# Compute int_0^(oo) (sin(x))/(xe^x)dx

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$\begin{array}{c}{\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}x}{x{e}^{x}}dx={\int }_{0}^{\mathrm{\infty }}\mathrm{sin}x{\int }_{1}^{\mathrm{\infty }}{e}^{-sx}dsdx={\int }_{1}^{\mathrm{\infty }}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{\int }_{0}^{\mathrm{\infty }}\mathrm{sin}\left(x\right){e}^{-sx}dx\phantom{\rule{thinmathspace}{0ex}}ds\hfill \\ \hfill ={\int }_{1}^{\mathrm{\infty }}\mathcal{L}\left[\mathrm{sin}\left(x\right)\right]\left(s\right)ds={\int }_{1}^{\mathrm{\infty }}\frac{ds}{{s}^{2}+1}=\frac{\pi }{2}-{\mathrm{tan}}^{-1}1=\frac{\pi }{4}\end{array}$
###### Did you like this example?
Jase Rocha
Let
$I\left(a\right)={\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}\left(x\right)}{x}{e}^{-ax}dx$
where your integral is I(1). I′(a) is then:
${I}^{\prime }\left(a\right)=-{\int }_{0}^{\mathrm{\infty }}\mathrm{sin}\left(x\right){e}^{-ax}dx$
which is the Laplace Transform of the sine function. Thus
${I}^{\prime }\left(a\right)=-\frac{1}{{a}^{2}+1}$
Integrating back we get:
$I\left(a\right)=-\mathrm{arctan}\left(a\right)+C$
To find the constant, we notice that
$I\left(0\right)={\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}\left(x\right)}{x}dx=\frac{\pi }{2}$
So we get that $C=\frac{\pi }{2}$ and that
$I\left(1\right)=-\mathrm{arctan}\left(1\right)+\frac{\pi }{2}=\frac{\pi }{4}$