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The following question was part of my complex analysis assignment and I am not able to prove it.

How to prove that $$\frac {\pi^2 } { \sin(\pi z)^2 } = \sum_{n=-\infty , n \neq0 }^{n=\infty} \frac{1}{(z-n)^2 }$$

I tried by using the identity $$\sin(\pi z) = \pi z \prod_{n=1}^{\infty} \left(1- \left(\frac zn \right)^2 \right) $$ But that is in product so can't be used here.

I don't have any other ideas except these. Actually the prof is known in department to be not good in teaching and online classes added more to it. In particularly in these type of questions where I have to prove some identity I am having very much difficulty.

It is my very humble request to you to help me.

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  • $\begingroup$ Look at encyclopediaofmath.org/wiki/Mittag-Leffler_theorem $\endgroup$
    – Gary
    Dec 16, 2020 at 9:34
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    $\begingroup$ Take the derivative of the $\log$ of the product. But the series is easier to prove: the RHS minus the LHS is an entire periodic function easily seen to be bounded, thus constant, and since it vanishes at $i\infty$.. $\endgroup$
    – reuns
    Dec 16, 2020 at 9:49
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    $\begingroup$ It is weird that $n=0$ is excluded in the summation, since the term corresponding to $n=0$ should also be included to make the identity hold true. Also, I suspect that this is a duplicate of the posting mention in Pedro Tamaroff's comment. $\endgroup$ Dec 16, 2020 at 10:22
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    $\begingroup$ Does this answer your question? $\sum_{n=-\infty}^{\infty}\frac{1}{(u +n)^2}=\frac{\pi^2}{(\sin \pi u)^2}$ $\endgroup$ Dec 16, 2020 at 10:24
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    $\begingroup$ As written, the identity is false. The $n=0$ term needs to be included to get $\frac{\pi^2}{\sin^2(\pi z)}$. Without that term, the sum is $\frac{\pi^2}3$ when $z=0$, and $\frac{\pi^2}{\sin^2(\pi z)}$ blows up at $z=0$. $\endgroup$
    – robjohn
    Oct 18, 2021 at 9:31

2 Answers 2

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A standard argument is as follows:

  1. Let $f$ denote your infinite sum, and let $g$ denote your function built out of the sine.
  2. Show that $f$ and $g$ are meromorphic functions with double poles only at the integers except zero. They also have no roots.
  3. This means that the difference $h=f-g$ is a holomorphic function, but it is also a function that satisfies $h(z)=h(z+1)$. Can you conclude?
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  • $\begingroup$ what is the contradiction in (3) can you please tell. I have understood all 3 points but can't conclude. $\endgroup$
    – Avenger
    Aug 18, 2021 at 14:00
  • $\begingroup$ @Avenger There's no contradiction, rather, you want to show that $h$ must be constant. $\endgroup$
    – Pedro
    Aug 18, 2021 at 15:25
  • $\begingroup$ How does h(z+1 ) =h(z) implies that h is a constant ? Sorry I am coming back so late but I have been not well $\endgroup$
    – Avenger
    Sep 10, 2021 at 4:08
  • $\begingroup$ Can you please reply? $\endgroup$
    – Avenger
    Sep 11, 2021 at 5:51
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    $\begingroup$ Show that $h=f-g$ is bounded on $0 \leq \Re z \leq 1$. Because of $h(z)=h(z+1)$ it is then bounded on the whole of the complex plane. Argue that $h$ is entire. Hence, by Liouville's theorem, $h$ is constant. $\endgroup$
    – Gary
    Sep 20, 2021 at 10:54
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This is a complex analysis problem.

This is a sum of a series by residues.

$$ \sum_{n = -\infty}^{\infty} f(n) = -\{ sum\, of\, residues\, of\, \pi\cot(\pi z)f(z)\, at\, all\, the\, poles\, of\, f(z) \} \tag{1}$$

Using $x$ to avoid confusion with $z$. $n$ is the variable of summation so replace it with $x$.

$$f(x) = \frac1{(z-x)^2} \tag{2}$$

$f(x)$ has a double pole at $x=z$.

$$ \sum_{n = -\infty}^{\infty} f(n) = - \pi \lim_{x \rightarrow z} \frac{d}{dx} \left[ (x-z)^2 \frac{\cot(\pi x)}{(z-x)^2}\right] \tag{3}$$

$$ \sum_{n = -\infty}^{\infty} f(n) = - \pi \lim_{x \rightarrow z} \frac{d}{dx} {\cot(\pi x)} \tag{4}$$

$$ \sum_{n = -\infty}^{\infty} f(n) = - \pi \lim_{x \rightarrow z} \frac{-\pi}{\sin^2(\pi x)} \tag{5}$$

$$ \sum_{n = -\infty}^{\infty} f(n) = \frac{\pi^2}{\sin^2(\pi z)} \tag{6}$$

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  • $\begingroup$ "$\sum_{n = -\infty}^{\infty} f(n) = -\{ sum\, of\, residues\, of\, \pi\cot(\pi z)f(z)\, at\, all\, the\, poles\, of\, f(z) \} $" Can you please write a proof of this statement or give some refence? $\endgroup$
    – Avenger
    Sep 21, 2021 at 16:21
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    $\begingroup$ @Avenger see for example this. $\endgroup$ Sep 21, 2021 at 18:42
  • $\begingroup$ Theorem 3.2. in @Evangelopoulos F reference. My reference is ancient: Schaum's outline Series, Theory and problems of complex variables - Murray R. Spiegel , 4th printing 1987, pg 187 Summation of series. $\endgroup$
    – user186104
    Sep 21, 2021 at 21:51
  • $\begingroup$ Take a look at Example 4.3. @Evangelopoulos F reference. Instead of $z \rightarrow \frac1{2}$ just leave it as $z$. $\endgroup$
    – user186104
    Sep 21, 2021 at 22:01
  • $\begingroup$ @arthur I am busy. But awarded you bounty. If I have a question I will comment. Thanks a lot. $\endgroup$
    – Avenger
    Sep 22, 2021 at 10:20

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