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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

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

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  • $\begingroup$ 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
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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 $2 a\sin(a x)\cos(a x) = a \sin(2 a x)$

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  • $\begingroup$ 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

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