Using residues to evaluate a sum involving the square of $\text{csch}$

Question

I have been trying to evaluate the following sum using residues:

$$ \sum_{n=1}^{\infty}\frac{1}{\sinh^{2}(\pi n)}=\frac{1}{6}-\frac{1}{2\pi}$$

I am mainly interested in using residues to do this. I can do it using real methods.

I tried using $\displaystyle \oint\frac{\pi\cot(\pi z)}{\sinh^{2}(\pi z)}dz$

The residue at $z=0$ is $\frac{-2}{3}$

The residue at $\displaystyle z=n, \;\ (n=\pm 1, \pm 2, \pm 3, ....)$ is $\displaystyle \lim_{z\to n}\frac{(z-n)\cos(\pi z)}{\sin(\pi z)\sinh^{2}(\pi z)}=\frac{1}{\sinh^{2}(\pi n)}$

The residue at $\displaystyle z=ni, \;\ (\pm 1, \pm 2, \pm 3, .....)$ is

$\displaystyle \lim_{z\to ni}\frac{(z-ni)\pi\cot(\pi z)}{\sinh^{2}(\pi z)}=\frac{1}{\sinh^{2}(\pi n)}$

So, by the residue theorem:

$\displaystyle \oint\frac{\pi \cot(\pi z)}{\sinh^{2}(\pi z)}dz=\frac{-2}{3}+4\sum_{n=1}^{N}\frac{1}{\sinh^{2}(\pi n)}$

As $N\to \infty$, the left side goes to 0, then solve for the sum at hand.

It would appear the 1/6 is in there, but I have failed to arrive at the correct solution.

Where does the $\frac{1}{2\pi}$ come into play?.

No doubt, I am doing it incorrectly. Can someone point me in the right direction?

Thanks a lot.

Answer 1

The inverse Mellin transform provides another approach that uses residues.

I had previously asked about evaluating a similar series using this method.

For $\operatorname{Re} (s)>2$, we have

$$ \begin{align} \left\{\mathcal{M} \ \frac{1}{\sinh^{2} (\pi x)} \right\}(s) &= \int_{0}^{\infty} \frac{x^{s-1}}{\sinh^{2}(\pi x)} \, dx \\ &= \pi^{-s}\int_{0}^{\infty} \frac{u^{s-1}}{\sinh^{2} u} \, du \\ &= 4 \pi^{-s} \int_{0}^{\infty} u^{s-1} \, \frac{e^{-2u}}{(1-e^{-2u})^2} \, du \\ &= 4 \pi^{-s} \int_{0}^{\infty} u^{s-1} \sum_{n=1}^{\infty} n e^{-2nu} \,du \\ &= 4 \pi^{-s} \sum_{n=1}^{\infty} n \int_{0}^{\infty} u^{s-1} e^{-2nu} \, du \\ &= 4 \pi^{-s} \sum_{n=1}^{\infty}n \, \frac{\Gamma(s)}{(2n)^{s}}\\ &= \pi^{-s} \ 2^{2-s} \ \Gamma(s) \zeta(s-1). \end{align}$$

So according to the Mellin inversion theorem,

$$\frac{1}{\sinh^{2} (\pi x)} = \frac{1}{2 \pi i} \int_{c - i \infty}^{c + i\infty} \pi^{-s} \ 2^{2-s} \ \Gamma(s) \zeta(s-1) x^{-s} \, ds \, , $$ where $c= \frac{5}{2}$.

(Since the Mellin transform of $\frac{1}{\sinh^{2}(\pi z)}$ converges absolutely in the right half-plane $\operatorname{Re}(s) >2$, we could have chosen any value of $c$ greater than $2$.)

