Integration of Bessel functions:Finding a suitable contour I have below function to integrate;
$$\int_{0}^{\infty} \frac{J_{0}(ax)x^3}{k^2-x^2} dx$$ here $a,k$ are constants.
Since this is an odd function, I am not allowed to extend the limits from negative infinity to positive infinity. Therefore, I am planning to have a contour in the first quadrant, and apply the residue theorem. In that case, I have no idea how the Bessel functions behave in that region of the contour. Can someone please throw some light on this ?
 A: Bessel functions on the imaginary line behave as the modified Bessel functions $(I_0(x))$ which are roughly exponentially increasing, so you'd have to take the lower quadrant to even have a chance for the far part of the contour to vanish. You will also have problems with contour integration already because of the $x=k$ point. This is why I wouldn't recommend contour integration, it does not look like the right type of integral to do that. You'll see that the integral over the far contour wouldn't vanish anyway:
The integral won't converge in the traditional sense. Asymptotically, you have
$$J_0(x)\asymp \sqrt{\frac{2}{\pi x}}\cos (x-\pi/4) $$
so your integrand behaves as
$$\sim\int^\infty \frac{x^{2.5}\times \text{oscillatory}}{k^2-x^2}dx\sim \int^\infty \sqrt{x}\times \text{oscillatory}\,dx$$
which is asymptotically something similar to adding a sum of the form $1-\sqrt{2}+\sqrt{3}-\sqrt{4}+\cdots$ (very roughly illustrated).
I'm not sure if one could regularize this integral to assign a finite value to it, but I'd be really careful where this even comes from.
