$\zeta(2)$ via partial fractions

Question

I was looking at old complex analysis exams, and there is one problem I can't figure out.

"Use the partial fraction expansion of $\frac{z}{e^z-1}$ to show $\sum_1^\infty 1/k^2=\frac{\pi^2}{6}$."

I recognize that as the generating function for the Bernoulli numbers, but I think the point of the problem is to solve it "from scratch", without that kind of knowledge.

The function has simple poles at $2\pi i k$ for $k\in\mathbb{Z}$, and residue $2\pi i k$ at $2\pi i k$ . Unfortunately, the obvious series, with terms of the form $\frac{2\pi i k}{z-2\pi i k}$, doesn't converge. Adding convergence terms (like in the proof of Mittag-Leffler's theorem) I get a series with terms of the form $\frac{z^2}{z^2-k^2}$, modulo some constants, but I don't see where to go from there, because it vanishes at 0. I think point is that we are supposed to evaluate the partial fraction decomposition at 0, as the function is clearly 1 there.

Thanks for the help.

As noted in the comments, there is an answer here that looks similar to what is intended: https://math.stackexchange.com/a/8373/1102 , however it seems much too involved for an exam setting, and is deliberately not rigorous. Would it be possible to modify it to be simpler and faster?

Edit: I managed to figure out a fairly slick solution that's much better than the accepted answer. I don't have time to write it up right now. If you read this and want to see it, ping me by posting a comment to this question.

Answer 1

Potato, this is just an idea, but note that

$$\int\limits_0^\infty {\frac{x}{{{e^x} - 1}}dx} = \frac{{{\pi ^2}}}{6}$$

With a change of variables one has that

$$\int\limits_0^\infty {\frac{x}{{{e^x} - 1}}dx} = - \int\limits_0^1 {\frac{{\log \left( {1 - x} \right)}}{x}dx} $$

Now use

$$ - \frac{{\log \left( {1 - x} \right)}}{x} = \sum\limits_{k = 1}^\infty {\frac{{{x^{k - 1}}}}{k}} \text{ ; } |x|&lt1$$

from where

$$ - \int\limits_0^x {\frac{{\log \left( {1 - t} \right)}}{t}dt} = \sum\limits_{k = 1}^\infty {\frac{{{x^k}}}{{{k^2}}}} $$

This means that

$$\int\limits_0^\infty {\frac{x}{{{e^x} - 1}}dx} = - \int\limits_0^1 {\frac{{\log \left( {1 - t} \right)}}{t}dt} = \sum\limits_{k = 1}^\infty {\frac{1}{{{k^2}}}} $$

So you might want to use residues (which I don't know about), to calculate

$$\int\limits_0^\infty {\frac{z}{{{e^z} - 1}}dz} $$

since it has singularities at every $z_k=2\pi ki$

---

Another known approach is given here, starting at $(35)$

$$\frac{z}{2} + \frac{z}{{{e^z} - 1}} = \frac{z}{2}\coth \frac{z}{2}$$

from where

$$\frac{z}{2}\coth \frac{z}{2} = \sum\limits_{n = 0}^\infty {{B_{2n}}\frac{{{z^{2n}}}}{{\left( {2n} \right)!}}} $$

and then

$$z\coth z = \sum\limits_{n = 0}^\infty {\frac{{{2^{2n}}{B_{2n}}}}{{\left( {2n} \right)!}}{z^{2n}}} $$

they then let $z=iz$, from where

$$z\cot z = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}\frac{{{2^{2n}}{B_{2n}}}}{{\left( {2n} \right)!}}{z^{2n}}} $$

They go on with partial fractions expansions, but I remember seeing elsewhere this:

$$\sin z = z\prod\limits_{n = 1}^\infty {\left( {1 - \frac{{{z^2}}}{{{n^2}{\pi ^2}}}} \right)} $$

then

$$\log \sin z = \sum\limits_{n = 1}^\infty {\log \left( {1 - \frac{{{z^2}}}{{{n^2}{\pi ^2}}}} \right)}+\log z $$

Differentiating and multiplying by $z$ gives

$$z\cot z = 1-2\sum\limits_{n = 1}^\infty {\dfrac{{ \dfrac{{{z^2}}}{{{n^2}{\pi ^2}}}}}{{1 - \dfrac{{{z^2}}}{{{n^2}{\pi ^2}}}}}} $$

but since

$$\frac{{\frac{{{z^2}}}{{{n^2}{\pi ^2}}}}}{{1 - \frac{{{z^2}}}{{{n^2}{\pi ^2}}}}} = \sum\limits_{k = 1}^\infty {\frac{1}{{{n^{2k}}{\pi ^{2k}}}}} {z^{2k}}$$

they write

$$z\cot z = 1- 2\sum\limits_{n = 1}^\infty {\sum\limits_{k = 1}^\infty {\frac{1}{{{n^{2k}}}}} \frac{{{z^{2k}}}}{{{\pi ^{2k}}}}} $$

then changing the order of the sum

$$z\cot z = 1 - 2\sum\limits_{k = 1}^\infty {\frac{{\zeta \left( {2k} \right)}}{{{\pi ^{2k}}}}{z^{2k}}} $$

Thus, since

$$z\cot z = 1 + \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{{{2^{2n}}{B_{2n}}}}{{\left( {2n} \right)!}}{z^{2n}}} $$

they argue that

$${\left( { - 1} \right)^{n + 1}}\frac{{{2^{2n - 1}}{B_{2n}}}}{{\left( {2n} \right)!}}{\pi ^{2n}} = \zeta \left( {2n} \right)$$