I am trying to find the inverse of the following kernel in 3 dimensions $$ \nabla^2-x^2, $$ where, $$ x^2=\vec{x}.\vec{x} $$ It seems quit simple and one would think there should already be solutions for it, but I could not find one and it turned out to be trickier than I thought.
$$ (\nabla^2-x^2)f(\vec{x})=\delta^3(\vec{x}-\vec{x}'), $$ performing the Fourier transform will give us the exact same kernel $$ (\nabla^2-q^2)\tilde{f}(\vec{q})=e^{i \vec{q}.\vec{x}'} $$ Assuming due to symmetry that $\vec{x}'$ is located on the 3rd axis, and taking advantage of spherical coordinates $$ (\nabla^2-q^2)\tilde{f}(\vec{q})=e^{i qx'\cos\theta}=\sum_{l=0}^{\infty}i^l(2l+1)\sqrt{\frac{\pi}{2x'q}}J_{l+\frac{1}{2}}(x'q)P_l(\cos\theta) $$ where $$ q^2=\vec{q} \cdot \vec{q},\quad x'^2=\vec{x}'\cdot \vec{x}' $$ where $P_l$ and $J_l$ are Legendre and Bessel function of the first kind of order $l$, respectively. Now we can write our $\tilde{f}$ in terms of Legndre polynomials as well $$ \tilde{f}(q,\cos \theta)=\sum_{l=0}^{\infty}a_l(q)P_l(\cos\theta) $$ after sum manipulation the equation simplifies to $$ \left(\frac{d^2}{dq^2}+\frac{2}{q}\frac{d}{dq}-\frac{l(l+1)}{q^2} -q^2\right)a_l(q)=i^l(2l+1)\sqrt{\frac{\pi}{2x'q}}J_{l+\frac{1}{2}}(x'q) $$ Here is where I hit the bump. I used Henkel Transform but it was hopeless. Does anybody have a solution or a hint? It will be greatly appreciated.
