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For $\int_{0}^{\infty}\frac{\sin^2(x)}{x^2}dx$. I considered using residue theorem. But since the function inside is holomorphic except for a removable singularity at the origin. So whatever contour I take, the integral would always be zero. Is there a way that we can get away with the situation and apply residue theorem to do the work?

Martin Sleziak
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jack
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    if u type ''integral sine squared'' in google, this is the FIRST link: http://math.stackexchange.com/questions/903117/integral-of-sinc-function-squared-over-the-real-line – tired Dec 03 '15 at 00:38
  • I see... Cuz I'm doing residue theorem recently, so when I see an integral like this, the first thing comes to my mind is solving by residue... – jack Dec 03 '15 at 00:45
  • https://math.stackexchange.com/questions/13344/proof-of-int-0-infty-left-frac-sin-xx-right2-mathrm-dx-frac-pi2?noredirect=1&lq=1 – Guy Fsone Jan 02 '18 at 10:26

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Using the Residue Theorem seems unnecessary.

$$I(a)=\int_{0}^{\infty}\frac{\sin^2(ax)}{x^2}dx$$

$$\implies \frac{dI}{da}=\int_{0}^{\infty}\frac{2\sin(ax)\cos(ax)}{x}dx=\int_{0}^{\infty}\frac{\sin(2ax)}{x}dx=\frac{\pi}{2}$$

$$\implies I(a)=\frac{\pi}{2}a+C$$

But $I(0)=0$, so $C=0$, so $I(1)=\frac{\pi}{2}$.

If you want to use the Residue Theorem, or feel that Dirichlet's Integral is too much to assume, try taking the contour as $Re^{i\theta}$, but rather than a circle with a diameter cutting through $(0,0)$ as part of the path, bring the path above or below the real line. This will circumvent the problems you're having.

Community
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Nicholas Pipitone
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