TITLE: Improper integral : $\int_0^{+\infty}\frac{x\sin x}{x^2+1}$
QUESTION [8 upvotes]: How to evaluate the following improper integral : $$\int_0^{+\infty}\frac{x\sin x}{x^2+1}\,dx$$
I have tried integration by parts and variable substitution, but it didn't work.

REPLY [8 votes]: Using the result from this OP: Integral evaluation $\int_{-\infty}^{\infty}\frac{\cos (ax)}{\pi (1+x^2)}dx$. We have
\begin{equation}
\int_0^\infty\frac{\cos ax}{1+x^2}\,dx=\frac{\pi}{2}e^{-|a|}
\end{equation}
Thus, our integration is simply
\begin{align}
\int_0^\infty\frac{x\sin x}{1+x^2}\,dx&=-\lim_{a\to1}\partial_a\int_0^\infty\frac{\cos ax}{1+x^2}\,dx\\
&=-\frac{\pi}{2}\lim_{a\to1}\partial_a\left[e^{-|a|}\right]\\
&=\frac{\pi}{2e}
\end{align}