Jtdylktuy

I need help calculation the integral $int_{-infty}^{infty} xsin(x)/(x^2+1)^2 dx$

I am getting confused on how to integrate around $pm i$. I think I'll have to use $gamma = Re^{itheta}$ where $|R|leq n$ for $ngeq1$

$$int_{-infty}^inftyfrac{xsin x}{(x^2+1)^2}mathrm dx=mathfrak Ileft{int_{-infty}^inftyfrac{xe^{ix}}{(x^2+1)^2}mathrm dxright}$$

Let us compute this integral. It is given by a contour in the complex plane that runs along the real axis. Since the exponential on top decays as $|x|toinfty$ in the upper half plane, this is where we choose to close the contour. Therefore we have a semicircular contour in the upper half plane, of the form $Re^{itheta}$ for $R>1$ and $thetain[0,pi]$. Let us denote this contour by $gamma_R$.

$$lim_{Rtoinfty}int_{gamma_R} frac{xe^{ix}}{(x^2+1)^2}mathrm dx=int_{-infty}^{infty}frac{xe^{ix}}{(x^2+1)^2}mathrm dx+lim_{Rtoinfty}int_0^pifrac{Re^{itheta}e^{iRe^{itheta}}}{(R^2e^{2itheta}+1)^2}iRe^{itheta}mathrm dthetatag1$$

Now we want to show that this second integral is zero in the limit as $Rtoinfty$.

$$begin{align}left|int_0^pi frac{Re^{itheta}e^{iRe^{itheta}}}{(R^2e^{2itheta}+1)^2}iRe^{itheta}mathrm dthetaright|&leint_0^pileft|frac{Re^{itheta}e^{iRe^{itheta}}}{(R^2e^{2itheta}+1)^2}iRe^{itheta}right|mathrm dtheta\&leint_0^pileft|frac{R^2e^{-Rsin theta}}{(R^2-1)^2}right|mathrm dthetaend{align}$$

$int_0^pi e^{-Rsintheta}mathrm dtheta$ is finite, and in the limit as $Rtoinfty$, the prefactor tends to zero, so the second integral in $(1)$ is zero.

So we have $$int_{-infty}^{infty}frac{xe^{ix}}{(x^2+1)^2}mathrm dx=lim_{Rtoinfty}int_{gamma_R} frac{xe^{ix}}{(x^2+1)^2}mathrm dxtag2$$

We evaluate the RHS of $(2)$ by residue calculus. The contour encloses a pole at $x=i$, of order $2$. We evaluate the residue at the pole.

$$text{Res}left(frac{xe^{ix}}{(x^2+1)^2},iright)=lim_{xto i}frac{d}{dx}left(frac{xe^{ix}(x-i)^2}{(x^2+1)^2}right)=lim_{xto i}frac{d}{dx}left(frac{xe^{ix}}{(x+i)^2}right)=frac1{4e}$$after some calculations.

Therefore the RHS of $(2)$ is $$2pi icdot text{Res}left(frac{xe^{ix}}{(x^2+1)^2},iright)=frac{pi i }{2e}$$

Since the integral we wish to calculate is the imaginary part of the LHS of $(2)$, the final answer is $$frac{pi}{2e}.$$

Let us define
$$ g(a) = int_{0}^{+infty}frac{xsin(ax)}{(x^2+1)^2},dx $$
for $ainmathbb{R}^+$. Our integral is clearly $2,g(1)$ by parity. We have
$$ mathcal L g(s) = int_{0}^{+infty}frac{x^2}{(s^2+x^2)(1+x^2)^2},dx = frac{pi}{4(1+s)^2} $$
by partial fraction decomposition, and
$$ g(a) = frac{pi a}{4}e^{-a} $$
by the inverse Laplace transform. It follows that the original integral equals $color{blue}{frac{pi}{2e}}$.