oliviayychengwh

Answered

2022-01-07

Compute the following integral:

${\int}_{0}^{\mathrm{\infty}}\frac{{e}^{x}\mathrm{sin}\left(x\right)}{x}dx$

Answer & Explanation

Mollie Nash

Expert

2022-01-08Added 33 answers

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,

$\int}_{0}^{\mathrm{\infty}}{e}^{-x}\frac{\mathrm{sin}\left(x\right)}{x}dx=\frac{\pi}{4$

Therefore,

Substituting r=1,

Virginia Palmer

Expert

2022-01-09Added 27 answers

Yet a different approach: parametric integration. Let

$F\left(\lambda \right)={\int}_{0}^{\mathrm{\infty}}\frac{{e}^{-\lambda x}\mathrm{sin}\left(x\right)}{x}dx,\text{}\lambda 0$

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$\int}_{0}^{\mathrm{\infty}}\frac{{e}^{-x}\mathrm{sin}\left(x\right)}{x}dx=F\left(1\right)=\frac{\pi}{4$

Then,

Integrating and taking into account that

and

star233

Expert

2022-01-11Added 238 answers

Another approach:

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