TITLE: Proving $\int_{-\infty}^{+\infty}{\mathrm dx\over x^2}\cdot\left(2-{\sin x\over \sin\left({x\over 2}\right)}\right)^{n+1}={2n\choose n}\pi$
QUESTION [6 upvotes]: Given integral $(1)$

$$\int_{-\infty}^{+\infty}{\mathrm dx\over x^2}\cdot\left(2-{\sin x\over \sin\left({x\over 2}\right)}\right)^{n+1}={2n\choose n}\pi\tag1$$
  Where $n\ge0$

How does one prove $(1)$?
An attempt:
Recall $$\sin(2x)=2\sin x\cos x$$
$(1)$ becomes
$$2^{n+1}\int_{-\infty}^{+\infty}{\mathrm dx\over x^2}\cdot\left(1-\cos\left({x\over 2}\right)\right)^{n+1}\tag2$$
Applying binomial series to $(2)$, then we have
$$2^{n+1}\sum_{k=0}^{n+1}(-1)^k{n+1\choose k}\int_{-\infty}^{+\infty}{\mathrm dx\over x^2}\cdot\cos^k\left({x\over 2}\right)\tag3$$
Not sure how to proceed...

REPLY [4 votes]: In this solution I will use the Fourier transform
\begin{equation*}
\hat{f}(\xi) = \int_{-\infty}^{\infty}e^{-i\xi x}f(x)\, dx.
\end{equation*}
It is well known that if $f(x) = \dfrac{\sin^2(x)}{x^2}$ then $\hat{f}(\xi) = \pi\left(1-\frac{|\xi|}{2}\right)$ if $|\xi|<2$ and $0$ otherwise.
Via $1-\cos\left(\frac{x}{2}\right) = 2\sin^2(\frac{x}{4})$, the substitution $x:=4x$ and the Euler formula $\sin(x) = \frac{1}{2i}(e^{ix}-e^{-ix})$ I can proceed from Bui's formula (3).
\begin{gather*}
I = 2^{n+1}\int_{-\infty}^{\infty}\dfrac{1}{x^2}\left(1-\cos\left(\frac{x}{2}\right)\right)^{n+1}\, dx =  4^{n}\int_{-\infty}^{\infty}\dfrac{\sin^{2}(x)}{x^2}\sin^{2n}(x)\, dx =\\[2ex] \int_{-\infty}^{\infty}\dfrac{\sin^{2}(x)}{x^2}(-1)^{n}\left(e^{ix}-e^{-ix}\right)^{2n}\, dx.
\end{gather*}
But (cf. Jack D'Aurizio)
\begin{equation*}
\left(e^{ix}-e^{-ix}\right)^{2n}= \sum_{k=0}^{2n}\binom{2n}{k}(-1)^{2n-k}e^{-i(2n-2k)x} = \sum_{k=-n}^{n}\binom{2n}{n-k}(-1)^{n+k}e^{-i2kx}.
\end{equation*}
Thus
\begin{gather*}
I = \sum_{k=-n}^{n}\binom{2n}{n-k}(-1)^{2n+k}\int_{-\infty}^{\infty}\dfrac{\sin^{2}(x)}{x^2}e^{-i2kx}\, dx = \sum_{k=-n}^{n}\binom{2n}{n-k}(-1)^{2n+k}\hat{f}(2k) = \\[2ex]
\binom{2n}{n}\hat{f}(0) = \pi\binom{2n}{n}.
\end{gather*}