Replacing $x$ with $n$ and summing both sides, we get

$$\begin{align} \sum_{n=1}^{\infty} \frac{1}{\sinh^{2} (\pi n)} &= \frac{1}{2 \pi i} \int_{5/2 - i \infty}^{5/2 + i\infty} \pi^{-s} \ 2^{2-s} \ \Gamma(s) \zeta(s-1) \zeta(s) \, ds \\ &= \frac{1}{2 \pi i} \int_{5/2 - i\infty}^{5/2+ i\infty} f(s) \, ds. \end{align}$$

Since the magnitude of $\Gamma(z)$ decays exponentially fast to $0$ as $\Im(z) \to \pm \infty$, we can shift the contour to the line $\operatorname{Re}(s) =1$ and conclude that $$\sum_{n=1}^{\infty} \frac{1}{\sinh^{2} (\pi n)} = \frac{1}{2 \pi i} \left( \operatorname{PV} \int_{-\infty}^{\infty} f(1+it) i \, dt + \pi i \, \text{Res}[f,1] + 2 \pi i \, \operatorname{Res}[f,2] \right),$$ where

$$\begin{align} \operatorname{Res} [f(s),1] &= \lim_{z \to 1} (s-1) \zeta(s) \pi^{-s} 2^{2-s} \Gamma(s) \zeta(s-1) \\ &= (1) \left( \frac{1}{\pi} \Big)(2)(1)\Big(-\frac{1}{2} \right) \\ &= - \frac{1}{\pi } \end{align}$$

and

$$ \begin{align} \operatorname{Res}[f(s),2] &= \lim_{z \to 2} (s-2) \zeta(s-1) \pi^{-s} 2^{2-s} \Gamma(s) \zeta(s) \\ &= (1) \left(\frac{1}{\pi^{2}} \Big) (1)(1) \Big(\frac{\pi^{2}}{6} \right) \\ &= \frac{1}{6}. \end{align}$$

But it can be shown using the functional equation of the Riemann zeta function that $f(s)$ is odd (and purely imaginary) along the line $\operatorname{Re}(s)=1$.

Specifically, $$f(1+it) = 4i \cosh \left(\frac{\pi t}{2} \right)\operatorname{csch}(\pi t) \left|\zeta(it) \right|^{2} \, , \quad t \in \mathbb{R}. $$

Therefore, $$\operatorname{PV} \int_{-\infty}^{\infty} f(1+it) i \, dt =0 $$ and the result follows.

EDIT:

Alternatively, it can be shown that $$\sum_{n=1}^{\infty} \frac{1}{\sinh^{2} (\pi n)} = 4 \sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1}. $$

See this answer. This is the series I had asked about in my question.

Also see this paper for the evaluation of similar sums using the inverse Mellin transform.

Answer 2

There are two problems. One is that you dropped the factor $2\pi\mathrm i$ in the residue theorem. The other is that the left-hand side doesn't go to zero as $N\to\infty$.

If we integrate over a quadratic contour at half-integer coordinates, opposite sides yield the same contributions, so we need twice the sum of the contributions from one horizontal segment and one vertical segment. The contribution from the vertical segments goes to $0$, since the denominator decays exponentially. However, the contribution from the horizontal segments doesn't go to zero; it is, with $a=2k\pi +\pi/2$,

$$ \begin{align} \int_{(-a+\mathrm ia)/\pi}^{(a+\mathrm ia)/\pi}\frac{\pi\cot(\pi z)}{\sinh^2(\pi z)}\mathrm dz &= \int_{-a+\mathrm ia}^{a+\mathrm ia}\frac{\cot z}{\sinh^2z}\mathrm dz \\ &= \int_{-a}^a\frac{\cos x\cosh a-\mathrm i\sin x\sinh a}{\sin x\cosh a+\mathrm i\cos x\sinh a}\frac1{(\sinh x\cos a+i\cosh x\sin a)^2}\mathrm dx \\ &= -\int_{-a}^a\frac{\cos x\cosh a-\mathrm i\sin x\sinh a}{\sin x\cosh a+\mathrm i\cos x\sinh a}\frac1{\cosh^2x}\mathrm dx\;. \end{align} $$

With $a\to\infty$, both $\cosh a$ and $\sinh a$ are asymptotic to $\mathrm e^a$, so the first fraction goes to $-\mathrm i$, and we're left with twice

$$ \mathrm i\int_{-\infty}^\infty\frac1{\cosh^2 x}\mathrm dx=2\mathrm i\;. $$

This contribution of $4\mathrm i$, divided by the $4$ in front of your sum and the factor $2\pi\mathrm i$ in the residue theorem, yields the term $1/(2\pi)$; the minus sign arises because I integrated clockwise.

