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.