Using complex contour integration, show that an integral is...

Question

Show that the $\displaystyle \int_0^\infty \dfrac{x \sin(ax)}{(x^2 + b^2)}= \dfrac{\pi}{2}e^{−ab}$.

I have no idea where to start.

Answer 1

Note that we can write

$$I(a,b)=\int_{0}^\infty\frac{x\sin(ax)}{x^2+b^2}\,dx=\text{Im}\left(\frac12\int_{-\infty}^\infty\frac{xe^{iax}}{x^2+b^2}\,dx\right)$$

We assume without loss of generality that $a>0$ and $b>0$. For $a<0$, the odd symmetry in $a$ shows that $I(-a,b)=-I(a,b)$. For $b<0$, the even symmetry in $b$ shows that $I(a,-b)=I(a,b)$. The cases for which $a=0$, or $b=0$ or $a=b=0$ are trivial.

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Now, we move to the complex plane. From the residue theorem we see that the integral $\oint_C \frac{ze^{iaz}}{z^2+b^2}\,dz$ is given by

$$\begin{align} \oint_C \frac{ze^{iaz}}{z^2+b^2}\,dz&=2\pi i \text{Res}\left(\frac{ze^{iaz}}{z^2+b^2}, z=ib\right)\\\\ &=2\pi i \frac{ib e^{-ab}}{2ib}\\\\ &=\pi e^{-ab}\tag 1 \end{align}$$

where $C$ is the contour comprised of the (i) line segment from $-R$ to $R$ along the real axis and (ii) semi-circular arc in the upper-half plane with center at $z=0$ and radius $R>b$.

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We can also write the integral on the left-hand side of $(1)$ as

$$\oint_C \frac{ze^{iaz}}{z^2+b^2}\,dz=\int_{-R}^R \frac{xe^{iax}}{x^2+b^2}\,dx+\int_0^\pi\frac{Re^{i\phi}e^{iaRe^{i\phi}}}{(Re^{i\phi})^2+b^2}\,iRe^{i\phi}\,d\phi\tag 2 $$

Taking the limit as $R\to 0$, it is straightforward to show that the contribution from the second integral on the right-hand side of $(2)$ tends to $0$. The first integral on the right-hand side of $(2)$ approaches $\int_{-\infty}^\infty \frac{xe^{iax}}{x^2+b^2}\,dx$.

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Putting it all together yields the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\int_{0}^\infty\frac{x\sin(ax)}{x^2+b^2}\,dx=\frac{\pi e^{-ab}}{2}}$$