Expert answers to "Finding value of ∫0∞xsin⁡x(x2+1)3dx"

Berasiniz1

Answered question

2022-01-24

Finding value of

\int_{0}^{\infty} \frac{x \sin x}{{(x^{2} + 1)}^{3}} d x

Answer & Explanation

enveradapb

Beginner2022-01-25Added 13 answers

$I = \int_{0}^{\infty} \frac{x \sin x}{{(1 + x^{2})}^{3}} d x = - {\frac{14}{\underset{⏟}{\frac{\sin x}{{(1 + x^{2})}^{2}} ∣_{0}^{\infty}}}}_{= 0} + {\frac{14}{\int}}_{0}^{\infty} \frac{\cos x}{{(1 + x^{2})}^{2}} d x$
Now we have that $I = - \frac{1}{4} J'$ (1) because we can take:
$J (a) = \int_{0}^{\infty} \frac{\cos x}{a + x^{2}} d x \Rightarrow J^{'} (a) = - \int_{0}^{\infty} \frac{\cos x}{{(a + x^{2})}^{2}} d x$
Thus all we need to do is to find J(a) then take a derivate and set a=1.
$J (a) = \frac{π}{2 \sqrt{a}} e^{- \sqrt{a}} \Rightarrow J^{'} (a) = \frac{π}{2} (- {\frac{12}{a}}^{- \frac{3}{2}} e^{- \sqrt{a}} - \frac{1}{\sqrt{a}} e^{- \sqrt{a}} \frac{1}{2 \sqrt{a}})$
$\Rightarrow I = - {\frac{14}{J}}^{'} (1) = \frac{π}{8} ({\frac{12}{e}}^{- 1} + \frac{1}{2} e^{- 1}) = \frac{π}{8 e}$

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