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

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?

• You cannot have $-\sin^2 x/x$ after integration by parts since the definite integral is not a function of $x$. You have to evaluate it at the endpoints.
– Gary
Mar 3, 2021 at 17:33
• @Gary right right, sorry. Mar 3, 2021 at 18:49
• So that part becomes $0$ and you are left with the integral of $\sin(2x)/x$. Now make the change of integration variables $t=2x$.
– Gary
Mar 3, 2021 at 18:54
• @Gary indeed. Thanks. Mar 3, 2021 at 19:13

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

Let $$u = 2x$$.

• The downvote doesnt make sense,it seems the downvoter overlooked OP's work+1 Mar 3, 2021 at 16:10

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

• OP has already shown the second integral as $\int_{-\infty}^{\infty} \frac{sin 2x}{x} dx$ Mar 3, 2021 at 16:08
• You are right, very sorry.
– user65203
Mar 3, 2021 at 16:09
• But isn't this term -$$-\dfrac{\sin^2(x)}{x}$$ still there ? Mar 3, 2021 at 16:43
• @user215805 no, I actually didn't write correctly (sorry), see my edit. And actually I should have write that using limits. And then that'd go to $0$. Mar 4, 2021 at 2:35
• @Veronika Rmz, Yes, that looks better now. Mar 4, 2021 at 4:55