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So, I have to solve the following integral $$ \int_{0}^{\infty} \frac{\sin^{2}{x}}{x^{2}} dx$$

I'm aware that this has been asked about before on this site, but I want comments on my attempted solution.

for $x \in \mathbb{R}$, the principal part of

$$ Re\int_{0}^{\infty} \frac{1- e^{2ix}}{x^{2}}dx = \int_{0}^{\infty}\frac{1-\cos{2x}}{x^2} dx= \int_{0}^{\infty}2\frac{\sin^{2}{x}}{x^{2}}dx$$

So, I'm going to integrate $$\int_{\Gamma} \frac{1- e^{2iz}}{z^{2}}dz $$

where, $\Gamma = \gamma_{1}+\gamma_{2}+\gamma_{3}+\gamma_{4}$. $\gamma_{1} = [-R,-r], \gamma_{2} = C_{r}, \gamma_{3} = [r, R]$, and $\gamma_{4}= C_{R}$. $r,R > 0$, $r < R$, and $C_{r}$ is a clockwise oriented semicircle of radius $r$ and $C_{R}$ is a counterclockwise oriented semicircle of radius $R$.

$$ \int_{\Gamma} f(z) dz = 0 $$ by residue theorem. Furthermore,

$$ \int_{\gamma_{3}} f(z) dz = -2\pi i (-2i)= -4\pi $$. I got this from calculating the residue:

$$ \frac{1- e^{2iz}}{z^{2}} = \frac{1-(1+2iz+...}{z^{2}} = -\frac{2i}{z}+... $$

and $$ \int_{\gamma_{4}} f(z) dz = 0 $$ by Jordan's lemma and estimation. Hence, as $r \rightarrow 0, R \rightarrow \infty$, we get:

$$ 0 = \int_{-\infty}^{0} f(x)dx -4\pi + \int_{0}^{\infty} f(x) dx +0$$

$$ \int_{-\infty}^{\infty} f(x) dx = 4\pi $$.

So, $$ \int_{0}^{\infty} \frac{\sin^{2}{x}}{x^{2}} dx = \frac{1}{4} \int_{-\infty}^{\infty} f(x) dx = \frac{1}{4}(4\pi) = \pi$$. But, Wolfram Alpha says that the answer should be $\frac{\pi}{2}$. Where did I go wrong?

thecatinfinity
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  • Can you expand your line calculating the integral on $\gamma_3$ as $-4 \pi$? I presume you mean to say the integral around $\gamma_2$, i.e. the shrinking semicircle which will contribute half? Your contours seem correct. My guess would be that you are not accounting for the fact that the contour about the origin is only half a circle, and thus only contributes half of its residue. If I'm not mistaken, this makes the contribution $-2\pi$ so the last line becomes $$\int_{0}^{\infty} \frac{\sin^{2}{x}}{x^{2}}dx=\frac{1}{4} \int_{-\infty}^{\infty} f(x)dx = \frac{1}{4}(2\pi) = \frac{\pi}{2}$$ – Brevan Ellefsen Oct 05 '20 at 23:26
  • Fair warning that I only briefly skimmed your post, so I might have misinterpreted you somewhere. If I am blatantly wrong, let me know. – Brevan Ellefsen Oct 05 '20 at 23:30
  • @BrevanEllefsen You are correct I meant $\gamma_{2}$. I was not aware that the semicircle would only have the contribution. My impression was that as long as we have a bounded domain, the residue theorem would apply as usual. But, I suppose your idea makes sense because if I do it with a full circle, the integral should be twice as what I get with the semicircle. – thecatinfinity Oct 05 '20 at 23:38
  • See here for an explanation: https://math.stackexchange.com/q/319959/269764 – Brevan Ellefsen Oct 06 '20 at 00:08

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