Inspired by this answer, I'm trying to show that $$\sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1} = \frac{1}{24} - \frac{1}{8 \pi}$$ using the inverse Mellin transform.

But the answer I get is twice as much as it should be, and I don't understand why.


With Marko Riedel's help, I corrected the error in my evaluation.

Since $$ \left\{ \mathcal{M} \ \frac{x}{e^{2\pi x}-1} \right\}(s) = \int_{0}^{\infty} \frac{x^{s}}{e^{2 \pi x}-1} \, dx = (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1) $$ for $\operatorname{Re}(s) >1$,

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

Replacing $x$ with $n$ and summing both sides, we get $$ \begin{align} \sum_{n=1}^{\infty}\frac{n}{e^{2\pi n}-1} &= \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}(2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1)\zeta(s)\, ds \\&= \frac{1}{2 \pi i} \int_{c-i\infty}^{c+i\infty} f(s) \, ds. \end{align} $$

The integrand has simple poles at $s=-1, 0$, and $1$.

The fact that $\left|\Gamma(s)\right|$ decays exponentially fast to $0$ as $\text{Im} (s) \to \pm \infty$ allows us to shift the contour to the left.

I originally shifted the contour all the way to negative infinity.

But as Marko Riedel explains below, we want to shift the contour to the imaginary axis since the integrand is odd there.

Indeed, using the functional equation of the Riemann zeta function, we get $$ f(it) = \frac{it}{2 \pi} \sinh \left(\frac{\pi t }{2} \right) \operatorname{csch}(\pi t) \left|\zeta(1+it)\right|^{2}, \quad t \in \mathbb{R}.$$


$$ \int_{c-i \infty}^{c+i \infty} f(s) \, ds = 2 \pi i \ \text{Res}[f,1] + \pi i \ \text{Res}[f,0] ,$$


$$ \begin{align} \text{Res}[f,0] &= \lim_{s \to 0} s (2 \pi)^{-(s+1)} \Gamma(s+1) \zeta(s+1) \zeta(s) \\ &= \lim_{s\to 0} s\zeta(s+1) (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s) \\ &= 1\left(\frac{1}{2 \pi} \right)(1)\left(- \frac{1}{2} \right) \\ &=-\frac{1}{4 \pi} \end{align} $$


$$ \begin{align} \text{Res}[f,1] &= \lim_{s \to 1} (s-1) (2 \pi)^{-(s+1)} \Gamma(s+1) \zeta(s+1) \zeta(s) \\ &= \lim_{s\to 1}(s-1)\zeta(s) (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1) \\&= 1\left(\frac{1}{4 \pi^{2}}\right)(1)\left(\frac{\pi^{2}}{6}\right) \\ &=\frac{1}{24} . \end{align} $$

The result then follows.

  • $\begingroup$ $+1$. Nice. Wow !!!, $\endgroup$ Jan 31, 2021 at 19:18

2 Answers 2


In response to your query on the other thread -- I don't have the time to typeset this properly, here are some comments. The effort looks good and you should keep working on it.

  1. Don't extract any terms out from the transform function in front of the integral. As you have noticed yourself these reappear later on in the computation, reducing readability.

  2. Now this is the important part -- when evaluating harmonic sums you get an asymptotic expansion about zero when shifting to the left and about infinity when shifting to the right. So it is no surprise that you do not get the right value -- the expansion does not converge there (at $x=1$).

  3. With $g(s)$ being the transform of your sum we have $$ g(s) = \left( 2\,\pi \right) ^{-s-1}\Gamma \left( s+1 \right) \zeta \left( s+1 \right) \zeta \left( s \right).$$ The residues are $$\operatorname{Res}(g(s)/x^s; s=1) = \frac{1}{24x},$$ $$\operatorname{Res}(g(s)/x^s; s=0) = -\frac{1}{4\pi},$$ $$\operatorname{Res}(g(s)/x^s; s=-1) = \frac{x}{24}.$$ Now I suggest you shift from $\Re(s)=3/2$ to $\Re(s)=0$, indenting around the pole at zero by making a half-circle of radius $\epsilon$ around said pole at $s=0.$ This only picks up half the residue, so that your result is $$\frac{1}{24\times 1} + \frac{1}{2}\left(-\frac{1}{4\pi} \right) = \frac{1}{24} - \frac{1}{8\pi}.$$ You still need to verify that $g(s)$ is odd on the imaginary axis when $x=1$ by simplifying with the functional equation of the Riemann Zeta function.

That is all for now. I hope there are no mistakes.

