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$$\sum_{k=-\infty}^\infty \frac{1}{(k+\alpha)^2} = \frac{\pi^2}{\sin^2\pi \alpha}$$

I have studied complex analysis.

My plan :

Let $g(z) = \frac{\pi^2}{\sin^2 \pi z} - \sum_{n=-\infty}^{n=\infty} \frac{1}{z-n}$

  1. Show that $g$ is entire with period 1
  2. Show that $g \to 0$ as $|z| \to \infty$.

Could you help me?

Seongqjini
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2 Answers2

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The problem can be solved using the following method: Let $C_N$ be the circle centered at the origin of radius $R_N = N + 1/2$, where $N$ is an integer and $N \geq \vert \alpha \vert$. Use the residue theorem to show that $$\frac{1}{2 \pi i}\int_{C_N} \frac{\pi \cot \pi z}{(z + \alpha)^2} = \sum_{k = -N}^{N}\frac{1}{(k + \alpha)^2} - \frac{\pi^2}{\sin^2\pi\alpha}.$$ The result then follows once you prove that the limit as $N \to \infty$ of the integral is $0$. This is just a matter of applying the usual estimate in terms of of curve length for complex line integrals (the tricky part is bounding $\cot \pi z$ on $C_N)$.

Note that you also need the hypothesis that $\alpha$ is not an integer.

Ethan Alwaise
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  • Since $\displaystyle f(z) = \sum_{k= -\infty}^\infty \frac{1}{(z-k)^2}$ converges compactly on $\mathbb{C}\setminus \mathbb{Z}$, as its term-by-term derivative, it is straightforward to show that $f(z)$ is holomorphic everywhere except at $z \in \mathbb{Z}$ where it has poles of order $2$, and is $1$-periodic and bounded as $Im(z) \to \pm \infty$ .

  • Since $\sin(z) = z (1+h(z))$ where $h(z)$ is analytic at $z=0$ and $h(0) = 0$, you have that $\frac{\pi^2}{\sin^2(\pi z)}- \frac{1}{z^2}$ is holomorphic around $z= 0$, and hence $g(z) = \frac{\pi^2}{\sin^2(\pi z)}-f(z)$ is holomorphic around any $z=n$. Since $ \frac{\pi^2}{\sin^2(\pi z)},f(z)$ have no other singularities, it means $g(z)$ is entire.

  • Since $ \frac{\pi^2}{\sin^2(\pi z)}$ is $1$-periodic, it also means that $g(z)$ is $1$-periodic, so it is bounded on the real line and on $|Im(z)| < T$, and since $f(z)$ and $ \frac{\pi^2}{\sin^2(\pi z)} = \mathcal{O}(e^{-|Im(z)|})$ both are bounded as $Im(z) \to \pm\infty$, it mans $g(z)$ is bounded. Hence by Liouville's theorem it is constant.

  • As $Im(z) \to \infty$ : $\frac{\pi^2}{\sin^2(\pi z)} \to 0$, while the decrease-ness of $\frac{1}{|z-n|^2} $and the absolute convergence gives $f(z) \to 0$, so that $g(z) \to 0$ and $$\frac{\pi^2}{\sin^2(\pi z)}= \sum_{k= -\infty}^\infty \frac{1}{(z-k)^2}$$

reuns
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