Berasiniz1

2022-01-24

Finding value of ${\int }_{0}^{\mathrm{\infty }}\frac{x\mathrm{sin}x}{{\left({x}^{2}+1\right)}^{3}}dx$

Expert

$I={\int }_{0}^{\mathrm{\infty }}\frac{x\mathrm{sin}x}{{\left(1+{x}^{2}\right)}^{3}}dx=-{\frac{14}{\underset{⏟}{\frac{\mathrm{sin}x}{{\left(1+{x}^{2}\right)}^{2}}{\mid }_{0}^{\mathrm{\infty }}}}}_{=0}+{\frac{14}{\int }}_{0}^{\mathrm{\infty }}\frac{\mathrm{cos}x}{{\left(1+{x}^{2}\right)}^{2}}dx$
Now we have that $I=-\frac{1}{4}J\prime$(1) because we can take:
$J\left(a\right)={\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{cos}x}{a+{x}^{2}}dx⇒{J}^{\prime }\left(a\right)=-{\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{cos}x}{{\left(a+{x}^{2}\right)}^{2}}dx$
Thus all we need to do is to find J(a) then take a derivate and set a=1.
$J\left(a\right)=\frac{\pi }{2\sqrt{a}}{e}^{-\sqrt{a}}⇒{J}^{\prime }\left(a\right)=\frac{\pi }{2}\left(-{\frac{12}{a}}^{-\frac{3}{2}}{e}^{-\sqrt{a}}-\frac{1}{\sqrt{a}}{e}^{-\sqrt{a}}\frac{1}{2\sqrt{a}}\right)$
$⇒I=-{\frac{14}{J}}^{\prime }\left(1\right)=\frac{\pi }{8}\left({\frac{12}{e}}^{-1}+\frac{1}{2}{e}^{-1}\right)=\frac{\pi }{8e}$

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