Tricky contour integral resulting from the integration of $\sin ax / (x^2+b^2)$ over the positive halfline I am trying to evaluate the definite integral
$$\int_0^\infty \frac{\sin ax\ dx}{x^2+b^2}$$
where $a,b>0$. This is a problem on an assignment for a class in complex variables. I understand the mechanics of contour integration, but I am stuck. (I have spoken to four classmates who are also stuck.) I would appreciate a small hint, such as a hint to point me toward the right contour, or a suggestion for how to modify one of my abandoned ideas (below) to make it work. (But please do not work out the whole integral; I want to do that myself.)
Here is what I have tried so far:


*

*The integral from $-\infty$ to $\infty$ is zero because the function is odd, so the semicircular contour in the upper half-plane won't do anything.

*Viewing the integral as 
$$
\mathrm{Im}\, \int_0^\infty \frac{e^{aiz}dz}{z^2+b^2},
$$ 
I tried integrating around the contour from $0$ to $R>0$, along the quarter-circle to $iR$, and back down to $0$, with an indentation at the pole at $bi$. The pole is simple so I can get the contribution from the indentation via the residue, and the contribution from the quarter-circle $\to 0$ as $R\to \infty$, but the problem is that I believe the integral along the segment of the imaginary axis from $iR$ to $0$ is imaginary (so contributes to the imaginary part) and blows up around the pole, and generally seems harder to deal with than the original integral.

*Viewing the integrand with either $\sin az$ or $e^{iaz}$ in the numerator, I tried integrating along the rectangle with vertices $0,R,R+ib,ib$, with an indentation at the vertex $ib$ due to the pole. Again, I don't know how to get control of the integral along the imaginary axis.

*I tried using integration by parts to get a cosine to come out, but the integrand remains odd so the upper half-plane semicircular contour still does no good.

*I had a crazy idea that I was unable to carry out. There exists some path thru the origin along which the integrand 
$$
\frac{e^{aiz}dz}{z^2+b^2}
$$ 
is pure real.  This path is the solution to an ordinary differential equation with boundary condition $y(0)=0$. If my calculations are right the equation is 
$$
y'=\frac{2xy\cos ax-(b^2+x^2-y^2)\sin ax}{2xy\sin ax+(b^2+x^2-y^2)\cos ax}.
$$
The idea was to integrate along $0$ to $R$, counterclockwise along the circle with radius $R$ till it hits this curve, and back to the origin along this curve. By construction, the integral along this last part doesn't contribute anything to the imaginary part; meanwhile the part of the circle in the upper half plane $\to 0$ as $R\to \infty$. The curve must lie entirely outside the upper half-plane or this would show that the original integral I'm trying to evaluate is zero (since the imaginary part of $2\pi i$ times the residue is zero), which isn't plausible. Evaluating the desired integral then rests on: 

(a) whether the pole in the lower half-plane ever gets enclosed by this contour (and I'm pretty sure it doesn't), and 

(b) if I can figure out what is going on with the integral along the part of the circle $|z|=R$ in the lower half-plane before it hits the curve; in particular, what its imaginary part is doing asymptotically as $R\to \infty$. 
However, I wasn't sure how to pursue these goals any further.
Update: As it turns out, the assignment contained a typo and the integral was supposed to be
$$\int_0^\infty \frac{\sin ax\ dx}{x(x^2+b^2)}$$
This makes the integrand even, so it is done easily with the half-circle contour in the upper half-plane.  I definitely learned more because of the typo.  Thanks all for your answers and comments.
 A: You can actually see directly that there is no elementary expression for the integral, using contour integration. For your integral is the same as 
$$\mathrm{Im}\int_0^\infty {e^{iax} \over x^2 + b^2}\mathrm dx$$
Note that
$${1 \over x^2 + b^2} = {1 \over 2ib}\left({1 \over x - ib} -  {1 \over x + ib}\right)$$
So it suffices to evaluate the integrals
$$\int_0^\infty {e^{iax} \over x \pm bi}\,\mathrm dx$$
These integrals converge despite the denominator having only degree one because the exponential factor modulates it, similar to for $\dfrac{\sin(x)}{x}$. You can use your quarter circle idea on each of these two terms (without even having to worry about indentations)... the upper quarter circle for $+bi$ and the bottom quarter circle for $-bi$. For example you get that the first one is equal to 
$$\int_0^\infty {e^{-ax} \over x + b}\,\mathrm dx$$
Letting $x = z - b$ this becomes
$$e^{ab}\int_b^\infty {e^{-az} \over z}\,\mathrm dz$$
Then letting $z = \dfrac{w}{a}$ this turns into
$$e^{ab}\int_{ab}^\infty {e^{-w} \over w}\,\mathrm dw$$
$$= e^{ab}\mathrm{Ei}(-ab)$$
Here $\mathrm{Ei}$ is the exponential function which is defined in terms of the above integral. So unless the exponential functions from the two quarter circles somehow cancel out (which I seriously doubt in view of Sasha's answer), you won't be able to get an elementary expression for the integral.
A: First, the Mellin transform convolution theorem (aka Slater theorem) states that if $\mathcal{M}_s(f) = F(s)$ and $\mathcal{M}_s(g) = G(s)$, then $\int_0^\infty x^{\alpha-1} f(x) g(x) \mathrm{d} x = \mathcal{M}^{-1}\left( F(s) G(s-\alpha) \right)$, where $\mathcal{M}^{-1}$ stands for the inverse Mellin transform. 
Now:
$$
  \mathcal{M}_s(\sin(a x)) = \int_0^\infty x^{s-1} \sin(a x) \mathrm{d} x = \frac{\sqrt{\pi}}{2} \left( \frac{a}{2} \right)^{-s} \frac{\Gamma( \frac{1}{2} + \frac{s}{2} )}{\Gamma(1- \frac{s}{2})}
$$
and
$$
  \mathcal{M}_s\left(\frac{1}{x^2 + b^2}\right) = \frac{b^{s-2}}{2} \Gamma\left(\frac{s}{2}\right)\Gamma\left(1-\frac{s}{2}\right)
$$
Therefore, by the Mellin convolution theorem:
$$ \begin{eqnarray}
   \int_0^\infty \frac{\sin(a x)}{x^2+b^2} \mathrm{d} x &=& \mathcal{M}^{-1}\left( 
    \frac{\sqrt{\pi}}{2} \left( \frac{a}{2} \right)^{-s} \frac{\Gamma( \frac{1}{2} + \frac{s}{2} )}{\Gamma(1- \frac{s}{2})} 
   \frac{b^{-s-1}}{2} \Gamma\left(\frac{1}{2}-\frac{s}{2}\right)\Gamma\left(\frac{1}{2}+\frac{s}{2}\right)
  \right) \\
  &=& \frac{\sqrt{\pi}}{2 b} \, G_{1,3}^{2,1}\left(\frac{a^2 b^2}{4}\left|
\begin{array}{c}
 \frac{1}{2} \\
 \frac{1}{2},\frac{1}{2},0
\end{array}\right.
\right) \\
  &=& \frac{\mathrm{Shi}(a b) \cosh (a b)-\mathrm{Chi}(a b) \sinh (a b)}{b}
 \end{eqnarray}
$$
Where $\mathrm{Shi}$ and $\mathrm{Chi}$ stand for hyperbolic sine and cosine integrals, and $ G_{1,3}^{2,1}(z)$ stands for Meijer G-function.
