxed{\coth(x \pi) = \frac{2}{\pi}\left[\frac{1}{2 x} + \sum_{n=1}^{\infty} \frac{x}{(x^2+n^2)}\right]\quad x\neq \pm i, \pm 2i, \pm 3i,..} \tag{3}\]
3. Another consequence is the expansion in zeta numbers of $\coth(x)$ and $\cot(x)$:

Suppose that $|x|\lt 1$, using (2) we have \begin{align*} \cot(x \pi) =& \frac{1}{\pi x}- \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{x}{(n^2-x^2)} \\ = &\frac{1}{\pi x}- \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{x}{n^2\left(1-\left(\frac{x}{n}\right)^2\right)} \\ = &\frac{1}{\pi x}- \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{x}{n^2} \sum_{j=0}^{\infty} \left(\frac{x}{n}\right)^{2j}\\ =&\frac{1}{\pi x}- \frac{2}{\pi}\underbrace{\sum_{j=0}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n^{2j+2}} x^{2j+1}}_{\textrm{Rearrangement}} \\ =&\frac{1}{\pi x}- \frac{2}{\pi}\sum_{j=0}^{\infty}\zeta(2j+2) x^{2j+1} \\ =&\frac{1}{\pi x}- \frac{2}{\pi}\sum_{k=1}^{\infty}\zeta(2k) x^{2k-1} \\ \end{align*} The rearrangement of the sums is possible due to the fact that the series \[ \sum_{j=0}^{\infty} \left(\frac{x}{n}\right)^{2j} \] is absolutely convergent and the series \[ \zeta(v) = \sum_{n=1}^{\infty} \frac{1}{n^v} \] is convergent. In a very similar way we can show the proof of $\coth(x)$ Therefore \[\boxed{ \cot(\pi x) =\frac{1}{\pi x}- \frac{2}{\pi}\sum_{k=1}^{\infty}\zeta(2k) x^{2k-1} \quad |x|\lt 1 } \tag{4}\] \[\boxed{ \coth(\pi x) = \frac{1}{\pi x}- \frac{2}{\pi}\sum_{k=1}^{\infty}(-1)^k\zeta(2k) x^{2k-1} \quad |x|\lt 1} \tag{5}\]
4. We will apply the residue theorem


It is easy to show that if $\displaystyle f(z) = \frac{\pi \operatorname{coth}(\pi z) }{z^{4m-1}}$ \[\lim_{N \to \infty} \oint_{C_{N}} \pi\cot(\pi z)f(z) dz = 0 \] where $C_{N}$ be a square with vertices $\left(N+\frac{1}{2}\right)(1+i), \left(N+\frac{1}{2}\right)(-1+i), \left(N+\frac{1}{2}\right)(-1-i), \left(N+\frac{1}{2}\right)(1-i)$

Therefore, we can apply the following result : \[\sum_{\substack{n=-\infty \\ n\neq 0}}^{\infty}\frac{\pi\coth(\pi n)}{n^{4m-1}} = -\sum \left\{\textrm{Residues of } f(z)\pi\cot(\pi z) \textrm{ at the sungularities of } f \right\}\] Given that the function $f(z)$ has a singularity at $z=0$ and a countable number of singularities at $\displaystyle z=in \quad n\in\mathbb{N}\setminus\left\{0\right\}$

We have: \[\sum_{\substack{n=-\infty \\ n\neq 0}}^{\infty}\frac{\pi\coth(\pi n)}{n^{4m-1}} = -\sum_{n=1}^{\infty}\operatorname{Res}\left(\pi\cot(\pi z),in\right) -\sum_{n=1}^{\infty}\operatorname{Res}\left(f(z)\pi\cot(\pi z),-in\right) -\operatorname{Res}\left(f(z)\pi \cot(\pi z),0\right)\] With the L'Hopital rule can be proven that \[\lim_{z \to \pm in} (z\mp in)\cot(\pi z) \coth (\pi z) = \mp \frac{i\coth(n \pi)}{\pi} \quad \forall n>0 \] \[ \therefore \operatorname{Res}\left(f(z)\pi\cot(\pi z),\mp in\right) = \lim_{z \to \pm in} \frac{(z\mp in)\pi^2 \cot(\pi z)\coth(\pi z)}{ z^{4m-1}} = \frac{\pi \coth(n \pi)}{ n^{4m-1}} \] Therefore \[\sum_{\substack{n=-\infty \\ n\neq 0}}^{\infty}\frac{\pi\coth(\pi n)}{n^{4m-1}} = -\sum_{n=1}^{\infty}\frac{\pi\coth(\pi n)}{n^{4m-1}} -\sum_{n=1}^{\infty}\frac{\pi\coth(\pi n)}{n^{4m-1}} -\operatorname{Res}\left(f(z)\pi \cot(\pi z),0\right)\] \[2\sum_{\substack{n=-\infty \\ n\neq 0}}^{\infty}\frac{\pi\coth(\pi n)}{n^{4m-1}} = -\operatorname{Res}\left(f(z)\pi \cot(\pi z),0\right) \]
We just have to find the residue at $z = 0$.

