lunja55
2022-09-24
Answered

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

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Guadalupe Reid

Answered 2022-09-25
Author has **8** answers

$$\begin{array}{c}{\int}_{0}^{\mathrm{\infty}}\frac{\mathrm{sin}x}{x{e}^{x}}dx={\int}_{0}^{\mathrm{\infty}}\mathrm{sin}x{\int}_{1}^{\mathrm{\infty}}{e}^{-sx}dsdx={\int}_{1}^{\mathrm{\infty}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{\int}_{0}^{\mathrm{\infty}}\mathrm{sin}(x){e}^{-sx}dx\phantom{\rule{thinmathspace}{0ex}}ds\hfill \\ \hfill ={\int}_{1}^{\mathrm{\infty}}\mathcal{L}[\mathrm{sin}(x)](s)ds={\int}_{1}^{\mathrm{\infty}}\frac{ds}{{s}^{2}+1}=\frac{\pi}{2}-{\mathrm{tan}}^{-1}1=\frac{\pi}{4}\end{array}$$

Jase Rocha

Answered 2022-09-26
Author has **2** answers

Let

$$I(a)={\int}_{0}^{\mathrm{\infty}}\frac{\mathrm{sin}(x)}{x}{e}^{-ax}dx$$

where your integral is I(1). I′(a) is then:

$${I}^{\prime}(a)=-{\int}_{0}^{\mathrm{\infty}}\mathrm{sin}(x){e}^{-ax}dx$$

which is the Laplace Transform of the sine function. Thus

$${I}^{\prime}(a)=-\frac{1}{{a}^{2}+1}$$

Integrating back we get:

$$I(a)=-\mathrm{arctan}(a)+C$$

To find the constant, we notice that

$$I(0)={\int}_{0}^{\mathrm{\infty}}\frac{\mathrm{sin}(x)}{x}dx=\frac{\pi}{2}$$

So we get that $C=\frac{\pi}{2}$ and that

$$I(1)=-\mathrm{arctan}(1)+\frac{\pi}{2}=\frac{\pi}{4}$$

$$I(a)={\int}_{0}^{\mathrm{\infty}}\frac{\mathrm{sin}(x)}{x}{e}^{-ax}dx$$

where your integral is I(1). I′(a) is then:

$${I}^{\prime}(a)=-{\int}_{0}^{\mathrm{\infty}}\mathrm{sin}(x){e}^{-ax}dx$$

which is the Laplace Transform of the sine function. Thus

$${I}^{\prime}(a)=-\frac{1}{{a}^{2}+1}$$

Integrating back we get:

$$I(a)=-\mathrm{arctan}(a)+C$$

To find the constant, we notice that

$$I(0)={\int}_{0}^{\mathrm{\infty}}\frac{\mathrm{sin}(x)}{x}dx=\frac{\pi}{2}$$

So we get that $C=\frac{\pi}{2}$ and that

$$I(1)=-\mathrm{arctan}(1)+\frac{\pi}{2}=\frac{\pi}{4}$$

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