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    $\begingroup$ What do you mean by "shifting?" We're trying to evaluate the integral of $g(s)$ along the vertical line $\Re (s)=\frac{3}{2}$ by closing the contour in such a way that the integral along the rest of the contour evaluates to zero (or evaluates to some finite value we can determine), right? I originally assumed that it would evaluate to zero along a half-circle of infinite radius opening to the left. That turned out to be an incorrect assumption. And when you say I should shift from $\Re (s)=\frac{3}{2}$ to $\Re (s)=0$ does that mean to use the rectangular contour with those sides? $\endgroup$ May 16, 2013 at 13:54
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    $\begingroup$ Yes, shifting refers to rectangular contours where the verb "shift" means to create a rectangle by moving the original Mellin integral to the left or to the right. As for question 297900, you need the half-plane of convergence of the transform term that is farthest to the right, giving $2(s-1)>1$ or $s>3/2.$ $\endgroup$ May 16, 2013 at 19:30
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    $\begingroup$ Remember that we are dealing with asymptotic expansions here -- these are not like a Taylor series that eventually converges to the exact value. You are right about $\pi^2/12-1/24$ being off the mark. This is because we are evaluating the sum $$\sum_{n\ge 0}\frac{x(2n+1)}{e^{x(2n+1)}-1}$$ at $x=1.$ The closer you get to $x=0$ the better the approximation $$\pi^2/12/x-x/24$$ gets. This just means that the integral on the left does not vanish. In the other example we had an integral on the left at $\Re(s)=0$ that vanished (!), so of course we used that to get an exact value. $\endgroup$ May 16, 2013 at 21:40
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    $\begingroup$ This is the type where we can get an exact answer. If we recast it in the mechanics that we have been talking about these past few days the transform of the relevant sum is $$\Gamma(s)(s-1/4)\zeta(2s).$$ The Mellin integral is on the line $\Re(s)=1$, we shift to $\Re(s)=1/4$ where the integrand vanishes, picking up the residue of the pole at $s=1/2$, which is $1/8\sqrt{\pi/x},$ getting $1/8$ when $x=\pi$ as an exact answer. The base function here is $e^{-x}(x-1/4).$ The old version that you refer to does it a bit differently. $\endgroup$ May 16, 2013 at 22:33
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    $\begingroup$ Yes of course. Fortunately Maple says that the Mellin inversion integrand is odd on the line $\Re(s)=1/4,$ so everything works out (the integral vanishes when $x=\pi$). You might want to verify this yourself and post your work here (shouldn't be more than a few lines). $\endgroup$ May 17, 2013 at 20:35

There is another missing piece (in response to the comments) that must be added here, and that is the proof of the claim that $$\frac{1}{\pi^s}\Gamma(s)(s-1/4)\zeta(2s)$$ is odd and purely imaginary on the line $\Re(s) = 1/4.$ This is due to a deep connection to the functional equation of the Riemman zeta function. We will show that $$\frac{1}{\pi^s}\Gamma(s)\zeta(2s)$$ is even and real when $\Re(s) = 1/4.$ The initial claim then follows.

We have from the proof of the functional equation of the Riemann zeta function that $$ \frac{1}{\pi^s}\Gamma(s)\zeta(2s) = -\frac{1}{2s(1-2s)} + \int_1^\infty (x^{s-1} + x^{-s-1/2}) \left(\sum_{q\ge 1} e^{-\pi q^2 x}\right)dx.$$ (The first term in the integral here is a straightforward Mellin transform while the second one requires more work.)

Putting $s=1/4+it$, we have that $$ \frac{1}{\pi^s}\Gamma(s)\zeta(2s) = -\frac{4}{1+16t^2} + \int_1^\infty (x^{-3/4+it} + x^{-3/4-it}) \left(\sum_{q\ge 1} e^{-\pi q^2 x}\right)dx.$$ But clearly $$x^{it} + x^{-it} = \exp(it\log x) + \exp(-it\log x) = 2\cos(t\log x)$$ is a real number and even in $t$ as is the fractional term in front and the result follows.

  • $\begingroup$ Thanks. Can I ask you another question? When we have to close the contour at negative infinity, how do we know how close we have to be to $x=0$ for the asymptotic expansion to be valid? In the other two examples we looked at, $x=1$ was close enough in one case and too far away in the other. $\endgroup$ May 20, 2013 at 19:26
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    $\begingroup$ @RandomVariable There is really nothing to stop you from studying additional harmonic sums to learn more. E.g. Maple will tell you what the Mellin transform of the sum is (though I like to prove these myself in SE posts) and it will compute the residues at the poles for you. You might want to attack a sum that has five poles instead of three in the left half plane and examine what the optimal location is for shifting the integral to the left. $\endgroup$ May 20, 2013 at 20:01
  • $\begingroup$ Another relevant concept is that of the fundamental strip of the transform of the base function, which is documented in the links I have posted. $\endgroup$ May 20, 2013 at 20:04
  • $\begingroup$ Yeah, I read about how two functions can have the same Mellin transform but for different strips like $e^{-x}$ and $e^{-x}-1$. I've been thinking of these strips as simply the values of $s$ on the complex plane where $\int_{0}^{\infty} f(x) x^{s-1} \ dx$ exists/converges. So we want $c$ to be in that region. That might be oversimplifying things. $\endgroup$ May 20, 2013 at 20:26
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    $\begingroup$ @RandomVariable see mathoverflow.net/questions/242907/… it is exactly about that. two different functions can have the same Mellin transform but with different domain of convergence $\endgroup$
    – reuns
    Jul 21, 2016 at 16:36

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