My actual goal is to prove the Lipschitz formula $$\sum_{n\in \mathbb{Z}}\frac{1}{(z+n)^k}=\frac{(-2\pi i)^k}{(k-1)!}\sum_{r=1}^\infty r^{k-1}e^{2\pi i r z}$$ with the help of the Poisson summation formula $$\sum_{n\in\mathbb{Z}}f(x+n)=\sum_{r\in\mathbb{Z}}\left(\int_\mathbb{R}f(t)e^{-2\pi i r t} dt \right)e^{2\pi i r x}.$$
If we use the Poisson summation formula with $f(x)=(x+iy)^{-k}$ the problem boils down to the Fourier transform of $f$ since $$\sum_{n\in \mathbb{Z}}f(x+n)=\sum_{n \in \mathbb{Z}}\frac{1}{(z+n)^k}=\sum_{r\in\mathbb{Z}}\left(\int_\mathbb{R}(t+iy)^{-k}e^{-2\pi i r t} dt \right)e^{2\pi i r x}.$$
Tackling the problem from the solution on, it must hold that $$\int_\mathbb{R}(t+iy)^{-k}e^{-2\pi i r t} dt= \begin{cases} 0, & r \leq 0 \\ \frac{(-2\pi i)^k}{(k-1)!}r^{k-1}e^{-2\pi ry}, & r>0\end{cases}.$$
So I went back trying to solve this integral. I started with the substitution $s=t+iy$ giving $$\int_\mathbb{R}(t+iy)^{-k}e^{-2\pi i r t} dt=\int_{-\infty+iy}^{\infty+iy}s^{-k}e^{-2\pi i r s}e^{-2\pi ry} ds.$$ As a consequence, we can already omit the $e^{-2\pi ry}$ since it's independent of the integral. So we are left with $$\int_{-\infty+iy}^{\infty+iy}\frac{1}{s^ke^{2\pi i rs}}ds.$$ Now, I had the idea to use the Residue theorem, however, I have no idea if it is allowed and how it can be justified. (I'll be glad if someone could clear things up). Luckily $$f(s)=\frac{1}{s^ke^{2\pi i rs}}$$ has only one pole at $0$ of order $k$. On wikipedia, I found the formula for the residue of a pole of order $k$. In my case I have $$res_0(f)=\frac{1}{(k-1)!}\lim_{s\to 0}\frac{\partial^{k-1}}{\partial s^{k-1}}[s^kf(s)]=\frac{(-2\pi i r)^{k-1}}{(k-1)!}.$$ So using the Residue theorem I get $$\int_{-\infty+iy}^{\infty+iy}\frac{1}{s^ke^{2\pi i rs}}ds=2\pi i \operatorname{res}_0(f)=\frac{2\pi i(-2\pi i )^{k-1}}{(k-1)!}r^{k-1}.$$
This looks pretty close, however, I am missing a $-1$ somewhere, because I need $(-2\pi i)^k$. Furthermore, I do not know how to make the use of the residue theorem rigorous. Moreover, I do not know how to show that the integral vanishes for $r\leq 0$. Thanks for your help!
It can be easier than that. Let us consider $$f\left(x\right)=\begin{cases} x^{k-1}e^{2\pi ixz}, & x>0\\ 0, & x\leq0. \end{cases}$$ Then, by the Poisson summation formula, we get
\begin{equation} \begin{split} \sum_{r=1}^{\infty}r^{k-1}e^{2\pi irz}= & = \sum_{r\in\mathbb{Z}}f\left(r\right) \\ & = \sum_{m\in\mathbb{Z}}\int_{\mathbb{R}}f\left(t\right)e^{2\pi imt}dt\\ & = \sum_{m\in\mathbb{Z}}\int_{0}^{\infty}t^{k-1}e^{2\pi it\left(m+z\right)}dt\\ & = \frac{\left(k-1\right)!}{\left(-2\pi i\right)^{k}}\sum_{m\in\mathbb{Z}}\frac{1}{\left(m+z\right)^{k}}. \end{split} \end{equation} It is clear that, with this techinque, the claim can be easily generalized.