Answer 3

I worked this out for a different problem in a bit more generality, so I will post the generalization here.

We are going to use the contour integral $$ \oint\pi\cot\left(\frac{2\pi}{x}z\right)\left(\mathrm{csch}^2(z)-\frac1{z^2}\right)\mathrm{d}z=-4\pi i\tag{1} $$ where the contours of interest are, for real $R\to\infty$ and integer $n\to\infty$, $$ \small\textstyle\color{#00A000}{[R,-R]+(n+\frac12)\pi i}\cup\color{#C00000}{-R+(n+\frac12)\pi i[1,-1]}\cup\color{#00A000}{[-R,R]-(n+\frac12)\pi i}\cup\color{#C00000}{R+(n+\frac12)\pi i[-1,1]} $$ The integral along the red paths becomes negligible as $R\to\infty$. Along the upper green path, where $\mathrm{Im}(z)\approx+\infty$, $\cot(z)\approx-i$. Along the lower green path, where $\mathrm{Im}(z)\approx-\infty$, $\cot(z)\approx+i$. Since $\mathrm{csch}^2(z+\frac\pi2i)=-\mathrm{sech}^2(z)$ is the derivative of $-\tanh(z)$, the integral along each of the green paths tends to $-2\pi i$. Therefore, the full integral is $-4\pi i$.

Since $$ \pi\cot\left(\frac{2\pi}{x}z\right)\text{ has residue }\frac x2\text{ at }z=\frac x2n\tag{2} $$ and $$ \mathrm{csch}^2(z)-\frac1{z^2}=-\frac13+O(z^2)\text{ at }z=0\tag{3} $$ the contribution from the singularities on the real axis is $$ 2\pi i\frac x2\left[2\sum_{n=1}^\infty\left(\mathrm{csch}^2\left(\frac x2n\right)-\frac4{x^2n^2}\right)-\frac13\right]\tag{4} $$ Since $$ \mathrm{csch}^2(z)=\frac1{(z-\pi in)^2}+O(1)\text{ at }z=\pi in\text{ for }n\ne0\tag{5} $$ and $$ \begin{align} \left.\frac{\mathrm{d}}{\mathrm{d}z}\pi\cot\left(\frac{2\pi}{x}z\right)\right|_{z=\pi in} &=\left.-\frac{2\pi^2}{x}\csc^2\left(\frac{2\pi}{x}z\right)\right|_{z=\pi in}\\ &=\frac{2\pi^2}{x}\mathrm{csch}^2\left(\frac{2\pi^2}{x}n\right)\tag{6} \end{align} $$ the contribution from the singularities on the imaginary axis is $$ 2\pi i\frac{2\pi^2}{x}\left[2\sum_{n=1}^\infty\mathrm{csch}^2\left(\frac{2\pi^2}{x}n\right)\right]\tag{7} $$ Putting together $(1)$, $(4)$, and $(7)$ gives $$ \frac x2\sum_{n=1}^\infty\mathrm{csch}^2\left(\frac x2n\right) +\frac{2\pi^2}{x}\sum_{n=1}^\infty\mathrm{csch}^2\left(\frac{2\pi^2}{x}n\right) =-1+\frac{\pi^2}{3x}+\frac x{12}\tag{8} $$ Setting $x=2\pi$ in $(8)$ yields $$ \sum_{n=1}^\infty\mathrm{csch}^2(\pi n)=\frac16-\frac1{2\pi}\tag{9} $$