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

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

• It can be a simple coincidence. Does this appear as a step in a proof ? Provide more context. – Yves Daoust Nov 27 '18 at 8:57

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