# How to prove this equality .

Let $f(z)=(\tau+z)^{-k}$. I want to apply the Poisson summation formula, to get that :

$$\sum_{n \in \mathbb{Z}} (\tau+n)^{-k} = \frac{(-2 \pi i)^{k}}{(k-1)!} \sum_{m=1}^{\infty} m^{k-1}e^{2 \pi i m \tau}.$$

But then how can I compute the Fourier transform?

I was thinking in residues but I don’t know which contour will be easiest for the computation, I mean to cancel integrals so that gives me the one I want and how to get that $\hat{f}(\zeta)=0$ for $\zeta <0$.

My attempt:

I choose $f(z)=\frac{e^{-2 \pi i z \zeta}}{(\tau +z)^k}$ and as our contour the lower-half semicircle union the real axis enclosing $-\tau$ our pole of order $k$, now, it is clear that the residue is

$$\frac{(-2 \pi i)^k}{(k-1)!} · \zeta^{k-1} e^{-2 \pi i (-\tau) \zeta}$$

Then the integral over the semicircle part is

$$\int_{\gamma_2} |f(z)| dz \le \frac{\pi |e^{-2 \pi i R \zeta}|}{(\tau +R)^{k}} \le \frac{\pi }{(\tau +R)^{k}} \to 0$$

as $R \to \infty$, so then:

$$\int_{-\infty}^{\infty}\frac{e^{-2 \pi i x \zeta}}{(\tau +x)^k}dx=\int_{-\infty}^{0}\frac{e^{-2 \pi i x \zeta}}{(\tau +x)^k}dx + \int_{0}^{\infty}\frac{e^{-2 \pi i x \zeta}}{(\tau +x)^k}dx= \frac{(-2 \pi i)^k}{(k-1)!} · \zeta^{k-1} e^{-2 \pi i (-\tau) \zeta}$$

With this observation we have two cases:

1) $\zeta <0 \Rightarrow \int_{-\infty}^{0}\frac{e^{-2 \pi i x \zeta}}{(\tau +x)^k}dx =\frac{(-2 \pi i)^k}{(k-1)!} · (-\zeta)^{k-1} e^{-2 \pi i (\tau) \zeta} \to 0$ as $\zeta \to \infty$

2) $\zeta >0 \Rightarrow \int_{0}^{\infty}\frac{e^{-2 \pi i x \zeta}}{(\tau +x)^k}dx= \frac{(-2 \pi i)^k}{(k-1)!} · \zeta^{k-1} e^{2 \pi i (\tau) \zeta}$

Then I conclude by the Poisson summation formula. Am I right?