I had to compute a series expansion of $1/(e^{x}-1)$ about $x=0$, and in the course of its derivation, I made a couple of manipulations that are not allowed mathematically. Still, comparing the final result against Maple showed that it was right.

The following is what I did:

\begin{equation} \begin{split} \frac{1}{e^{x}-1} &= \frac{e^{-x}}{1-e^{-x}} = \sum_{n=1}^{\infty} e^{-nx} \\ &= \sum_{n=1}^{\infty} \sum_{k=-\infty}^{\infty}\frac{(-nx)^{k}}{\Gamma(k+1)}\\ &= \sum_{k=-\infty}^{\infty}(-1)^{k}\frac{x^{k}}{\Gamma(k+1)}\sum_{n=1}^{\infty} n^{k}\\ &= \sum_{k=-\infty}^{\infty}(-1)^{k}\frac{\zeta(-k)}{\Gamma(k+1)} x^{k}\\ &= \sum_{k=-1}^{\infty}(-1)^{k}\frac{\zeta(-k)}{\Gamma(k+1)} x^{k}\\ \end{split} \end{equation} I naively exchanged the order of summation to obtain the third line. Then, I pretended that the summation over $n$ converged (i.e., as if $k$ were always smaller than -1), and replaced it by the Riemann zeta function. Notice that whenever $1/\Gamma(k+1) = 0$ , the coefficient of $x^{k}$ vanishes except when $k=-1$, for which the poles of gamma and zeta functions cancel out and give a finite value.

I suppose that there is a deeper reason that such unreliable steps led me to the correct result? My guess is that it has to do with the analytic structure of the summand as a function of $k$, but I haven't been able to figure out in detail why and when this works.

I'll appreciate any insights on this.

  • 11
    $\begingroup$ Perhaps you should post this on PHYS.SE? Physicists are skilled in the art of getting meaningful answers using mathematically non-kosher, but physically intuitive, techniques. :) $\endgroup$ – user_of_math Aug 12 '14 at 5:01
  • 1
    $\begingroup$ Naive question: why do you think the order of summation in this case is not commutive? $\endgroup$ – user117644 Aug 12 '14 at 5:48
  • 4
    $\begingroup$ I think it was a joke about phys.se ; but in any case this question certainly belongs on math.stackexchange. $\endgroup$ – littleO Aug 12 '14 at 5:54
  • 2
    $\begingroup$ Why in Frank's name did you expand $$e^z = \sum_{k=-\infty}^\infty \frac{z^k}{\Gamma(k+1)}\,?$$ $\endgroup$ – Daniel Fischer Aug 12 '14 at 23:41
  • 2
    $\begingroup$ You could post the question on PhysicsOverflow, see the link on my profile. $\endgroup$ – Dilaton Aug 13 '14 at 0:11

Let us assume ${\rm Re}(x)>0$ so that the geometric series is convergent. Perhaps the most intuitive way to justify OP's calculation is to use the Mellin transform. The Mellin integration coutour $\gamma$ in the complex $s$-plane is a vertical line ${\rm Re}(s)=c$ directed upward. The positive constant $c>1$ is chosen positive enough to justify switching the order of $s$-integration and $n$-summation below. Then we can mimick OP's calculation as follows.

\begin{equation} \begin{split} \frac{1}{e^{x}-1} ~&=~ \frac{e^{-x}}{1-e^{-x}} ~\stackrel{{\rm Re}(x)>0}{=}~ \sum_{n\in\mathbb{N}} e^{-nx} \\ ~&=~\sum_{n\in\mathbb{N}} \sum_{k\in\mathbb{N}_0} \frac{(-nx)^k}{k!}\\ ~&=~\sum_{n\in\mathbb{N}} \sum_{k\in\mathbb{N}_0} {\rm Res}\left(s\mapsto\Gamma(s)(nx)^{-s} ,-k\right) \\ ~&=~\sum_{n\in\mathbb{N}}\int_{\gamma}\! \frac{\mathrm{d}s}{2\pi i} \Gamma(s)(nx)^{-s} \\ ~&=~\int_{\gamma}\! \frac{\mathrm{d}s}{2\pi i} \sum_{n\in\mathbb{N}}\Gamma(s)(nx)^{-s} \\ ~&\stackrel{c>1}{=}~\int_{\gamma}\! \frac{\mathrm{d}s}{2\pi i}\Gamma(s)\zeta(s)x^{-s} \\ ~&=~ \sum_{k=-1}^{\infty} {\rm Res}\left(s\mapsto \Gamma(s)\zeta(s)x^{-s} ,-k\right) \\ ~&=~\frac{1}{x}+ \sum_{k\in\mathbb{N}_0} \frac{\zeta(-k)(-x)^k}{k!}.\\ \end{split} \end{equation}

Here we have (among other things) used:

  1. that the Gamma function $\Gamma(s)$ has poles at the non-positive integers $s=-k$, $k\in\mathbb{N}_0$, with residue ${\rm Res}(\Gamma,-k)=\frac{(-1)^k}{k!}$, and

  2. that the Riemann zeta function $\zeta(s)$ has a pole at $s=1$ with residue ${\rm Res}(\zeta,1)=1$.

| cite | improve this answer | |
  • $\begingroup$ Thank you very much for such a crystal-clear explanation! $\endgroup$ – higgsss Aug 16 '14 at 12:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.