The Poisson summation formula is $$\sum_{n \in \mathbb{Z}} f(2n\pi) = \dfrac{1}{2\pi} \sum_{n \in \mathbb{Z}} \widehat{f}(n)$$ where $\widehat{f}$ is the Fourier transform of $f$.
I have to apply the Poisson summation formula to $g(x) = (1-|x|)1_{[-1,1]}$, for all $x \in \mathbb{R} / \mathbb{Z}$ in order to prove $$\sum_{k \in \mathbb{Z}} \dfrac{1}{(x+k)^2} = \left(\dfrac{\pi}{\sin(\pi x)}\right)^2 .$$
Then, $\sum g(2n \pi) = \sum (1-|2n \pi|)1_{[-1,1]} = \sum (1-2\pi |n|) 1_{[-1,1]} = \dfrac{1}{2\pi} \sum \widehat{g}(n)$ but I don't know how to calculate $\widehat{g}(x) = \int_{\mathbb{R}} g(t) e^{-itx} dt$ and I don't know how to find this result.
Could someone help me ?
Hint
The idea is to consider the integral piece-wise so that each on each interval the expression of $g$ is nicer: \begin{align} \widehat{g}(x)&=\int_{\Bbb R} g(t) e^{-itx} dt = \int_{-\infty}^{-1}g(t) e^{-itx} dt+\int_{-1}^0 g(t) e^{-itx} dt + \int_0^1 g(t) e^{-itx} dt+\int_1^{+\infty} g(t) e^{-itx} dt\\ &=0+\int_{-1}^0 (1+t) e^{-itx} dt + \int_0^1 (1-t) e^{-itx} dt+0 \end{align} computing the remaining integrals: $$\widehat{g}(x)=\frac{2(1-\cos(x))}{x^2}$$