2
$\begingroup$

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.

$\endgroup$
4
  • 2
    $\begingroup$ 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. $\endgroup$
    – Gary
    Mar 3, 2021 at 17:33
  • $\begingroup$ @Gary right right, sorry. $\endgroup$ Mar 3, 2021 at 18:49
  • 1
    $\begingroup$ So that part becomes $0$ and you are left with the integral of $\sin(2x)/x$. Now make the change of integration variables $t=2x$. $\endgroup$
    – Gary
    Mar 3, 2021 at 18:54
  • $\begingroup$ @Gary indeed. Thanks. $\endgroup$ Mar 3, 2021 at 19:13

2 Answers 2

2
$\begingroup$

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

Let $u = 2x$.

$\endgroup$
1
  • 3
    $\begingroup$ The downvote doesnt make sense,it seems the downvoter overlooked OP's work+1 $\endgroup$ Mar 3, 2021 at 16:10
0
$\begingroup$

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

$\endgroup$
7
  • 2
    $\begingroup$ OP has already shown the second integral as $\int_{-\infty}^{\infty} \frac{sin 2x}{x} dx$ $\endgroup$
    – Z Ahmed
    Mar 3, 2021 at 16:08
  • 1
    $\begingroup$ You are right, very sorry. $\endgroup$
    – user65203
    Mar 3, 2021 at 16:09
  • $\begingroup$ But isn't this term -$$-\dfrac{\sin^2(x)}{x}$$ still there ? $\endgroup$
    – user215805
    Mar 3, 2021 at 16:43
  • $\begingroup$ @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 $0$. $\endgroup$ Mar 4, 2021 at 2:35
  • $\begingroup$ @Veronika Rmz, Yes, that looks better now. $\endgroup$
    – Z Ahmed
    Mar 4, 2021 at 4:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .