# Solution of the integral $\int_0^{\infty} \frac{e^{- k \, z}\, k}{k^3-a}J_0\left(k \, r\right) dk$ involving Bessel functions

I am trying to evaluate the following integral:

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

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

this question is closely related to another one:

solution for integral $\int_0^{\infty} \frac{k}{k^3-a}J_0\left(k \, r\right) dk$ involving Bessel function (Hankel transform)

Please not that this integral can be seen as an Hankel transform of the function $\frac{e^{- k \, z}\,}{k^3-a}$.

I have done some steps into the solution, you can find them below:

A possible solution goes through first the complex partial fraction decomposition of the integrand into:

$\int_0^{\infty} e^{- k \, z} \left[ \frac{A}{k-a^{1/3}}+\frac{B}{k+(-1)^{1/3}a^{1/3}}+\frac{C}{k-(-1)^{2/3}a^{1/3}} \right]J_0\left(k \, r\right) \, dk$

where the constants $A$, $B$, and $C$ have to be found matching through partial fraction decomposition.

we can now look for a solution of each integral. Please note also that $-(-1)^{2/3}a^{1/3}$ is the complex conjugate of $(-1)^{1/3}a^{1/3}$.

Taking for simplicity the first integral:

$\int_0^{\infty} e^{- k \, z} \frac{A}{k-a^{1/3}}J_0\left(k \, r\right) \, dk$

one could proceed using a property of Laplace (and inverse Laplace) transformation with respect to the variable $k$, which gives you:

$\int_0^{\infty} e^{- k \, z} \frac{A}{k-a^{1/3}}J_0\left(k \, r\right) \, dk = \int_0^{\infty} \mathcal{L}\left(J_0\left(k \, r\right) \right) \mathcal{L}^{-1} \left( e^{- k \, z} \frac{A}{k-a^{1/3}} \right) ds$

now, I know that (see the link at the top):

$\mathcal{L}\left(J_0\left(k \, r\right) \right)= \frac{1}{\sqrt{r^2+s^2}}$

and:

$\mathcal{L}^{-1} \left( e^{- k \, z} \frac{A}{k-a^{1/3}} \right)= A \, e^{-a^{1/3}(s-z)}\, u(s-z)$

where $u(x)$ is the heaviside step function.

it means that we only have to evaluate the following integral:

$\int_0^{\infty} A \, u(s-z) \, \frac{e^{-a^{1/3}(s-z)}}{\sqrt{r^2+s^2}}ds$

and by performing a change of variable we get:

$$\int_0^{\infty} A \,\frac{e^{-a^{1/3}t}}{\sqrt{r^2+(t+z)^2}}dt$$

However I am not able to solve and integral like the one above, do you have any reference for that? or show me the steps? Also Mathematica is not helping with the solution.

I believe the solution to the integral is going to be something like: $- \frac{\pi}{2}A \left(H_0(a^{1/3} \, r)+Y_0(a^{1/3} \, r) \right) e^{a^{1/3} z}$

• shifting property of laplace transform? – tired Feb 6 '17 at 13:52
• mmmm not sure how to do that, I was thinking more of something like a change of variable... something like $x = s-z$ do you think that might be useful? – SSC Napoli Feb 6 '17 at 14:03
• that is more or less what i was suggesting ;) – tired Feb 6 '17 at 14:06
• i don't clearly see how the integral becomes simpler tho, with that change of variable – SSC Napoli Feb 6 '17 at 14:13
• There is a missing parenthesis: $$\mathcal{L}^{-1}\left(e^{-kz}\frac{A}{k-a^{1/3}}\right)= A\cdot\mathbb{1}_{s\geq z}\exp\left(a^{1/3}(s-z)\right)$$ hence the key integral is: $$A \int_{0}^{+\infty}\frac{\exp\left(-a^{1/3}t\right)}{\sqrt{r^2+(t+z)^2}}\,dt$$ – Jack D'Aurizio Feb 6 '17 at 18:04