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

  • $\begingroup$ By the tagging I'm afraid not, but: can you use complex analysis? In particular, complex integration? $\endgroup$ – DonAntonio Jan 2 '18 at 8:26
  • 1
    $\begingroup$ You didn't understand: I meant *complex analysis, like with "complex numbers"...:) $\endgroup$ – DonAntonio Jan 2 '18 at 8:39
  • $\begingroup$ @DonAntonio Oh yes complex analysis is required, but I think that will be complicated so any idea to develop the result using series? $\endgroup$ – Chen Guo Jan 2 '18 at 8:41
  • $\begingroup$ @DonAntonio Yes i accepted the answer below $\endgroup$ – Chen Guo Jan 2 '18 at 8:42
  • $\begingroup$ Well, that means you've already solved your problem with Guy's answer. Good for you. $\endgroup$ – DonAntonio Jan 2 '18 at 8:43

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 $\int_0^\infty \frac{\sin x} x \ dx = \frac \pi 2$?

  • $\begingroup$ 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. $\endgroup$ – DonAntonio Jan 2 '18 at 8:32
  • $\begingroup$ @Guy Fsone Can you elaborate the reason for asign(pi/2) how did u get that? $\endgroup$ – Chen Guo Jan 2 '18 at 8:33
  • $\begingroup$ @DonAntonio it takes the value 1 for $a>0$ and -1 for $a<0$ and 0 for $a=0$ $\endgroup$ – Guy Fsone Jan 2 '18 at 8:37
  • $\begingroup$ @GuyFsone I know that, but then "a sign(a)" is exactly the same as $\;|a|\;$ ...and indeed the solution to this integral is $\;\cfrac{\pi|a|}2\;$ ... $\endgroup$ – DonAntonio Jan 2 '18 at 8:38
  • $\begingroup$ @DonAntonio thanks very much for the remark it is true $\endgroup$ – Guy Fsone Jan 2 '18 at 8:40


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

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

  • $\begingroup$ 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? $\endgroup$ – DonAntonio Jan 2 '18 at 8:35
  • $\begingroup$ @Gribouillis I done till this, but after this i need to prove Guy Fsone's answer $\endgroup$ – Chen Guo Jan 2 '18 at 8:35
  • $\begingroup$ It is one change of variable away from a known integral. $\endgroup$ – Tob Ernack Jan 2 '18 at 8:40
  • $\begingroup$ @TobErnack If you mean the integral of $\;\sin x/x\;$ , the question is: "known integral"...for whom? Apparently not the OP... $\endgroup$ – DonAntonio Jan 2 '18 at 8:42
  • $\begingroup$ Right I guess I assumed they knew that one and wanted to reduce their problem to it. $\endgroup$ – Tob Ernack Jan 2 '18 at 8:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.