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. 

I don't see how this integral can be evaluated using complex analysis. At some point, you're going to need a circular path with $r \rightarrow \infty$ to go to zero, and the numerator has:
$$
\sin \left(r e^{i\theta}\right) = \frac{1}{2i}\left[\exp\left(i r \cos \theta\right) \exp\left(- r \sin \theta\right) - \exp\left(-i r \cos \theta\right)\exp\left(r \sin \theta\right)\right].
$$
You might look at that and think you can break the integral up into two pieces: the first closed above the $x$ axis so that $\sin \theta > 0$ and the second closed below so that $\sin \theta < 0$. But as you noted, you have to integrate along the positive real axis only (the entire real axis will yield 0), which means you have to use a circular path at $r \rightarrow \infty$ with $\theta$ from $0$ to $2 \pi$.
