I have to show summability, then compute the following integral:
$$\int\limits_{-\infty}^{+\infty} \frac{\sin(\pi\,x)}{\prod_{k = - n}^n (x - k)}\,dx = \frac{(-4)^n}{(2\,n)!}\,\pi $$
for every $n\in \mathbb{N}$. Is it possible to avoid the residue theorem?
I would be happy with a real-analytic solution, if possible. Thanks in advance.
A possible approach is the following: for first, prove that:
$$\frac{1}{\prod_{k=-n}^{n}(x-k)}=\frac{1}{(2n)!}\sum_{h=0}^{2n}(-1)^h \binom{2n}{h}\frac{1}{x-n+h}\tag{1}$$ through the residue theorem or partial fraction decomposition, then recall that: $$ \forall m\in\mathbb{Z},\qquad \int_{-\infty}^{+\infty}\frac{\sin(\pi x)}{x-m}\,dx = (-1)^m \int_{-\infty}^{+\infty}\frac{\sin(\pi x)}{x}\,dx = \pi(-1)^m \tag{2}$$ and at last, combine $(1)$ and $(2)$ to prove your claim: $$ \int_{-\infty}^{+\infty}\frac{\sin(\pi x)}{\prod_{k=-n}^{n}(x-k)}\,dx = \frac{\pi (-1)^n}{(2n)!}\sum_{h=0}^{2n}\binom{2n}{h} = \color{red}{\frac{\pi(-4)^n}{(2n)!}}.\tag{3}$$
Notice that by $(1)$ and $(2)$, summability is granted due to the (Riemann-)summability of $\frac{\sin x}{x}$, that follows from integration by parts or the Dirichlet's criterion.