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I'm reading a paper by Riesel and Gohl; in it they say that "partial fraction expansion" of $1/(e^x-1)$ is $$\frac{1}{e^x-1}=\frac{1}{x}-\frac{1}{2}+2x \sum_{k=1}^{\infty} \frac{1}{x^2+4\pi^2k^2}$$ I tried using Euler's formula as $2\pi k$ are the roots of sine, but it leads nowhere. As the only root of $e^x-1$ is at $x=0$ and I fail to see how I can get this result with partial fractions. The Taylor expansion gives the first two terms correctly, but the rest is a power series involving Bernoulli numbers. So ho do I use partial fractions to expand $1/(e^x-1)$?

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Consider $f(z)=\frac{1}{e^z-1}$ as a meromorphic function on the complex plane and apply the residue theorem. $f(z)$ has simple poles for $z\in 2\pi i\mathbb{Z}$.

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    $\begingroup$ One thing this seems to leave out is how one gets the constant $1/2$ term. Taking the limit of $f(z)-1/z$ as $z\to 0$ works, but I'm forgetting if there's a slicker way. $\endgroup$ Sep 24 '14 at 17:53
  • $\begingroup$ @Semiclassical: maybe exploiting the fact that $\frac{x}{e^x-1}+\frac{x}{2}$ is an even function - in general, this is used to show that all the Bernoulli numbers with odd index, except $B_1=-1/2$, are zero. $\endgroup$ Sep 24 '14 at 17:57
  • $\begingroup$ That's a nice idea. A simple application of that: $$\frac{1}{2}(f(z)+f(-z))=\frac{1}{2}[\frac{1}{e^z-1}+\frac{1}{e^{-z}-1}]=-1/2.$$ So the even part is just a constant term. (Can't figure out why that Mathjax isn't displaying in comments.) $\endgroup$ Sep 24 '14 at 18:06
  • $\begingroup$ @JackD'Aurizio I don't have much experience in complex analysis. Could you detail your answer a bit. $\endgroup$
    – Nabigh
    Sep 25 '14 at 4:52
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$~\quad~$ Alternately, start by differentiating the natural logarithm of Euler's infinite product for the sine function, and then use the well-known relations between hyperbolic and trigonometric functions to establish the desired result.

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  • $\begingroup$ Did you mean the hyperbolic sine $\sinh(x)=x \prod_{k=1}^{\infty} \left(1+\frac{x^2}{\pi^2k^2} \right)$? Because if I use $\sin(x)=x \prod_{k=1}^{\infty} \left(1-\frac{x^2}{\pi^2k^2} \right)$, sign in summand is negative. $\endgroup$
    – Nabigh
    Sep 25 '14 at 4:01
  • $\begingroup$ After differentiating $\ln \sinh (x)$ and setting $x \rightarrow x/2$ I get $\frac{\cosh(x/2)}{\sinh(x/2)}=\frac{e^x}{e^x-1}+\frac{1}{e^x-1}=2/x+4x \sum_{k=1}^{\infty} (x^2+4\pi^2 k^2)^{-1}$. After this I'm stuck $\endgroup$
    – Nabigh
    Sep 25 '14 at 4:22
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    $\begingroup$ @Nabigh: Concerning your first comment, we have $\sin x=-i\sinh ix$, which follows from Euler's formula. As regards your second observation, all that's left to do now is proving that the two expressions for the value of our infinite series are equivalent, which is trivial. We have $a=b+c~S$ and $A=B+C~S$, which implies $\dfrac{a-b}c=\dfrac{A-B}C~$ $\endgroup$
    – Lucian
    Sep 25 '14 at 8:18

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