$$ \int_0^\infty\frac{1}{k^2-k_p^2}J_0\left(k\rho\right)\;k\,dk $$

Suppose that $k_p$ is in the first quadrant in the complex plane, and that $\rho$ is purely real. $J_0$ is the Bessel function of the first kind and order zero.

What is the right set of moves to calculate this integral? What are the complex analysis arguments you have to use (closing contours around the two poles, residues, etc.; are these applications of the Cauchy integral formula, or the residue theorem)? Do you have extend the integral to the whole real line and replace the Bessel function with a Hankel function? What's the final result? Mathematica is happy to tell me the answer is a specific Hankel function, but I would like to know how these things work for THIS particular problem.


We may exploit the following facts: $$\mathcal{L}(J_0(k\rho)) = \frac{1}{\sqrt{\rho^2+s^2}},\qquad \mathcal{L}^{-1}\left(\frac{k}{k^2-k_p^2}\right) = \cosh(k_p s). $$ If we temporarily assume that $k_p = ip$ and $p\in\mathbb{R}^+$, the original integral equals: $$ \int_{0}^{+\infty}\frac{\cos(ps)}{\sqrt{\rho^2+s^2}}\,ds = \int_{0}^{+\infty}\frac{\cos(\rho p s)}{\sqrt{s^2+1}} = K_0(\rho p)$$ by the integral definition of $K_0$. By analytic continuation, it is expected that the original integral, if converging, equals $\color{red}{K_0(-i\rho k_p)}$.

  • $\begingroup$ I must admit I don't see how those Lapalace transforms were used in the integral in question; however, I do know that you are right for the value of the integral. Can you be a little more explicit about the steps in between? $\endgroup$ – rajb245 Jun 2 '16 at 1:47
  • $\begingroup$ Someone just upvoted the question, so I'm here again and want to ask: is $\int_0^\infty a(k) b(k)\;dk$ = $\int_0^\infty \mathcal{L}[a(k)]\mathcal{L}^{-1}[b(k)]\;ds$ some general trick you can do, or are you using some other deep mathematics that makes it look like that's what you did? In the former case, I can't quite work out how that is so. Otherwise, can you clarify? $\endgroup$ – rajb245 Jan 13 '17 at 21:10
  • $\begingroup$ Have a look at the Wikipedia page about the Laplace transform, integrals along the positive real axis. I am just using that property. $\endgroup$ – Jack D'Aurizio Jan 13 '17 at 22:21
  • $\begingroup$ The details of this answer really eluded me until your last comment, thank you! $\endgroup$ – rajb245 Jan 15 '17 at 1:28

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