For that end, we will make use of the expansion in zeta numbers of $\cot$ and $\coth$

Note \begin{align*} \coth(\pi z)\cot(\pi z) =& \left[ \frac{1}{\pi z}- \frac{2}{\pi}\sum_{k=1}^{\infty}(-1)^k\zeta(2k) z^{2k-1}\right]\left[\frac{1}{\pi z}- \frac{2}{\pi}\sum_{k=1}^{\infty}\zeta(2k) z^{2k-1} \right] \quad (\textrm{from (4) and (5)})\\ &= \frac{1}{\pi^2 z^2} -\frac{2}{\pi^2} \sum_{k=1}^{\infty}\zeta(2k) z^{2k-2} - \frac{2}{\pi^2} \sum_{k=1}^{\infty}(-1)^k\zeta(2k) z^{2k-2} + \frac{4}{\pi^2} \underbrace{\sum_{k=1}^{\infty}\sum_{j=1}^{k} (-1)^{j}\zeta(2j)\zeta(2k-2j+2)z^{2k}}_{\textrm{Cauchy product}}\\ &= \frac{1}{\pi^2 z^2} -\frac{4}{\pi^2} \sum_{k=1}^{\infty}\zeta(4k) z^{4k-2} - \frac{4}{\pi^2} \sum_{k=1}^{\infty}\sum_{j=1}^{k} (-1)^{j}\zeta(2j)\zeta(2k-2j+2)z^{2k}\\ \end{align*} Hence \[ \frac{\pi^2 \cot(\pi z)\cot(\pi z)}{z^{4m-1}} = z^{-4m-1-2}- 4\sum_{k=1}^{\infty}\zeta(4k) z^{4k-4m-1} + 4 \sum_{k=1}^{\infty}\sum_{j=1}^{k} (-1)^{j}\zeta(2j)\zeta(2k-2j+2)z^{2k-4m+1} \] The residue at $z=0$ implies that \[ z^{4k-4m-1}= z^{-1} \Longrightarrow 4k-4m-1 = -1 \Longrightarrow k=m\] \[ z^{2k-4m+1} = z^{-1} \Longrightarrow 2k-4m+1 = -1 \Longrightarrow k=2m-1\] Therefore \[ \operatorname{Res}\left(\frac{\pi^2\cot(\pi z)\coth(\pi z)}{z^{4m-1}}, 0\right) = -4\zeta(4m) + 4\sum_{j=1}^{2m-1} (-1)^{j}\zeta(2j)\zeta(4m-2j) \] Therefore \[2\sum_{\substack{n=-\infty \\ n\neq 0}}^{\infty} \frac{\pi \coth(\pi n)}{n^{4m-1}} = 4\zeta(4m) + 4\sum_{j=1}^{2m-1} (-1)^{j-1}\zeta(2j)\zeta(4m-2j) \] Given that the function $\displaystyle f(z) = \frac{\pi \coth(\pi z)}{z^{4m-1}}$ is even we have \[4\sum_{n=1}^{\infty} \frac{\pi \coth(\pi n)}{n^{4m-1}} = 4\zeta(4m) + 4\sum_{j=1}^{2m-1} (-1)^{j-1}\zeta(2j)\zeta(4m-2j) \] Hence, we can conclude \[\boxed{\sum_{n=1}^{\infty} \frac{\pi \coth(\pi n)}{n^{4m-1}} = \zeta(4m) + \sum_{j=1}^{2m-1} (-1)^{j-1}\zeta(2j)\zeta(4m-2j) }\]