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.

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 03 '21 at 17:33
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 03 '21 at 18:54

score 3 · Answer 1 · answered Mar 03 '21 at 16:02

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

Let $u = 2x$.

answered Mar 03 '21 at 16:02

Spencer

12,271

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The downvote doesnt make sense,it seems the downvoter overlooked OP's work+1 – Albus Dumbledore Mar 03 '21 at 16:10

MathArt · Answer 2 · 2024-03-19T11:45:30.027

Since this $$\int_{-\infty}^{\infty} \frac{\sin \left( x\right )}{x} \mathrm{d}x=\int_{-\infty}^{\infty} \frac{\sin^2\left( x\right )}{x^2} \mathrm{d}x$$ is really beautiful, I'am not satisfied to be a bystander. The left one is mostly known and we can evaluate the right one by Laplace techniques since both are even. \begin{align} \int_{-\infty}^{\infty}\sin^2\left( x\right )\frac{1}{x^2}\mathrm{d}x&=2\int_0^{\infty}\mathcal{L}\left[\sin^2\left(t\right)\right](x)\mathcal{L}^{-1}\left[\frac{1}{s^2}\right](x)\mathrm{d}x\\ &=2\int_0^{\infty}\frac{2}{4x + x^3}x\mathrm{d}x\\ &=2\int_0^{\infty}\frac{2}{2^2+ x^2}\mathrm{d}x\\ &=\pi. \end{align}

score 0 · Accepted Answer · answered Mar 03 '21 at 16:06

0

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

answered Mar 03 '21 at 16:06

Z Ahmed

43,235

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OP has already shown the second integral as $\int_{-\infty}^{\infty} \frac{sin 2x}{x} dx$ – Z Ahmed Mar 03 '21 at 16:08
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You are right, very sorry. – Mar 03 '21 at 16:09
But isn't this term -$$-\dfrac{\sin^2(x)}{x}$$ still there ? – user215805 Mar 03 '21 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 $0$. – Verónica Rmz. Mar 04 '21 at 02:35
@Veronika Rmz, Yes, that looks better now. – Z Ahmed Mar 04 '21 at 04:55
If $a\le 0$, this is FALSE. – Mark Viola Jan 20 '24 at 16:58
Yes you are right. I need to mention $a>0$. – Z Ahmed Jan 20 '24 at 17:54
For $a>0$, $$\int_{-\infty}^{\infty} \frac{\sin ax}{x} dx=\int_{-\infty}^{\infty} \frac{\sin ax}{ax} \mathrm dax$$ – MathArt Mar 19 '24 at 11:50

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

3 Answers3