About $\int_{-\infty}^{\infty} \frac{\sin \left( x\right )}{x} \mathrm{d}x = \int_{-\infty}^{\infty} \frac{\sin ^ 2\left( x\right )}{x^2} \mathrm{d}x$

Question

How to show $\displaystyle\int_{-\infty}^{\infty} \frac{\sin \left( x\right )}{x} \mathrm{d}x = \int_{-\infty}^{\infty} \frac{\sin ^ 2\left( x\right )}{x^2} \mathrm{d}x$ ?

I've found this identity in this post. Following the comment by rnorris in that post "use integration by parts, starting from the right-hand side. Let $u=\sin^2(x)$, and $dv=x^{-2}dx$. Look at trig identities that involve the product $\sin(x)\cos(x)$."

I arrived to $$-\dfrac{\sin^2(x)}{x}\Big\vert_{-\infty}^{\infty}+\int_{-\infty}^{\infty}\dfrac{\sin(2x)}{x}\mathrm dx$$ where I used the trig identity $\sin(x+y)=\sin(x)\cos(x)+\cos(x)\sin(x)$ to change $\sin(x)\cos(x)$.

Could someone tell me what can be done next?

I don't see how I could modify the integral to have $\displaystyle\int_{-\infty}^{\infty} \frac{\sin \left( x\right )}{x} \mathrm{d}x $

Also, the trig identity $\sin(x+y)=\sin(x)\cos(x)+\cos(x)\sin(x)$, was that the one to be applied in the first place?

Thanks in advance.

Answer 1

Instead of using the trig identity you should do a change of variable.

Let $u = 2x$.

Answer 2

Let $ax=t$ $$I=\int_{-\infty}^{\infty} \frac{\sin ax}{x} dx=\int_{-\infty}^{\infty} \frac{\sin t}{t} dt$$ $I$ is independent of $a$.