oliviayychengwh

2022-01-07

Compute the following integral:
${\int }_{0}^{\mathrm{\infty }}\frac{{e}^{x}\mathrm{sin}\left(x\right)}{x}dx$

Mollie Nash

Using Laplace Transform,
$L\left(\mathrm{sin}\left(x\right)\right)=\frac{1}{{s}^{2}+1}$
$L\left(\frac{\mathrm{sin}\left(x\right)}{x}\right)={\int }_{r}^{\mathrm{\infty }}\frac{1}{{s}^{2}+1}ds=\frac{\pi }{2}-\mathrm{arctan}\left(r\right)$
Therefore,
${\int }_{0}^{\mathrm{\infty }}{e}^{-rx}\frac{\mathrm{sin}\left(x\right)}{x}dx=\frac{\pi }{2}-\mathrm{arctan}\left(r\right)$
Substituting r=1,
${\int }_{0}^{\mathrm{\infty }}{e}^{-x}\frac{\mathrm{sin}\left(x\right)}{x}dx=\frac{\pi }{4}$

Virginia Palmer

Yet a different approach: parametric integration. Let

Then,
${F}^{\prime }\left(\lambda \right)=-{\int }_{0}^{\mathrm{\infty }}{e}^{-\lambda x}\mathrm{sin}\left(x\right)dx=-\frac{1}{1+{\lambda }^{2}}$
Integrating and taking into account that $\underset{\lambda \to \mathrm{\infty }}{lim}F\left(\lambda \right)=0$ we have
$F\left(\lambda \right)=\frac{\pi }{2}-\mathrm{arctan}\lambda$
and${\int }_{0}^{\mathrm{\infty }}\frac{{e}^{-x}\mathrm{sin}\left(x\right)}{x}dx=F\left(1\right)=\frac{\pi }{4}$

star233

Another approach:
${\int }_{0}^{\mathrm{\infty }}dx\frac{{e}^{-x}\mathrm{sin}\left(x\right)}{x}\phantom{\rule{0ex}{0ex}}={\int }_{0}^{\mathrm{\infty }}dx\frac{{e}^{-x}}{x}\sum _{k=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{k}{x}^{2k+1}}{\left(2k+1\right)!}\phantom{\rule{0ex}{0ex}}=\sum _{k=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{k}}{\left(2k+1\right)!}{\int }_{0}^{\mathrm{\infty }}dx{x}^{2k}{e}^{-x}\phantom{\rule{0ex}{0ex}}=\sum _{k=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{k}}{\left(2k+1\right)!}\left(2k\right)!\phantom{\rule{0ex}{0ex}}=\sum _{k=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{k}}{2k+1}\phantom{\rule{0ex}{0ex}}=\frac{\pi }{4}$

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