# Compute the following integral: \int_0^\infty\frac{e^x\sin(x)}{x}dx

Compute the following integral:
$$\displaystyle{\int_{{0}}^{\infty}}{\frac{{{e}^{{x}}{\sin{{\left({x}\right)}}}}}{{{x}}}}{\left.{d}{x}\right.}$$

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Mollie Nash
Using Laplace Transform,
$$\displaystyle{L}{\left({\sin{{\left({x}\right)}}}\right)}={\frac{{{1}}}{{{s}^{{2}}+{1}}}}$$
$$\displaystyle{L}{\left({\frac{{{\sin{{\left({x}\right)}}}}}{{{x}}}}\right)}={\int_{{r}}^{\infty}}{\frac{{{1}}}{{{s}^{{2}}+{1}}}}{d}{s}={\frac{{\pi}}{{{2}}}}-{\arctan{{\left({r}\right)}}}$$
Therefore,
$$\displaystyle{\int_{{0}}^{\infty}}{e}^{{-{r}{x}}}{\frac{{{\sin{{\left({x}\right)}}}}}{{{x}}}}{\left.{d}{x}\right.}={\frac{{\pi}}{{{2}}}}-{\arctan{{\left({r}\right)}}}$$
Substituting r=1,
$$\displaystyle{\int_{{0}}^{\infty}}{e}^{{-{x}}}{\frac{{{\sin{{\left({x}\right)}}}}}{{{x}}}}{\left.{d}{x}\right.}={\frac{{\pi}}{{{4}}}}$$
###### Not exactly what youâ€™re looking for?
Virginia Palmer
Yet a different approach: parametric integration. Let
$$\displaystyle{F}{\left(\lambda\right)}={\int_{{0}}^{\infty}}{\frac{{{e}^{{-\lambda{x}}}{\sin{{\left({x}\right)}}}}}{{{x}}}}{\left.{d}{x}\right.},\ \lambda{>}{0}$$
Then,
$$\displaystyle{F}'{\left(\lambda\right)}=-{\int_{{0}}^{\infty}}{e}^{{-\lambda{x}}}{\sin{{\left({x}\right)}}}{\left.{d}{x}\right.}=-{\frac{{{1}}}{{{1}+\lambda^{{2}}}}}$$
Integrating and taking into account that $$\displaystyle\lim_{{\lambda\to\infty}}{F}{\left(\lambda\right)}={0}$$ we have
$$\displaystyle{F}{\left(\lambda\right)}={\frac{{\pi}}{{{2}}}}-{\arctan{\lambda}}$$
and $$\displaystyle{\int_{{0}}^{\infty}}{\frac{{{e}^{{-{x}}}{\sin{{\left({x}\right)}}}}}{{{x}}}}{\left.{d}{x}\right.}={F}{\left({1}\right)}={\frac{{\pi}}{{{4}}}}$$
star233

Another approach:
$$\int_0^\infty dx\frac{e^{-x}\sin(x)}{x}\\=\int_0^\infty dx\frac{e^{-x}}{x}\sum_{k=0}^\infty\frac{(-1)^k x^{2k+1}}{(2k+1)!} \\=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\int_0^\infty dxx^{2k}e^{-x} \\=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}(2k)! \\=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1} \\=\frac{\pi}{4}$$