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Problem of the Week #361 - October 6, 2020

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Euge

MHB Global Moderator
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Jun 20, 2014
1,901
Here is this week's POTW:

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Using the method of contour integration or otherwise, evaluate the improper integral $$\int_0^\infty \left(\frac{\sin ax}{x}\right)^2\, dx$$ where $a$ is a positive constant.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,901
No one answered this week's problem. You can read my solution below.

Consider the contour integral $$\oint_\Gamma \frac{1 - e^{2aiz}}{z^2}\, dz$$ where $\Gamma$ is a semicircular contour of radius $R$ with an $\epsilon$-bump above the origin. In the upper half plane, the integrand is $O(\vert z\rvert^{-2})$ so that the integral of $(1 - e^{2aiz})/z^2$ along the semicircular arc of radius $R$ is $O(1/R)$ as $R \to \infty$. Since the integrand has principal part $-2ai/z$ about the origin, its integral over the $\epsilon$-bump is $-\pi i(-2ai) + O(\epsilon) = -2\pi a + O(\epsilon)$ as $\epsilon \to 0$. By Cauchy's theorem the contour integral is trivial, so in the limit as $R \to \infty$ and $\epsilon \to \infty$ we obtain $$0 = \operatorname{P.V.} \int_{-\infty}^\infty \frac{1-e^{2aix}}{x^2}\, dx - 2\pi a$$ or $$\operatorname{P.V.}\int_{-\infty}^\infty \frac{1-e^{2aix}}{x^2}\, dx = 2\pi a$$ Taking the real part and dividing by two yields $$\int_{-\infty}^\infty \frac{1-\cos(2ax)}{2x^2}\, dx = \pi a$$ By the trig identity $(1 - \cos(2ax))/2 = \sin^2 (ax)$ and symmetry we deduce $$\int_0^\infty \left(\frac{\sin ax}{x}\right)^2\, dx = \frac{\pi a}{2}$$