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I am trying to figure out how to prove the following (where $a$ is a positive real number): $$\int_0^\infty \frac{x \sin ax}{x^2+1} dx=\frac{\pi}{2}e^{-a}$$

This integral comes up in solving the 3D screened Poisson equation, and the link there suggests it may be evaluated using a complex contour integral. But as best I can tell, a contour integral around the pole at $x=i$ would evaluate to $\frac{\pi}{2}(e^{-a}-e^a)$, and I can't see what contour to take that would allow cancellation of the part of the result proportional to $e^a$.

How can one perform this integral?

Matt Dickau
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1 Answers1

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Since your function is even, it suffices to compute $I = \int_{-\infty}^\infty \frac{x \sin ax}{x^2 + 1} dx$ and divide by $2$.

Use the upper semicircle $\Gamma$ consisting of $\Gamma_1: [-R, R]$, and $\Gamma_2: Re^{it}, 0 \leq t \leq \pi$.

$\sin az$ does not tend to $0$ on $\Gamma_2$. Instead, use $e^{iaz}$ and take imaginary part. By the residue theorem, \begin{align*} \int_{-R}^R \frac{xe^{iax}}{x^2 + 1} dx + \int_{\Gamma_2} \frac{ze^{iaz}}{z^2 + 1} dz& = \oint_\Gamma \frac{ze^{iaz}}{z^2 + 1} dz \\ & = 2\pi i \frac{e^{-a}}{2}\\ & = \pi i e^{-a}. \end{align*} The integral $\int_{\Gamma_2} \frac{ze^{iaz}}{z^2 + 1} dz$ tends to $0$ as $R \rightarrow 0$. So we get $$ \int_{-\infty}^\infty \frac{xe^{iax}}{x^2 + 1}dx = \pi i e^{-a}. $$ Once you take imaginary part and divide by $2$, you get $$\int_0^\infty \frac{x\sin ax}{x^2 + 1}dx = \pi\frac{e^{-a}}{2}.$$

nowhere
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