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.

  • 2
    $\begingroup$ By the way, since the terms decay very rapidly, this yields $$ \frac1{\sinh^2\pi}\approx\frac16-\frac1{2\pi}\;. $$ $\endgroup$
    – joriki
    Apr 1, 2013 at 10:41
  • $\begingroup$ besides the points that joriki makes in his answer, it should be noted that for the residues along the imaginary axis, you cannot play the same trick with L'Hospital that you can for the residues along the real axis. This is because the singularities on the imaginary axis are second order. The residues luckily come out the same, but more care should be taken. $\endgroup$
    – robjohn
    May 23, 2013 at 22:42
  • 1
    $\begingroup$ See this technique. $\endgroup$ May 24, 2013 at 5:49

3 Answers 3


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

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


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

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.


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.

  • $\begingroup$ Wow, thanks RV. You're a good egg. This Mellin tranforms can be mighty useful in evaluating series. May I ask something?. How do you determine what vertices of the rectangle to use?. I noticed on some you use 3/2, but on this one you used 5/2 and 1. $\endgroup$
    – Cody
    May 24, 2013 at 10:28
  • $\begingroup$ Thanks. I see. Nice method. $\endgroup$
    – Cody
    May 24, 2013 at 20:48

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.

  • $\begingroup$ Thanks very much, Joriki. I sure would not have spotted that. I was under the impression one did not have to multiply by $2\pi i$ in this case. I was thinking of $\sum_{-\infty}^{\infty}f(n)=-\text{(sum of residues of} \pi\cot(\pi z) f(z) \text{at the poles of f(z))}$. That is why I did not use it. Apparently, this is not applicable here because f(z) does not have a finite number of poles?. Thanks for your help. I reckon I do not know enough to see things like this. $\endgroup$
    – Cody
    Mar 30, 2013 at 22:29
  • $\begingroup$ Oh, btw, joriki. The -1/6 part is unaffected by the $2\pi i$?. $\endgroup$
    – Cody
    Mar 30, 2013 at 22:45
  • $\begingroup$ If the contour integral as $ N \to \infty$ approaches zero and you divide both sides of the equation by $ 2 \pi i $, the $ 2 \pi i$ effectively disappears. $\endgroup$ Mar 30, 2013 at 23:36
  • $\begingroup$ @Cody: The problem isn't that $f(z)$ doesn't have a finite number of poles; it's that that formula assumes that the integral vanishes as you expand the contour to enclose all poles. As RV remarked, if the integral does vanish, then you don't need the factor $2\pi\mathrm i$ since it's multiplied by zero; but the way you wrote it, with the integral still in there (and rightly so, since it has a finite limit), you do need the factor. If you try deriving your formula from the residue theorem, you'll see why the factor $2\pi\mathrm i$ is in front of the integral and not in front of the other terms. $\endgroup$
    – joriki
    Mar 31, 2013 at 0:03
  • 2
    $\begingroup$ Are you using a square with vertices at $\pm(2N + \frac{1}{2}) \pm i (2N+\frac{1}{2})$ and not a square with vertices at $\pm(N + \frac{1}{2}) \pm i (N+\frac{1}{2})$ because it makes the evaluation easier? $\endgroup$ Apr 1, 2013 at 1:51

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 $$ \sum_{n=1}^\infty\mathrm{csch}^2(\pi n)=\frac16-\frac1{2\pi}\tag{9} $$


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