Contour integration with absolute value I am trying to use contour integration to evaluate the integral
$$
\int_\mathbb{R}\frac{|x|e^{i\omega x}}{x^2+ a}dx \,,
$$
where $a>0$ and $\omega$ can take any real value. I am worried about the existence of the absolute value and I am not sure how to proceed. 
Any help appreciated.
EDIT: Contour integration might not be the way to go here (see comments). In any case, it is easy to see that the integral in question is also equal to
$$
2\int_0^\infty \frac{x \cos(\omega x)}{x^2+ a}dx \,,
$$
and I am wondering whether anyone can produce a solution regardless of the method used. 
 A: Let's treat it like a Fourier Transform, and use a number of table lookups and theorems.
Set $\omega = -2\pi s$, 
$$\begin{align*}\int_{-\infty}^\infty{\dfrac{|x|e^{i\omega x}}{x^2+a}}dx &= \int_{-\infty}^\infty{\dfrac{|x|e^{-2\pi is x}}{x^2+a}}dx \\
\\
&= \mathscr{F}\left\{\dfrac{|x|}{x^2+a}\right\} \\
\\
&= \mathscr{F}\left\{\dfrac{x\space\mathrm{sgn}(x)}{x^2+a}\right\}\\
\\
&= \mathscr{F}\left\{\dfrac{1}{x^2+a}\right\}*\mathscr{F}\left\{x\space\mathrm{sgn}(x)\right\} \\
\\
&= \dfrac{(2\pi)^2}{2\left(2\pi\sqrt{a}\right)}\mathscr{F}\left\{\dfrac{2\left(2\pi\sqrt{a}\right)}{(2\pi x)^2+\left(2\pi\sqrt{a}\right)^2}\right\}*-\dfrac{1}{2\pi i}\mathscr{F}\left\{-2\pi i x\space\mathrm{sgn}(x)\right\} \\
\\
&= \dfrac{2\pi}{2\sqrt{a}}e^{-2\pi\sqrt{a}|s|}*-\dfrac{1}{2\pi i}\dfrac{d}{ds}\dfrac{1}{i\pi s} \\
\\
&= \dfrac{1}{2\sqrt{a}}e^{-2\pi\sqrt{a}|s|}*\dfrac{d}{ds}\dfrac{1}{\pi s} \\
\\
&= \dfrac{1}{2\sqrt{a}}e^{-2\pi\sqrt{a}|s|}*-\dfrac{1}{\pi s^2} \\
\\
&= -\dfrac{1}{2\pi\sqrt{a}}\int_{-\infty}^{\infty} {\dfrac{e^{-2\pi\sqrt{a}|\tau|}}{(s-\tau)^2}}d\tau \\
\\
&= -\dfrac{1}{2\pi\sqrt{a}}\left[\int_{-\infty}^{0} {\dfrac{e^{2\pi\sqrt{a}\tau}}{(s-\tau)^2}}d\tau +\int_{0}^{\infty} {\dfrac{e^{-2\pi\sqrt{a}\tau}}{(s-\tau)^2}}d\tau\right] \\
\\
&= -\dfrac{1}{2\pi\sqrt{a}}\left[\int_{-\infty}^{-s} {\dfrac{e^{2\pi\sqrt{a}(t+s)}}{t^2}}dt +\int_{-s}^{\infty} {\dfrac{e^{-2\pi\sqrt{a}(t+s)}}{t^2}}dt\right] \\
\\
&= -\dfrac{1}{2\pi\sqrt{a}}\left[e^{2\pi\sqrt{a}s}\int_{-\infty}^{-s} {\dfrac{e^{2\pi\sqrt{a}t}}{t^2}}dt +e^{-2\pi\sqrt{a}s}\int_{-s}^{\infty} {\dfrac{e^{-2\pi\sqrt{a}t}}{t^2}}dt\right] \\
\\
&= -\dfrac{1}{2\pi\sqrt{a}}\left[e^{-\sqrt{a}\omega}\int_{-\infty}^{\frac{\omega}{2\pi}} {\dfrac{1}{t}\dfrac{e^{2\pi\sqrt{a}t}}{t}}dt +e^{\sqrt{a}\omega}\int_{\frac{\omega}{2\pi}}^{\infty} {\dfrac{1}{t}\dfrac{e^{-2\pi\sqrt{a}t}}{t}}dt\right] \\
\\
&= -\dfrac{1}{2\pi\sqrt{a}}\left[e^{-\sqrt{a}\omega}\int_{-\frac{\omega}{2\pi}}^\infty {\dfrac{1}{t}\dfrac{e^{-2\pi\sqrt{a}t}}{t}}dt +e^{\sqrt{a}\omega}\int_{\frac{\omega}{2\pi}}^{\infty} {\dfrac{1}{t}\dfrac{e^{-2\pi\sqrt{a}t}}{t}}dt\right] \\
\\
\end{align*}$$
So those integrals are looking really close to being an integration by parts resulting in an expression involving the Exponential Integral function, and Wolfram Alpha confirms the indefinite integral results include the Exponential Integral function.
I think getting to the final result from here is straightforward, but tedious.
