Is there a closed form for $\int_0^\infty \frac{\sin(x)}{x^2+a^2} dx$? As the title says, is there a closed form for
$$
\int_0^\infty \frac{\sin(x)}{x^2+a^2} dx \,?
$$
The one with $\cos(x)$ instead of $\sin(x)$ can be calculated via a simple application of the Residue theorem, but the computation uses the fact that the function is even, so the same trick does not work for this one.
Clarification In case this was unclear, I do not care about the method used to calculate this integral.
 A: Contour integration of $(e^{iz}-e^{-a})/(a^2+z^2)$ along the boundary of $|z|<R,0<\arg z<\pi/2$, with $z=ia$ a removable singularity (we assume $a>0$), yields $$\int_0^\infty\frac{\sin x}{a^2+x^2}\,dx=\int_0^\infty\frac{e^{-x}-e^{-a}}{a^2-x^2}\,dx,$$ expressible in terms of the exponential integral function (after doing partial fractions).
A: First, we enforce the substitution $x\mapsto ax$.  Then, we use the convolution theorem of the Fourier Transform to write
$$\begin{align}
I(a)&\equiv \int_0^\infty \frac{\sin(x)}{x^2+a^2}\,dx\\\\
&=\frac1a\int_0^\infty \frac{\sin(ax)}{x^2+1}\,dx\\\\
&=\frac1a \int_{-\infty}^\infty \frac{\sin(ax)}{x^2+1}\,H(x)\,dx\\\\
&=\frac1a\text{Im}\left(\int_{-\infty}^\infty \frac{H(x)}{x^2+1}e^{iax}\,dx\right)\\\\
&=\frac1{2\pi a} \text{Im}\left(\left(\pi e^{-|a|}\right)*\left(\pi \delta(a)+\text{PV}\left(\frac ia\right)\right)\right)\\\\
&=\frac1{2a}\text{PV}\left(\int_{-\infty}^\infty \frac{e^{-|a-a'|}}{a'}\,da'\right)\\\\
&=\frac{e^{-a}\text{Ei}(a)-e^{a}\text{Ei}(-a)}{2a}
\end{align}$$
where $\text{Ei}(a)=\int_{-\infty}^a \frac{e^{x}}{x}\,dt$ is the exponential integral, interpreted in the Cauchy Principal Value sense for $a>0$.
