# An outrageous way to derive a Laurent series: why does this work?

+ 8 like - 0 dislike
2336 views

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:

$$\begin{split} \frac{1}{e^{x}-1} &= \frac{e^{-x}}{1-e^{-x}} = \sum_{n=1}^{\infty} e^{-nx} \\ &= \sum_{n=1}^{\infty} \sum_{k=-\infty}^{\infty}\frac{(-nx)^{k}}{\Gamma(k+1)}\\ &= \sum_{k=-\infty}^{\infty}(-1)^{k}\frac{x^{k}}{\Gamma(k+1)}\sum_{n=1}^{\infty} n^{k}\\ &= \sum_{k=-\infty}^{\infty}(-1)^{k}\frac{\zeta(-k)}{\Gamma(k+1)} x^{k}\\ &= \sum_{k=-1}^{\infty}(-1)^{k}\frac{\zeta(-k)}{\Gamma(k+1)} x^{k}\\ \end{split}$$ 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.

This post imported from StackExchange Mathematics at 2014-08-13 07:21 (UCT), posted by SE-user higgsss
asked Aug 12, 2014
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. :)

This post imported from StackExchange Mathematics at 2014-08-13 07:21 (UCT), posted by SE-user user_of_math
Naive question: why do you think the order of summation in this case is not commutive?

This post imported from StackExchange Mathematics at 2014-08-13 07:21 (UCT), posted by SE-user mistermarko
@user_of_math I'll certainly do that. Thanks!

This post imported from StackExchange Mathematics at 2014-08-13 07:21 (UCT), posted by SE-user higgsss
I think it was a joke about phys.se ; but in any case this question certainly belongs on math.stackexchange.

This post imported from StackExchange Mathematics at 2014-08-13 07:21 (UCT), posted by SE-user littleO
@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.

This post imported from StackExchange Mathematics at 2014-08-13 07:21 (UCT), posted by SE-user higgsss
@littleO You are probably right. I guess I can first ask at Phys.SE.Meta whether it is off-topic or not.

This post imported from StackExchange Mathematics at 2014-08-13 07:21 (UCT), posted by SE-user higgsss
@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...

This post imported from StackExchange Mathematics at 2014-08-13 07:21 (UCT), posted by SE-user mistermarko
Even if I wear my ex-physicist's hat, this way of manipulating infinities looks a little bit too much ;-p

This post imported from StackExchange Mathematics at 2014-08-13 07:21 (UCT), posted by SE-user achille hui
Why in Frank's name did you expand $$e^z = \sum_{k=-\infty}^\infty \frac{z^k}{\Gamma(k+1)}\,?$$

This post imported from StackExchange Mathematics at 2014-08-13 07:21 (UCT), posted by SE-user Daniel Fischer
@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.

This post imported from StackExchange Mathematics at 2014-08-13 07:21 (UCT), posted by SE-user higgsss
So you did that "to make it work". Comforting. Still, it's interesting that it gives the right result.

This post imported from StackExchange Mathematics at 2014-08-13 07:21 (UCT), posted by SE-user Daniel Fischer

## 1 Answer

+ 1 like - 0 dislike

For $k=0,1,...$, let $a_k$ be the rational number defined by the serie expansion

$\frac{1}{e^x -1} = \frac{1}{x}+ \sum_{k=0}^\infty (-1)^k \frac{a_k}{\Gamma(k+1)} x^k$

The computation of the question simply shows that if there would be a "good value" to attach to $\sum_{n=1}^{+ \infty} n^k$, it should be $a_k$. The real question is why this "good value" is the same as the one obtained via the zeta function: $\zeta (-k)$. The answer is that the analytic continuation of the zeta function is defined by essentially the same computation.

The zeta function is defined by the formula $\zeta(s) = \sum_{n \geq 1} n^{-s}$ for $s \in \mathbb{C}$, $Re(s)>1$. To show that this function admits a meromorphic continuation to the all complex plane, a way to do is to use the trick to write

$n^{-s} = \frac{1}{\Gamma(s)} \int_{0}^{+\infty} e^{-nt} t^{s-1} dt$.

(Remark not useful for what follows: this trick is used very often in quantum field theory, where $t$ is called in this context a "Schwinger parameter" or the "Schwinger proper time").

Using this and summing the geometric serie, one finds for $Re(s)>1$.

$\zeta(s)= \frac{1}{\Gamma(s)} \int_{0}^{+\infty} \frac{t}{e^t - 1} t^{s-2} dt$.

It is possible to do an integration by part to obtain:

$- \frac{1}{(s-1) \Gamma(s)} \int_{0}^{+\infty} \frac{d}{dt}(\frac{t}{e^t - 1}) t^{s-1} dt$

The point is that this expression makes sense for $Re(s)>0$ so this formula defines a (and so the) analytic continuation of the zeta function to $Re(s)>0$. Doing successive integrations by part, one obtains the "full" zeta function and it is immediate from this that $\zeta(-k)=a_k$.

This computation defining the analytic continuation of the zeta function is really the same as (precisely the Mellin transform of) the OP's computation. It is one of the standard way to guess what $\sum_{n=1}^{+\infty} n^k$ "should be".

Remark: essentially the same computation appears in other ways to "regularize" $\sum_{n=1}^{+\infty} n^k$, for example the exponential regularization $\sum_{n=1}^{+\infty} n^k e^{-n \epsilon}$

answered Aug 13, 2014 by (5,120 points)
edited Aug 13, 2014 by 40227

## Your answer

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\varnothing$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.