Evaluate improper integral $\int_0^\infty \frac{x\sin x}{x^2+1}dx$ How to prove that $$\int_0^\infty \frac{x\sin x}{x^2+1}dx=\frac{\pi}{2e}$$
I've tried several basic approaches like substitution and IBP but can't move forward. 
 A: Suppose we are interested in 
$$J = \int_0^\infty \frac{x\sin(x)}{1+x^2} dx$$
In  view of  the  ML  bound that  we  will be  using  later it  proves
convenient to write this as
$$\left[ -\frac{x\cos(x)}{1+x^2} \right]_0^\infty
+ \int_0^\infty \frac{1-x^2}{(1+x^2)^2} \cos(x) dx
= \frac{1}{2} 
\int_{-\infty}^\infty \frac{1-x^2}{(1+x^2)^2} \cos(x) dx.$$
Observe that this is
$$\frac{1}{2} \Re \int_{-\infty}^\infty 
\frac{1-x^2}{(1+x^2)^2} \exp(ix) dx.$$
Therefore we integrate
$$f(z) = \frac{1-z^2}{(1+z^2)^2} \exp(iz)$$
around a semicircular  contour of radius $R$ in  the upper half plane,
which consists of a horizontal segment $\Gamma_1$ on the real axis and
a  semicircle $\Gamma_2$  in the  upper half  plane. Now  the integral
along $\Gamma_1$ is the integral we  seek to evaluate in the limit and
the integral along  $\Gamma_2$ vanishes. This is because  we have from
the  ML  bound   (parameterize  $\Gamma_2$  by  $R\exp(i\theta)$  with
$0\le\theta\le\pi$)
$$\lim_{R\rightarrow\infty} \pi R \times
\frac{(R^2+1) |\exp(iR\exp(i\theta))|}{(R^2-1)^2}
\\ = \lim_{R\rightarrow\infty} \pi R \times
\frac{(R^2+1) |\exp(iR\cos(\theta)+i^2R\sin(\theta))|}{(R^2-1)^2}
\\ = \lim_{R\rightarrow\infty} \pi R \times
\frac{(R^2+1) |\exp(-R\sin(\theta))|}{(R^2-1)^2}.$$
Now we  have $|\exp(-R\sin(\theta))|\le 1$  because $0\le\theta\le\pi$
so we  get two terms,  the first of  which is $O(1/R)$ and  the second
$O(1/R^3)$ so they both vanish  in the limit and the contribution from
the circular segment $\Gamma_2$ is zero.

This yields
$$J =  \frac{1}{2} \times \Re(2\pi i \; \mathrm{Res}_{z=i} f(z)).$$
We require the second term in the Taylor series of
$$\frac{1-z^2}{(z+i)^2} \exp(iz)$$
at $z=i$ 
which is 
$$\left.\left(\frac{-2z\exp(iz)+(1-z^2)i\exp(iz)}{(z+i)^2}
- 2 \frac{1-z^2}{(z+i)^3} \exp(iz)\right)\right|_{z=i}.$$
The first term here is zero and the second is
$$-2\frac{2}{(2i)^3} \exp(-1)
= \frac{1}{2} \frac{1}{i} \exp(-1).$$
Substitute this into the formula for $J$ to finally obtain
$$J=\frac{1}{2} \Re(2\pi i \exp(-1)/2/i)
= \frac{\pi}{2e}.$$
A: I am going to post this as a means of posting alternate solutions.
We know that $$\int_0^\infty \frac{\cos ax}{x^2+1}\text{ d}x=\frac{\pi e^{-|a|}}{2}$$
Taking the partial derivative on the variable a
$$\frac{\partial}{\partial a}\int_0^\infty \frac{\cos ax}{x^2+1}\text{ d}x=\frac{d}{da}\frac{\pi}{2e^{|a|}}=\frac{\pi}{2e^{|a|}}$$ 
And now plug in $a=1$
A: Here is a much simple solution. 
Since it is even function, we evaluate
$$
\int_{-\infty }^\infty \frac{x\sin x}{x^2+1}dx
$$
We consider the analytic function
$$
F(z)=\frac{ze^{iz}}{z^2+1}
$$
on the contour of $C=[-R,R]\cup C_R$, where $C_R$ is the upper half circle with radius of $R$. By Cauchy's residue theorem, we have
$$
\int_{-R}^RF(x)dx+\int_{C_R}F(z)dz=2\pi iRes(F,i)
$$
Note that only pole in the $C_R$ is $z=i$. Since
$$
Res(F,i)=\lim_{z\to i}(z-i)F(z)=\lim_{z\to i}\frac{ze^{iz}}{z+i}=\frac{e^{-1}}{2}
$$
We have
$$
\int_{-R}^RF(x)dx+\int_{C_R}F(z)dz=\frac{\pi i}{e}\tag1
$$
By Jordan lemma
$$
\int_{C_R}|e^{iz}||dz|=2R\int_0^{\pi/2}e^{-R\sin t}\:dt<\pi
$$
Hence
$$
\left|\int_{C_R}F(z)dz\right|\leqslant \frac{R}{R^2-1}\int_{C_R}|e^{iz}||dz|<\frac{\pi R}{R^2-1}\to0
$$
as $R\to\infty$. So from $(1)$
$$
\int_{-\infty}^{\infty}F(x)dx=\frac{\pi i}{e}
$$
And 
$$
\int_{-\infty }^\infty \frac{x\sin x}{x^2+1}dx=\Im{\int_{-\infty}^{\infty}F(x)dx}=\frac{\pi }{e}
$$
So 
$$
\int_{0}^\infty \frac{x\sin x}{x^2+1}dx=\frac{\pi }{2e}
$$
A: Here is a real method. We will use Feynman's trick by differentiating under the integral sign.
Let
$$I(a) = \int^\infty_0 \frac{x \sin (ax)}{1 + x^2} \, dx, \quad a > 0.$$
As this improper integral converges conditionally, we will rewrite it so that it converges absolutely.
\begin{align*}
I(a) &= \int^\infty_0 \frac{x^2 \sin (ax)}{x(1 + x^2)} \, dx = \int^\infty_0 \frac{[(1 + x^2) - 1] \sin (ax)}{x(1 + x^2)} \, dx\\
&= \int^\infty_0 \frac{\sin (ax)}{x} \, dx - \int^\infty_0 \frac{\sin (ax)}{x(1 + x^2)} \, dx.
\end{align*}
The first of these integrals, on making a substitution of $u = ax$, corresponds to the Dirichlet integral. Its value is well known, namely
$$\int^\infty_0 \frac{\sin (ax)}{x} \, dx = \int^\infty_0 \frac{\sin u}{u} \, du = \frac{\pi}{2}.$$
Thus
$$I(a) = \frac{\pi}{2} - \int^\infty_0 \frac{\sin (ax)}{x(1 + x^2)} \, dx.$$
Note that as $a \to 0^+$, $I(a) \to \frac{\pi}{2}$.
Now differentiating $I(a)$ with respect to $a$ we have
$$I'(a) = -\partial_a \int^\infty_0 \frac{\sin (ax)}{x(1 + x^2)} \, dx = -\int^\infty_0 \frac{\cos (ax)}{1 + x^2} \, dx.$$
Note that as $a \to 0^+$,
$$I'(a) \to -\int^\infty_0 \frac{dx}{1 + x^2} = -\frac{\pi}{2}.$$
Differentiating again gives
$$I''(a) = -\partial_a \int^\infty_0 \frac{\cos (ax)}{1 + x^2} \, dx = \int^\infty_0 \frac{x \sin (ax)}{1 + x^2} \, dx = I(a),$$
or $I''(a) - I(a) = 0$. On solving this second-order linear differential equation we have
$$I(a) = C_1 e^a + C_2 e^{-a},$$
where $C_1$ and $C_2$ are two constants to be found. As $a \to 0^+$, using $I(0^+) = \pi/2$ and $I'(0^+) = -\pi/2$ we find $C_1 = 0$ and $C_2 = \pi/2$. Thus
$$I(a) = \int^\infty_0 \frac{x \sin (ax)}{1 + x^2} \, dx = \frac{\pi}{2e^a},$$
while our integral is recovered on setting $a = 1$, namely
$$I(1) = \int^\infty_0 \frac{x \sin x}{1 + x^2} \, dx = \frac{\pi}{2e}.$$
