How to evaluate the following improper integral : ∫0+∞xsin⁡xx2+1dx

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How to evaluate the following improper integral :  ∫0+∞xsin⁡xx2+1dx

Navreaiw · Accepted Answer

According to the residue theory,  ∫0+∞1s2+x2dx=π2s, I(α)=∫0+∞xsin⁡αx1+x2dx  Laplace transform:  L[I(α)]=∫0+∞x1+x2⋅xs2+x2dx  =∫0+∞x2+1−11+x2⋅1s2+x2dx  =∫0+∞1s2+x2dx−∫0+∞11+x2⋅1s2+x2dx  =∫0+∞1s2+x2dx−1s2−1∫0+∞(11+x2−1s2+x2)dx  =π2⋅1s+1  Inverse transform:  L−1[I(α)]=π2e−α⇒I(1)=∫0+∞xsin⁡x1+x2d=π2e

Juan Spiller · Accepted Answer

Using the result from this OP: Integral evaluation ∫−∞∞cos⁡(ax)π(1+x2)dx. We have  ∫0∞cos⁡ax1+x2dx=π2e−|a|  Thus, our integration is simply  ∫0∞xsin⁡x1+x2dx=−lima→1da∫0∞cos⁡ax1+x2dx  =−π2lima→1da[e−[a]]  =π2e

star233 · Accepted Answer

Since your integrand is even, this integral is equal to one half of ∫−∞∞xsin⁡xx2+1dx This integral is the imaginary part of ∫−∞∞x⋅eixx2+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 e−1πi. So taking half the imaginary part, we get π2e.

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

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