Can anyone prove the identity $\sum_{m=-\infty}^\infty (z+\pi m)^{-2} = (\sin z)^ {-2} $

Question

I came across this identity in a paper on elliptic curves, and the proof wasn't provided. It really irked me, and I couldn't find an explanation anywhere else. Can anyone shed some light?

$$\sum_{m=-\infty}^\infty (z+\pi m)^{-2} = (\sin z)^ {-2} $$

Answer 1

Hint: Differentiate twice the natural logarithm of Euler's infinite product expression for the sine function.

Answer 2

I don't have a solution, but I have a simplification.

I'd start with: $$\frac{d}{dx} -\cot(z) = \sin^{-2}z$$ $$\int \frac{1}{(z + \pi m)^2}dz = -\frac{1}{z + \pi m}$$

So then you would just need to prove that $\cot(z)$ is equal to $\sum_m \frac{1}{1+m\pi}$.

Answer 3

Show that $f(z)=\sum_{m=-\infty}^\infty (z+\pi m)^{-2} -(\sin z)^ {-2}$ is a bounded holomorphic function in $\Bbb C$, whence it is constant. Since it is $0$ in one point (for example at $i\infty$), it is identically $0$. To bound $f$ consider vertical strips (motivated by the series and the periodicity of the sine. To show it is holomorphic note the poles of both functions are of the same order and cancel throughout. Use then Riemann's continuation theorem.

Answer 4

Another way (using properties of polygamma). Let

\begin{align} A&\equiv\sum_{m=0}^\infty\frac{1}{(z/\pi+m)^2}=\Psi^{(1)}\!\left(1,\frac{z}{\pi}\right)\quad\text{and} \\ B&\equiv\sum_{m=-\infty}^0\frac{1}{(z/\pi+m)^2}=\sum_{m=0}^\infty\frac{1}{(-z/\pi+m)^2}=\Psi^{(1)}\!\left(1,-\frac{z}{\pi}\right) \\ &=-\Psi^{(1)}\!\left(1,\frac{z}{\pi}\right)+\frac{\pi^2}{z^2}+\pi^2\!\left(1+\cot(z)^2\right). \end{align}

Then

$$ \pi^{-2}\sum_{m=-\infty}^{\infty}\frac{1}{(z/\pi+m)^2}=\frac{A+B}{\pi^2}-\frac{1}{z^2}=\frac{1}{\sin^2(z)}. $$

Answer 5

Following the suggestion by @Lucian, we use the Euler's infinite product representation of the sine function and write

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

Next, taking the logarithmic derivative of $(1)$ yields

$$\cot z=\frac1z-2z\sum_{n=1}^{\infty}\frac{1}{n^2\pi^2-z^2} \tag 2$$

And then, taking the derivative of $(2)$ and multiplying by $-1$ yields

$$\begin{align} \csc^2z&=\frac{1}{z^2}+2\sum_{n=1}^{\infty}\frac{1}{n^2\pi^2-z^2} +4z^2\sum_{n=1}^{\infty}\frac{1}{(n^2\pi^2-z^2)^2} \\\\ &=\frac{1}{z^2}+2\sum_{n=1}^{\infty}\frac{(n^2\pi^2+z^2)}{(n^2\pi^2-z^2)^2}\\\\ &=\frac{1}{z^2}+\sum_{n=1}^{\infty}\left(\frac{1}{(n\pi+z)^2}+\frac{1}{(n\pi-z)^2}\right)\\\\ &=\frac{1}{z^2}+\sum_{n=1}^{\infty}\frac{1}{(n\pi+z)^2}+\sum_{n=-\infty}^{-1}\frac{1}{(n\pi+z)^2}\\\\ &=\sum_{n=-\infty}^{\infty}\frac{1}{(n\pi+z)^2} \end{align}$$

as was to be shown.