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I'm trying to use the residue theorem to find the limit of $$\sum_{n=1}^{\infty} \frac{1}{n^4}.$$

So I am considering the function $$f(z) = \frac{\pi \cos(\pi z)}{\sin (\pi z)z^4}$$ on a square contour.

Now I am trying to find the residues of this function but am having some difficulty. I can see that we have singularities at $z = n$ for $n \in \mathbb{Z}$ which are simple except at $z=0$ and that is the residue I am having difficulty with.

Do I have no choice but to calculate the Laurent series here (I would need to take quite a lot of terms because of the $z^4$ in the denominator) or is there a better method that I haven't spotted.

Thanks

Wooster
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  • The only singularity in $\mathbb{C}$ that is not a simple pole is $0$, which is a pole of order $5$. You need to compute a bit of the Laurent series there. – Daniel Fischer Jun 13 '14 at 14:09
  • Have you seen this? – Antonio Vargas Jun 13 '14 at 14:10
  • Ah I meant to write $z=0$. I haven't seen that, it seems useful, strangely not in my lecture notes. So I guess the best option is just to calculate the laurent series then (or use that formula) – Wooster Jun 13 '14 at 14:11
  • @Wooster I wouldn't recommend using that formula, unless masochism is your thing. – Git Gud Jun 13 '14 at 14:12
  • Yes, it doesn't look like it would be pretty. But calculating the Laurent series to 4 terms doesn't seem too nice either – Wooster Jun 13 '14 at 14:12
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    @Wooster It's not really bad, you need the $z^4$ term of $$\frac{1- \frac{\pi^2z^2}{2} + \frac{\pi^4z^4}{24}}{1 - \frac{\pi^2z^2}{6}+\frac{\pi^4z^4}{120}}.$$ – Daniel Fischer Jun 13 '14 at 14:16
  • @DanielFischer Shouldn't it be the $z^3$? – Git Gud Jun 13 '14 at 14:17
  • @GitGud No, I pulled out a factor of $\pi z$ from $\sin \pi z$, so the factor for that fraction is $\frac{1}{z^5}$. – Daniel Fischer Jun 13 '14 at 14:18
  • @DanielFischer I did check for that trick, but missed something in my head. Thanks. – Git Gud Jun 13 '14 at 14:20
  • You can find the solutions in http://math.stackexchange.com/questions/28329/nice-proofs-of-zeta4-pi4-90 – xpaul Jun 13 '14 at 14:22
  • Okay great, I never want to work out these series if I can avoid it, I guess I just need to get on and do it! Thanks for the link, I can see it's been discussed quite a bit already. – Wooster Jun 13 '14 at 14:24
  • @Wooster, does it have to be with residues? Any reason why not to use Fourier Series, say? – DonAntonio Jun 13 '14 at 15:24
  • I can see it would probably be more straightforward with fourier series, but unfortunately this is revision for a complex analysis exam! – Wooster Jun 13 '14 at 21:20

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