Evaluating the integral: $\int\limits_{0}^{\infty}\left(\frac{\sin(ax)}{x}\right)^2 dx , a \neq 0 $

Question

Integrate from 0 to infinity
$$\int\limits_{0}^{\infty}\left(\frac{\sin(ax)}{x}\right)^2 dx , a \neq 0 $$

I tried evaluating the indefinite integral of that function using Sine integral. But I failed to do it. I am having no idea to evaluate the definite integral. I seek for help

By the tagging I'm afraid not, but: can you use complex analysis? In particular, complex integration? — DonAntonio, Jan 02 '18 at 08:26
You didn't understand: I meant *complex analysis, like with "complex numbers"...:) — DonAntonio, Jan 02 '18 at 08:39
@DonAntonio Oh yes complex analysis is required, but I think that will be complicated so any idea to develop the result using series? — Chen Guo, Jan 02 '18 at 08:41
Well, that means you've already solved your problem with Guy's answer. Good for you. — DonAntonio, Jan 02 '18 at 08:43

score 2 · Accepted Answer · edited Jan 17 '18 at 09:11

2

First from here https://math.stackexchange.com/q/2508827 it is easy to prove using the changes $u=2y$ that $$\int_0^\infty \frac{\sin(u)}{u}dx= \int_0^\infty\left(\frac{\sin(u)}{u}\right)^2dx$$

Inserting the change of variables $u=ax$ one gets

$$\int_0^\infty\left(\frac{\sin(ax)}{x}\right)^2dx = a\int_0^\infty\left(\frac{\sin(ax)}{ax}\right)^2d(ax)\\= a \operatorname{sign}(a)\int_0^\infty\left(\frac{\sin(u)}{u}\right)^2dx =|a|\int_0^\infty \frac{\sin(u)}{u}dx =\color{blue}{\frac{|a|π}{2} }$$

this comes from the Dirichlet integral Evaluating the integral $\int_0^\infty \frac{\sin x} x \ dx = \frac \pi 2$?

edited Jan 17 '18 at 09:11

user153012

12,240

answered Jan 02 '18 at 08:28

Guy Fsone

23,903

Why "a sign(a)" instead of simply $,|a|;$ ? And the rightmost integral looks as messy as the original one. I mean, if the OP knows that then the integral in the question doesn't look that hard. – DonAntonio Jan 02 '18 at 08:32
@Guy Fsone Can you elaborate the reason for asign(pi/2) how did u get that? – Chen Guo Jan 02 '18 at 08:33
@DonAntonio it takes the value 1 for $a>0$ and -1 for $a<0$ and 0 for $a=0$ – Guy Fsone Jan 02 '18 at 08:37
@GuyFsone I know that, but then "a sign(a)" is exactly the same as $;|a|;$ ...and indeed the solution to this integral is $;\cfrac{\pi|a|}2;$ ... – DonAntonio Jan 02 '18 at 08:38
@DonAntonio thanks very much for the remark it is true – Guy Fsone Jan 02 '18 at 08:40
@Guy Fsone Thank you for wonderfully analysing the problem – Chen Guo Jan 02 '18 at 08:42

score 1 · Answer 2 · answered Jan 02 '18 at 08:26

1

Hint

$$\int\frac{\sin^2(ax)}{x^2} d x = - \frac{\sin^2(a x)}{x} + \int \frac{2a \sin(a x)\cos(a x)}{x} d x$$

and $2 a\sin(a x)\cos(a x) = a \sin(2 a x)$

answered Jan 02 '18 at 08:26

Gribouillis

14,188

How is it going to help the OP to attempt to solve $;\int\limits_0^\infty\frac{a\sin2ax}x\mathrm dx;$ ? Is this integral easier in any way than the original one? – DonAntonio Jan 02 '18 at 08:35
@Gribouillis I done till this, but after this i need to prove Guy Fsone's answer – Chen Guo Jan 02 '18 at 08:35
It is one change of variable away from a known integral. – Tob Ernack Jan 02 '18 at 08:40
@TobErnack If you mean the integral of $;\sin x/x;$ , the question is: "known integral"...for whom? Apparently not the OP... – DonAntonio Jan 02 '18 at 08:42
Right I guess I assumed they knew that one and wanted to reduce their problem to it. – Tob Ernack Jan 02 '18 at 08:43
@TobErnack That's probably true, though. Observe the OP's already accepted the first answer... – DonAntonio Jan 02 '18 at 08:44
Well, at least this explain why the original integral reduces to Dirichlet's integral, which solution is already on this site. – Gribouillis Jan 02 '18 at 08:47

Evaluating the integral: $\int\limits_{0}^{\infty}\left(\frac{\sin(ax)}{x}\right)^2 dx , a \neq 0 $

2 Answers2