TITLE: Evaluating $\int_0^\infty\frac{\sin(x)}{x^2+1}\, dx$
QUESTION [20 upvotes]: 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.

REPLY [4 votes]: Partial fractions to the rescue!
$$ \frac{1}{x^2 + 1} = \frac{i}{2} \left( \frac{1}{x+i} - \frac{1}{x-i} \right) $$
Then, the angle addition formulas to match the arguments to the denominators
$$ \sin(x) = \sin(x+i) \cos(i) - \sin(i) \cos(x+i) $$
$$ \sin(x) = \sin(x-i) \cos(i) + \sin(i) \cos(x-i) $$
And we can compute
$$ \int_0^\infty \frac{\sin(x+i)}{x+i} \, dx = \int_i^\infty \frac{\sin(x)}{x} \, dx = \frac{\pi}{2} -\text{Si}(i) $$
and similar. Therefore,
$$ \int_0^\infty \frac{\sin x}{x+i} \, dx = \left(\frac{\pi}{2}-\text{Si}(i)\right) \cos(i) +
\sin(i) \text{Ci}(i) $$
$$ \int_0^\infty \frac{\sin x}{x-i} \, dx = \left(\frac{\pi}{2}-\text{Si}(-i)\right) \cos(i) - 
\sin(i) \text{Ci}(-i) $$
Therefore,
$$ \int_0^\infty \frac{\sin(x)}{x^2 + 1}
= \frac{i}{2} \left( \left(-\text{Si}(i) + \text{Si}(-i) \right) \cos(i) + (\text{Ci}(i) + \text{Ci}(-i)) \sin(i) \right) $$