Evaluate the following improper integral

Question

Evaluate the improper integral $\displaystyle\int_{0}^{\infty}\frac{x\sin{ax}}{1+x^2}dx$, if $a\neq 0$.

Answer 1

The following integral

$$F(a) = \int_{0}^{\infty}\frac{\cos(ax)}{1+x^2}dx = \frac{\pi}{2}e^{-|a|}$$

is on of the standard integrals in complex analysis books. The integral can be solved by contour integration by considering the function

$$G(z) = \frac{e^{iaz}}{z^2+1}$$

Using a half a circle in the upper half plane.

You can refer to the post Computing $\int_{-\infty}^{\infty} \frac{\cos x}{x^{2} + a^{2}}dx$ using residue calculus

Hence by differentiation

$$F'(a) = \int_{0}^{\infty}\frac{x\sin(ax)}{1+x^2}dx = \frac{\pi a}{2|a|}e^{-|a|}$$

Note for differentiation you can consider two cases $a>0$ and $a<0$.

Answer 2

By Fourier transform, Let $$F(x)=\left\lbrace\begin{array}{c l}e^x,&x\leqslant0\\e^{-x},&x>0\end{array}\right.$$‎ so with the Fourier integral we have \begin{eqnarray} a(\omega)&=&\frac{1}{\pi}\int_{-\infty}^\infty F(x)\cos\omega x~dx\\ &=&\frac{1}{\pi}\int_{-\infty}^0 e^x\cos\omega x~dx+\frac{1}{\pi}\int_0^\infty e^{-x}\cos\omega x~dx\\&=&\frac{2}{\pi}\frac{1}{\omega^2+1}\\ b(\omega)&=&\frac{1}{\pi}\int_{-\infty}^\infty F(x)\sin\omega x~dx\\ &=&0 \end{eqnarray} Then \begin{eqnarray} f(x)&=&\int_{0}^\infty a(\omega)\cos\omega x+b(\omega)\sin\omega x~d\omega\\ \int_{0}^\infty \frac{2}{\pi}\frac{1}{\omega^2+1}\cos\omega x+0~d\omega&=&\left\lbrace\begin{array}{c l}e^x,&x<0\\e^{-x},&x>0\end{array}\right.\\ \int_{0}^\infty \frac{\cos\omega x}{\omega^2+1}~d\omega&=&\left\lbrace\begin{array}{c l}\frac{\pi}{2}e^x,&x<0\\\frac{\pi}{2}e^{-x},&x>0\end{array}\right.\\ \end{eqnarray} With derivative say: \begin{eqnarray} \int_{0}^\infty \frac{x\sin\omega x}{\omega^2+1}~d\omega&=&\left\lbrace\begin{array}{c l}-\frac{\pi}{2}e^x,&x<0\\\frac{\pi}{2}e^{-x},&x>0\end{array}\right.\\ \end{eqnarray}