during my research I am facing for the first time integrals involving Bessel functions. In particular i need to evaluate the following integral:

$\int_0^{\infty} \frac{k}{k^3-a}J_0\left(k \, r\right) dk $

with $a$ and $r$ being two real positive numbers. $J_0$ is Bessel function of the first kind and order zero.

I know this can be seen as an Hankel transform of the function $\frac{1}{k^3-a}$ however I was not able to find reference for this transformation. Maybe it is a known one.

Unfortunately Mathematica is not helping in finding the solution to the problem. Any help or hint is appreciated.

Maybe a way of solving this could be using a complex decomposition of the fraction?


We may go through the Laplace transform and partial fraction decomposition: $$\mathcal{L}\left(J_0(kr)\right) = \frac{1}{\sqrt{r^2+s^2}},\qquad \mathcal{L}^{-1}\left(\frac{1}{k+b}\right)=e^{-bs}\tag{1}$$ leads to: $$ \int_{0}^{+\infty}\frac{J_0(kr)}{k+b}\,dk = \int_{0}^{+\infty}\frac{e^{-bs}}{\sqrt{r^2+s^2}}\,ds \tag{2}$$ where the RHS of $(2)$ is a multiple of the difference between a Bessel $Y_0$ and a Struve $H_0$ function.

  • $\begingroup$ What about if we had the Bessel function of the second order $J_{2}$ in the integral? Do you think it has a closed form? $\endgroup$ – SSC Napoli Feb 3 '17 at 14:14
  • 1
    $\begingroup$ @SSCNapoli: if you are ready to accept a combination of Struve and Bessel functions as a closed form, sure. The Laplace transform of $J_2$ is similar to the Laplace transform of $J_0$ and the same argument applies. $\endgroup$ – Jack D'Aurizio Feb 3 '17 at 14:23
  • $\begingroup$ many thanks for the quick answer, you might be interested to this other question i have about integrals involving Bessel functions: math.stackexchange.com/questions/2125838/… $\endgroup$ – SSC Napoli Feb 3 '17 at 15:49
  • $\begingroup$ I apologise for bombarding you with comments, however I was also interested to the case when the function to integrate was $\frac{e^{-k z }k }{k^3-a}$ (with $z$ a positive real number). I guess the steps would be exactly the same, but then the final integral would look like $\int_{0}^{\infty} \frac{e^{-b s -z}}{\sqrt{r^2+s^2}} u(s-z) ds$ where $u(x)$ is the heaviside step function (ref:wikipedia). Am I correct? can this be expressed as Struve or Bessel functions? $\endgroup$ – SSC Napoli Feb 6 '17 at 11:25
  • 1
    $\begingroup$ Yes you are. For more details, you may post a separate question and I will be glad to provide them. $\endgroup$ – Jack D'Aurizio Feb 6 '17 at 11:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.