Evaluating $\int_0^\infty\frac{\sin(x)}{x^2+1}\, dx$ I have seen $$\int_0^\infty \frac{\cos(x)}{x^2+1} \, dx=\frac{\pi}{2e}$$ evaluated in various ways. 
It's rather popular when studying CA. 
But, what about $$\int_0^\infty \frac{\sin(x)}{x^2+1} \, dx\,\,?$$
This appears to be trickier and more challenging.
I found that it has a closed form of 
$$\cosh(1)\operatorname{Shi}(1)-\sinh(1)\text{Chi(1)}\,\,,\,\operatorname{Shi}(1)=\int_0^1 \frac{\sinh(x)}{x}dx\,\,,\,\, \text{Chi(1)}=\gamma+\int_0^1 \frac{\cosh(x)-1}{x} \, dx$$
which are the hyperbolic sine and cosine integrals, respectively.
It's an odd function, so 
$$\int_{-\infty}^\infty \frac{\sin(x)}{x^2+1} \, dx=0$$
But, does anyone know how the former case can be done? Thanks a bunch. 

Here is another solution:
Consider the integral
$$I(\alpha) = \int_{0}^{\infty} \frac{\sin (\alpha x)}{1+x^2} \, dx = \int_{0}^{\infty} \frac{\alpha \sin x}{\alpha^2+x^2} \, dx.$$
Differentiating $I(\alpha)$ with the first equality, we have
\begin{align*}
I'(\alpha)
&= \int_{0}^{\infty} \frac{x \cos (\alpha x)}{1+x^2} \, dx
 = \int_{0}^{\infty} \frac{x \cos x}{\alpha^2+x^2} \, dx.
\end{align*}
Differentiating once again, we have
\begin{align*}
I''(\alpha)
&= -\int_{0}^{\infty} \frac{2\alpha x \cos x}{(\alpha^2+x^2)^2} \, dx
 = \left[ \frac{\alpha \cos x}{\alpha^2+x^2} \right]_{0}^{\infty} + \int_{0}^{\infty} \frac{\alpha \sin x}{\alpha^2+x^2} \, dx \\
&= -\frac{1}{\alpha} + I(\alpha).
\end{align*}
Thus $I$ satisfies the differential equation
$$ I'' - I = -\frac{1}{\alpha}. \tag{1}$$
To solve this equation, we let
$$ I(\alpha) = u e^{\alpha}. $$
Plugging this to $(1)$ and multiplying $e^{\alpha}$ to both sides, we obtain
$$ (u'e^{2\alpha})' = -\frac{1}{\alpha}e^{\alpha}. $$
Thus integrating both sides, we have
$$ u'e^{2\alpha} = -\mathrm{Ei}(\alpha) - \frac{c_{1}}{2}, $$
where
$$\mathrm{Ei}(\alpha) = PV \int_{-\infty}^{\alpha} \frac{e^{t}}{t} \, dt$$
is the exponential integral function. Then
$$ u' = -e^{-2\alpha}\mathrm{Ei}(\alpha) - \frac{c_{1}}{2}e^{-2\alpha} $$
and hence
\begin{align*}
u
&= \int \left( -e^{-2\alpha}\mathrm{Ei}(\alpha) - \frac{c_{1}}{2}e^{-2\alpha} \right) \, d\alpha \\
&= \frac{1}{2}e^{-2\alpha} \mathrm{Ei}(\alpha) - \int \frac{e^{-\alpha}}{2\alpha} \, d\alpha + c_{1}e^{-2\alpha} + c_{2} \\
&= \frac{1}{2}e^{-2\alpha} \mathrm{Ei}(\alpha) - \frac{1}{2}\mathrm{Ei}(-\alpha) + c_{1}e^{-2\alpha} + c_{2}.
\end{align*}
Therefore it follows that
$$ I(\alpha) = \frac{e^{-\alpha} \mathrm{Ei}(\alpha) - e^{\alpha}\mathrm{Ei}(-\alpha)}{2} + c_{1}e^{-\alpha} + c_{2} e^{\alpha} $$
for some $c_1$ and $c_2$. To determine $c_1$ and $c_2$, observe that
$$\mathrm{Ei}(\alpha) \sim c + \log |\alpha|$$
near $\alpha = 0$. (In fact, we have $c = \gamma$.) Thus taking $\alpha \to 0$,
$$ 0 = I(0) = c_1 + c_2. $$
This shows that we may write
$$ I(\alpha) = \frac{e^{-\alpha} \mathrm{Ei}(\alpha) - e^{\alpha}\mathrm{Ei}(-\alpha)}{2} + c \sinh \alpha. $$
But L'hospital's rule shows that
$$ \mathrm{Ei}(\alpha) \sim \frac{e^{\alpha}}{\alpha} $$
as $|\alpha| \to \infty$. Thus $ I(\alpha) \sim c \sinh \alpha$ as $\alpha \to \infty$. But it is clear that $I(\alpha)$ is bounded:
$$ \left|I(\alpha)\right| \leq \int_{0}^{\infty} \frac{1}{1+x^2} \, dx = \frac{\pi}{2}. $$
Therefore $c = 0$ and we have
$$ \int_{0}^{\infty} \frac{\sin (\alpha x)}{1+x^2} \, dx = \frac{e^{-\alpha} \mathrm{Ei}(\alpha) - e^{\alpha}\mathrm{Ei}(-\alpha)}{2}. $$
