Is $\int_{0}^{\infty} \frac{\sin^2 x }{x^2}dx$ equal to $\int_{0}^{\infty} \frac{\sin x }{x}dx$?

Question

I saw this in a solution to a problem and can't see why this would be the case. Tried substition and other techniques.

Answer 1

$$ \int_{\mathbb{R}}\frac{\sin^{2}x}{x^{2}}dx = \left[\sin^{2}x\left(-\frac{1}{x}\right)\right]^{\infty}_{-\infty} - \int_{\mathbb{R}} 2\sin x\cos x \left(-\frac{1}{x}\right)dx \\ =\int_{\mathbb{R}}\frac{\sin 2x}{x}dx = \int_{\mathbb{R}}\frac{\sin y}{y}dy $$ Here we use integration by parts and the substitution $y = 2x$.