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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.

Nate Eldredge
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higgsss
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    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. :) – user_of_math Aug 12 '14 at 05:01
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    Naive question: why do you think the order of summation in this case is not commutive? –  Aug 12 '14 at 05:48
  • @user_of_math I'll certainly do that. Thanks! – higgsss Aug 12 '14 at 05:53
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    I think it was a joke about phys.se ; but in any case this question certainly belongs on math.stackexchange. – littleO Aug 12 '14 at 05:54
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    @mistermarko I'm not well-versed with mathematics, but from what I've looked up in the internet, it seems that Fubini-Tonelli theorem is relevant here. What is obvious is that the $n$-summation in the third line is divergent, which automatically makes it nonsensical. – higgsss Aug 12 '14 at 05:57
  • @littleO You are probably right. I guess I can first ask at Phys.SE.Meta whether it is off-topic or not. – higgsss Aug 12 '14 at 06:03
  • @higgsss But it can't be automatic because it works! I've never worked with this kind of thing before but I sense a mystery... –  Aug 12 '14 at 07:39
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    Even if I wear my ex-physicist's hat, this way of manipulating infinities looks a little bit too much ;-p – achille hui Aug 12 '14 at 23:16
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    Why in Frank's name did you expand $$e^z = \sum_{k=-\infty}^\infty \frac{z^k}{\Gamma(k+1)},?$$ – Daniel Fischer Aug 12 '14 at 23:41
  • @DanielFischer What I really did at first was to calculate the Taylor series of $1/(e^{x}+1)$. For this case, the expansion $e^{x} = \sum_{k=0}^{\infty}\frac{x^{k}}{k!}$ sufficed to give the right result, and the only outrage was "evaluating" the divergent $n$-summation using the Riemann zeta function. Then I wondered whether a similar procedure would work for the Laurent series of $1/(e^{x}-1)$, and found that extending the lower limit of the $k$-summation to $-\infty$ using the gamma function does the job. – higgsss Aug 12 '14 at 23:49
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    So you did that "to make it work". Comforting. Still, it's interesting that it gives the right result. – Daniel Fischer Aug 12 '14 at 23:52
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    You could post the question on PhysicsOverflow, see the link on my profile. – Dilaton Aug 13 '14 at 00:11
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    To me, this naively looks similar to zeta function regularization, as is used in QFT a lot. – Danu Aug 13 '14 at 10:18
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    The question has now obtained a quite detailled answer on PhysicsOverflow. – Dilaton Aug 13 '14 at 11:55
  • @mistermarko I think it is clear that the order of summation cannot be changed, as $$\sum_{n=0}^\infty n^k$$ diverges for all $k\geq 0$. – E.P. Aug 13 '14 at 17:06

1 Answers1

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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 ${\rm Res}(\zeta,1)=1$.

Qmechanic
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