1. ## Poisson summation formula

I just have no idea how to apply Poisson summation formula. Can someone give some help?
a) Let $\tau$ be fixed with $Im(\tau)>0$. Apply the Poisson summation formula to $f(z)=(\tau +z)^{-k}$ where $k \geq 2$ to obtain
$\sum_{n=-\infty}^{\infty} \frac {1}{(\tau+n)^k}= \frac{(-2\pi i)^k}{(k-1)!} \sum_{m=1}^{\infty} m^{k-1} e^{2 \pi im \tau}$
b) Does this formula still hold whenever $\tau$ is any complex number that is not an integer?

2. Originally Posted by namelessguy
I just have no idea how to apply Poisson summation formula. Can someone give some help?
a) Let $\tau$ be fixed with $Im(\tau)>0$. Apply the Poisson summation formula to $f(z)=(\tau +z)^{-k}$ where $k \geq 2$ to obtain
$\sum_{n=-\infty}^{\infty} \frac {1}{(\tau+n)^k}= \frac{(-2\pi i)^k}{(k-1)!} \sum_{m=1}^{\infty} m^{k-1} e^{2 \pi im \tau}$
b) Does this formula still hold whenever $\tau$ is any complex number that is not an integer?
As for how to apply Poisson sommation formula, everything is given in the text: they even give the function to use. The formula says:
$\sum_{n\in\mathbb{Z}} f(n)=\sum_{n\in\mathbb{Z}} \widehat{f}(n)$,
provided the second sum is absolutely convergent. And if the Fourier transform is defined as $\widehat{f}(\xi)=\int e^{-2i\pi x\xi}f(x)dx$.

The left-hand side is exactly what you want, so you have to check that the right-hand side matches the formula you're given. In other words, you have to compute the Fourier transform of $f$.

This can be tedious, but there are tricks:
first, notice that $f(x)=\frac{(-1)^{k-1}}{(k-1)!}\frac{d^{k-1}}{dx^{k-1}}\left(\frac{1}{\tau+x}\right)$, hence $\widehat{f}(\xi)=\frac{(-1)^{k-1}}{(k-1)!}(2i\pi\xi)^{k-1}\mathcal{F}\left(\frac{1}{\tau+x}\right)(\xi)$. You have to be careful here, since $x\mapsto \frac{1}{\tau+x}$ is not integrable, but only square-integrable.
Then you can notice that (by easy integration) $\mathcal{F}^{-1}\left(H(\xi)e^{2i\pi \xi\tau}\right)(x)=\frac{1}{-2i\pi(\tau+x)}$ where $H$ is the Heavyside function ( $H(x)=1$ if $x\geq 0$ and $H(x)= 0$ else). As a consequence, using Fourier inversion formula (in $L^2$), $\mathcal{F}\left(\frac{1}{\tau+x}\right)(\xi)=(-2i\pi)H(\xi)e^{2i\pi \xi\tau}$.
We deduce $\widehat{f}(\xi)=\frac{(-1)^{k-1}}{(k-1)!}(2i\pi\xi)^{k-1}(-2i\pi)e^{2i\pi\xi\tau}H(\xi)$ in $L^2$ (hence for almost every $\xi$), and hence for every $\xi$ because both functions are continuous (even for $\xi=0$ because $k\geq 2$).

As a consequence, $\widehat{f}(n)=\frac{(-2i\pi)^k}{(k-1)!}n^{k-1}e^{2i\pi n\tau}$ if $n> 0$ and 0 otherwise.

The series $\sum_n \widehat{f}(n)$ is absolutely convergent because $|e^{2i\pi n\tau}|=e^{{\rm Re}(2i\pi n\tau)}=e^{-2\pi n {\rm Im}(\tau)}$ and ${\rm Im}(\tau)>0$. That's why we can apply the Poisson sommation formula. And it gives exactly what is expected.

By the way, if ${\rm Im}(\tau)\leq 0$, the series would diverge.

3. Thanks a lot for your help Laurent. I really need to review this Fourier Transform materials. I wasn't able to apply this concept to show that $\sum_{n=-\infty}^{\infty} \frac {1}{(\tau+n)^2}=\frac{\pi^2}{sin^2(\pi \tau)}$. I plug in $k=2$ on both sides of the formula, but got nowhere.
I then tried an alternative way suggested by my friend. I integrate $f(z)=\frac{\pi cot \pi z}{(\tau + n)^2}$ over the circle $\mid z \mid=R_N$. I almost got the result (still working on showing $cot \pi z$ is bounded in that circle).

4. Originally Posted by namelessguy
Thanks a lot for your help Laurent. I really need to review this Fourier Transform materials. I wasn't able to apply this concept to show that $\sum_{n=-\infty}^{\infty} \frac {1}{(\tau+n)^2}=\frac{\pi^2}{sin^2(\pi \tau)}$. I plug in $k=2$ on both sides of the formula, but got nowhere.
For this, you should compute $\sum_{m=1}^\infty m x^m$ where $x=e^{2i\pi\tau}$. This sum is obtained by differentiating $\sum_{m=0}^\infty x^m=\frac{1}{1-x}$.