(Dis)Prove that the two integrals are equal

Question

(Dis)Prove that $$\int_0^{\infty}\frac{\sin^2x}{x^2}dx=\int_0^{\infty}\frac{\sin x}{x}dx$$

I don't know the approach for these kind of problems involving improper integrals. Like what should I do? Should I try to evaluate them by Feynman's trick or is there any other easy method?

Any help is greatly appreciated.

Answer 1

I think the easiest method is integration by parts: $$\int_0^{\infty}\frac{\sin^2x}{x^2}dx=-\frac{\sin^2x}{x}|_{x=0}^{\infty}+\int_0^{\infty}\frac{2\sin(x)\cos(x)}{x}dx=-\frac{\sin^2x}{x}|_{x=0}^{\infty}+2\int_0^{\infty}\frac{\sin(2x)}{2x}dx$$$$=-\frac{\sin^2x}{x}|_{0}^{\infty}+\int_0^{\infty}\frac{\sin(t)}{t}dt$$ Note that $\lim_{x\to 0}\frac{\sin^2x}{x}=0$ and $\lim_{x\to\infty}\frac{\sin^2x}{x}=0$, therefore: $$-\frac{\sin^2x}{x}|_{x=0}^{\infty}=0\Rightarrow \int_0^{\infty}\frac{\sin^2x}{x^2}dx=\int_0^{\infty}\frac{\sin(x)}{x}dx$$

Answer 2

Using the Trapezoidal Formula;

$${\int_0^\infty \left( \frac {\sin x}{x} \right)^p dx = \frac{\pi}{2^p(p-1)!} \left( p^{p-1} - { p \choose 1 } (p-2)^{p-1} + { p \choose 2 } (p-4)^{p-1} - \cdots \right)}$$