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Another exam problem I'm looking at is to evaluate the following integral. $$ \int_{-\infty}^{\infty} \frac{x\hspace{-0.04 in}\cdot\hspace{-0.04 in}\sin(x)}{x^{\hspace{.02 in}2}+1} dx $$ This is a complex analysis exam, so the solution probably involves contours. $\:$ Since the integrand is an even function, one could potentially simplify by changing one endpoint to $0$. However, I have no idea how to make a contour work since the absolute value of the integrand grows exponentially away from the real axis. What does one need to do to evaluate that integral?

Ayman Hourieh
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    Consider the imaginary part of the same integral in $z$ where the $sin(z)$ has been replaced by $exp(\mathrm{i}z)$. Form a semicircular contour around the upper half plane. The pole at $z=\mathrm{i}$ is simple, and the Residue theorem should give you the result. – Bennett Gardiner Sep 12 '13 at 07:49
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    In addition to @BennettGardiner's hint, you also need some version of Jordan's lemma to control the contribution from the semi-circle. – mrf Sep 12 '13 at 08:37

2 Answers2

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Define

$$f(z)=\frac{ze^{iz}}{z^2+1}\;,\;\;C_R:=\{z=Re^{it}\in\Bbb C\;;\;R,t\in\Bbb R\;,\;0\le t\le \pi\}\;,\;$$

$$\gamma_r:=[-R,R]\cup C_R\;,\;\;\text{positively oriented}$$

For $\;R>1\;$ , our function has one unique simple pole at $\;z=i\;$ within $\;\gamma_R\;$ , with residue

$$\text{Res}_{z=i}(f)=\lim_{z\to i}\;(z-i)f(z)=\frac{ie^{-1}}{2i}=\frac1{2e}$$

So by Cauchy's Theorems

$$\frac{2\pi i}{2e}=\frac{\pi i}e=\oint\limits_{\gamma_R}f(z)dz=\int\limits_{-R}^R\frac{xe^{ix}}{x^2+1}dx+\int\limits_{C_R}f(z)dz$$

But by Jordan's Lemma

$$\left|\;\int\limits_{\gamma_R} f(z)dz\;\right|\xrightarrow [R\to\infty]{}0\;$$

So we get

$$\frac{\pi i}e=\int\limits_{-\infty}^\infty\frac{x(\cos x+i\sin x)}{x^2+1}dx$$

Take now just the imaginary parts in both sides to get the result.

DonAntonio
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0

By using Fourrier Transform

$$ \int^\infty_0 \frac{x \sin (ax)}{1 + x^2} \, dx= -\partial_a \int^\infty_0 \frac{\cos (ax)}{1 + x^2} \, dx = -\frac{1}{2}\partial_a \int_\Bbb R \frac{\cos (ax)}{1 + x^2} \, dx\\=-\frac{1}{2}\partial_a Re\left(\int_\Bbb R \frac{e^{iax}}{1 + x^2} \, dx\right)=-\frac{1}{2}\partial_a \left[Re\mathcal{F}^{-1}\left(\frac{1}{1 + x^2} \right)(a)\right]\\=\color{red}{-\frac{1}{2}\partial_a \left(\pi e^{-|a|}\right) =\frac{sign(a)\pi}{2} e^{-|a|}} $$

See all the explanatory details below

Recall that, if we consider the Fourier transform $$\mathcal Ff (a) =\int_\Bbb R e^{-ia x}f(x)dx$$ then its Fourier inverse is defined as $$\mathcal F^{-1}f (x) =\frac{1}{2\pi}\int_\Bbb R e^{it x}f(t)dt.$$

But we have, \begin{split} \mathcal F(e^{-|t|})(x) = \int_{-\infty}^{\infty}e^{-|t|}e^{-ix t}\,dt &=&\int_{-\infty}^{0}e^{t}e^{-ix t}\,dt+\int_{0}^{\infty}e^{-t}e^{-ix t}\,dt\\ &=&\left[ \frac{e^{(1-ix)t}}{1-ix} \right]_{-\infty}^0-\left[\frac{e^{-(1+ix)t}}{1+ix} \right]_{0}^{\infty}\\ &=&\frac{1}{1-ix}+\frac{1}{1+ix}\\ &=&\color{red}{\frac{2}{x^2+1}.} \end{split} Then, $$ \begin{align} e^{-|a|}=\mathcal F^{-1}\left( \frac{2}{x^2+1}\right)(a) &=\frac{1}{2\pi}\int_\Bbb R \frac{2}{x^2+1}e^{ix a}\,dx = \frac{1}{\pi}\int_\Bbb R\frac{e^{ix a}}{x^2+1}\,dx \\&=\frac{1}{\pi}\int_\Bbb R\frac{\cos a x}{x^2+1}\,dx = \frac{2}{\pi}\int_0^\infty\frac{\cos ax}{x^2+1}\,dx \end{align} $$ Given that, as $x\mapsto\sin ax $ is an old function we have, $$\int_\Bbb R \frac{\sin{a x}}{x^2+1}dx= 0.$$

Thus we have, $$ \int_0^\infty\frac{\cos ax}{x^2+1}\,dx =\frac{\pi}{2}e^{-|a|} $$

Guy Fsone
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