I want to prove the identity
$$ \int_{-\infty}^{+\infty}\frac{\sin x }{x}\frac{\sin(x /2)}{x / 2}dx = \pi $$
I think I can use here the following proposition, that can be easily proved using Fubini theorem: $$f, g \in L_1(\Bbb{R}) : \int_{-\infty}^{+\infty}gdx = 1 \Rightarrow \int_{-\infty}^{+\infty}(f * g)(x)dx = \int_{-\infty}^{+\infty}f(x)dx$$
We know that $$\int_{-\infty}^{+\infty}\frac{\sin x}{x}dx = 2 \int_{0}^{+\infty}\frac{\sin x}{x}dx = \pi$$ So can we find function $$g \in L_1(\Bbb{R}): (f * g)(x) = \frac{\sin x}{x}\frac{\sin (x / 2)}{x/2}, \: \int_{-\infty}^{+\infty}gdx = 1 ?$$
Maybe Fourier transform could be useful here ?
Integrate by parts,
$$ \int_{-\infty}^{+\infty}\frac{\sin x }{x}\frac{\sin\frac x2}{\frac x2}dx $$ $$ =-4\int_{0}^{\infty}\sin x \sin \frac x2 d(\frac1x) $$ $$ =\int_{0}^{\infty}\frac{4\cos x\sin\frac x2+2\sin x\cos \frac x2 }{x}dx $$
$$ =\int_{0}^{\infty} \frac{3 \sin\frac{3x}2 - \sin \frac x2 }{x}dx $$ $$ =(3-1)\int_{0}^{\infty}\frac{\sin t }{t}dt=\pi $$ where $\int_{0}^{\infty}\frac{\sin t }{t}dt=\frac\pi2$ is used.