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?


1 Answer 1


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, 2017 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. $\endgroup$ Feb 3, 2017 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, 2017 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, 2017 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. $\endgroup$ Feb 6, 2017 at 11:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .