How to evaluate the following improper integral : \int_0^{+\infty}\frac{x\sin x}{x^2+1}dx

elvishwitchxyp

elvishwitchxyp

Answered question

2022-01-06

How to evaluate the following improper integral :
0+xsinxx2+1dx

Answer & Explanation

Navreaiw

Navreaiw

Beginner2022-01-07Added 34 answers

According to the residue theory,
0+1s2+x2dx=π2s, I(α)=0+xsinαx1+x2dx
Laplace transform:
L[I(α)]=0+x1+x2xs2+x2dx
=0+x2+111+x21s2+x2dx
=0+1s2+x2dx0+11+x21s2+x2dx
=0+1s2+x2dx1s210+(11+x21s2+x2)dx
=π21s+1
Inverse transform:
L1[I(α)]=π2eαI(1)=0+xsinx1+x2d=π2e
Juan Spiller

Juan Spiller

Beginner2022-01-08Added 38 answers

Using the result from this OP: Integral evaluation cos(ax)π(1+x2)dx. We have
0cosax1+x2dx=π2e|a|
Thus, our integration is simply
0xsinx1+x2dx=lima1da0cosax1+x2dx
=π2lima1da[e[a]]
=π2e
star233

star233

Skilled2022-01-11Added 403 answers

Since your integrand is even, this integral is equal to one half of
xsinxx2+1dx
This integral is the imaginary part of
xeixx2+1dx
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 e1πi. So taking half the imaginary part, we get π2e.

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