Integral identity involving sin(x)/x

Question

Prove or disprove

$$\displaystyle\int_{-\infty}^{\infty} \frac{3 \sin \left( x\right )}{x} \mathrm{d}x = \int_{-\infty}^{\infty} \frac{4 \sin ^ 3\left( x\right )}{x^3} \mathrm{d}x$$

Answer 1

Any integral of the form $$ \int_{\mathbb{R}}\left(\frac{\sin x}{x}\right)^n\,dx $$ can be computed through integration by parts: it is enough to compute the Fourier sine series of $\sin^n x$, differentiate it $(n-1)$ times and exploit: $$ \int_{\mathbb{R}}\frac{\sin(mx)}{x}\,dx = \pi.$$ In this case, the result directly follows from the identity: $$ \sin(3x)=3\sin x-4\sin^3 x,$$ leading to: $$ 4\int_{\mathbb{R}}\left(\frac{\sin x}{x}\right)^3\,dx = \int_{\mathbb{R}}\frac{3\sin x-\sin(3x)}{x^3}\,dx =\frac{1}{2}\int_{\mathbb{R}}\frac{9\sin(3x)-3\sin x}{x}\,dx=3\pi.$$