elvishwitchxyp

2022-01-06

How to evaluate the following improper integral :
${\int }_{0}^{+\mathrm{\infty }}\frac{x\mathrm{sin}x}{{x}^{2}+1}dx$

Navreaiw

According to the residue theory,

Laplace transform:
$L\left[I\left(\alpha \right)\right]={\int }_{0}^{+\mathrm{\infty }}\frac{x}{1+{x}^{2}}\cdot \frac{x}{{s}^{2}+{x}^{2}}dx$
$={\int }_{0}^{+\mathrm{\infty }}\frac{{x}^{2}+1-1}{1+{x}^{2}}\cdot \frac{1}{{s}^{2}+{x}^{2}}dx$
$={\int }_{0}^{+\mathrm{\infty }}\frac{1}{{s}^{2}+{x}^{2}}dx-{\int }_{0}^{+\mathrm{\infty }}\frac{1}{1+{x}^{2}}\cdot \frac{1}{{s}^{2}+{x}^{2}}dx$
$={\int }_{0}^{+\mathrm{\infty }}\frac{1}{{s}^{2}+{x}^{2}}dx-\frac{1}{{s}^{2}-1}{\int }_{0}^{+\mathrm{\infty }}\left(\frac{1}{1+{x}^{2}}-\frac{1}{{s}^{2}+{x}^{2}}\right)dx$
$=\frac{\pi }{2}\cdot \frac{1}{s+1}$
Inverse transform:
${L}^{-1}\left[I\left(\alpha \right)\right]=\frac{\pi }{2}{e}^{-\alpha }⇒I\left(1\right)={\int }_{0}^{+\mathrm{\infty }}\frac{x\mathrm{sin}x}{1+{x}^{2}}d=\frac{\pi }{2e}$

Juan Spiller

Using the result from this OP: Integral evaluation ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\mathrm{cos}\left(ax\right)}{\pi \left(1+{x}^{2}\right)}dx$. We have
${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{cos}ax}{1+{x}^{2}}dx=\frac{\pi }{2}{e}^{-|a|}$
Thus, our integration is simply
${\int }_{0}^{\mathrm{\infty }}\frac{x\mathrm{sin}x}{1+{x}^{2}}dx=-\underset{a\to 1}{lim}{d}_{a}{\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{cos}ax}{1+{x}^{2}}dx$
$=-\frac{\pi }{2}\underset{a\to 1}{lim}{d}_{a}\left[{e}^{-\left[a\right]}\right]$
$=\frac{\pi }{2e}$

star233

Since your integrand is even, this integral is equal to one half of
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x\mathrm{sin}x}{{x}^{2}+1}dx$
This integral is the imaginary part of
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x\cdot {e}^{ix}}{{x}^{2}+1}dx$
This can be solved as the contour integral with contour a half disc of radius R with base on the real axis, letting R go to infinity. The integral along the arc goes to 0 ( needs some showing ). The contour integral is equal to the residue at $x=i$, which is ${e}^{-1}\pi i$. So taking half the imaginary part, we get $\frac{\pi }{2e}.$

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