I've recently been trying to analytically solve an improper integral $I$ which contains a Bessel function of the first kind $J_0$, an exponential function, and the reciprocal of a linear polynomial. It looks like this:
$I(a,b,k)=\int_0^\infty \frac{J_0(ax)}{b+x}e^{-kx} dx,\quad a,b,k>0.$
I've tried two approaches, both with somewhat disappointing conclusions.
Approach 1
The Bessel function can be expanded in a series as
$J_0(ax)=\sum_{n=0}^{\infty}\frac{(-1)^n}{4^nn!^2}\left(ax\right)^{2n},$
such that the integral becomes
$I(a,b,k)=\sum_{n=0}^{\infty}\frac{(-1)^na^{2n}}{4^nn!^2}\int_0^\infty \frac{x^{2n}}{b+x}e^{-kx} dx=\sum_{n=0}^{\infty}\frac{(-1)^na^{2n}}{4^nn!^2}I_n(b,k).$
This integral in the sum can be solved by Wolfram Alpha by framing it as a Laplace Transform, so
$I_n(b,k)=\int_0^\infty \frac{x^{2n}}{b+x}e^{-kx} dx=(2n)!e^{bk}b^{2n}\Gamma\left(-2n,bk\right)=\frac{(2n)!e^{bk}}{k^{2n}}E_{2n+1}(bk),$
where $E_{2n+1}(bk)=(bk)^{2n}\Gamma(-2n,bk)$ is a generalized exponential integral.
Inserting the expression for $I_n$ in $I$ and rearranging then yields
$I(a,b,k)=e^{bk}\sum_{n=0}^{\infty}(-1)^n\left(\frac{a}{k}\right)^{2n}\frac{(2n)!}{4^nn!^2}E_{2n+1}(bk)=\underline{\underline{\frac{e^{bk}}{\sqrt{\pi}}\sum_{n=0}^{\infty}(-1)^n\left(\frac{a}{k}\right)^{2n}(n+1)_{-1/2}E_{2n+1}(bk)}},$
where the Pochhammer symbol $(n+1)_{-1/2}$ arises from
$\frac{(2n)!}{4^nn!^2}=\frac{1}{\sqrt{\pi}}\frac{\Gamma(n+1/2)}{\Gamma(n+1)}=\frac{1}{\sqrt{\pi}}(n+1)_{-1/2}.$
I now have a way to evaluate the integral, however if $a>k$ the $\left(\frac{a}{k}\right)^{2n}$ coefficients will grow very fast to the point where I get overflow errors on my computer. I'm not sure if the sum will theoretically end up converging in such cases, but as it stands, evaluating $I$ like this is not practical.
Approach 2
I've read here that an integral over the positive real axis can be rewritten using the Laplace transform as
$\int_0^{\infty}f(x)g(x) dx=\int_0^{\infty}\mathcal{L}_x(f(x))(s)\mathcal{L}_x^{-1}(g(x))(s) ds,$
so we can rewrite $I$ as
$I(a,b,k)=\int_0^\infty \frac{J_0(ax)}{b+x}e^{-kx} dx=\int_0^\infty \mathcal{L}_x\left(J_0(ax)e^{-kx}\right)(s)\mathcal{L}_x^{-1}\left(\frac{1}{b+x}\right)(s)ds.$
We then have
$\mathcal{L}_x\left(J_0(ax)e^{-kx}\right)(s)=\frac{1}{\sqrt{a^2+(s+k)^2}},\quad\mathrm{and}\quad\mathcal{L}_x^{-1}\left(\frac{1}{b+x}\right)(s)=e^{-bs},$
which can be inserted into the expression for $I$, so
$I(a,b,k)=\int_0^\infty \frac{1}{\sqrt{s^2+2ks+k^2+a^2}}e^{-bs} ds $
I haven't been able to make any meaningful progress from this point. Essentially, what's inside the square root is a second degree polynomial on the form $x^2+c_1x+c_0$. Looking through Gradshteyn and Ryzhik's Table of Integrals, I've found an identity of the form
$\int_0^\infty\frac{1}{\sqrt{x^2+c_1x)}}e^{-kx}=e^{c_1k/2}K_0\left(c_1k/2\right),$
which is very close to what I need except that it is missing the constant term $c_0$ in the polynomial.
If anyone has any advice or tricks on how to proceed and evaluate $I$, I'd appreciate the help. Either way, thank you for your time.
$\int_0^\infty\dfrac{e^{-kx}}{b+x}J_0(ax)~dx$
$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^na^{2n}x^{2n}e^{-kx}}{4^n(n!)^2(b+x)}~dx$
$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^na^{2n}(bx)^{2n}e^{-kbx}}{4^n(n!)^2(b+bx)}~d(bx)$
$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^na^{2n}b^{2n}x^{2n}e^{-bkx}}{4^n(n!)^2(1+x)}~dx$
$=\int_1^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^na^{2n}b^{2n}e^{bk}(x-1)^{2n}e^{-bkx}}{4^n(n!)^2x}~dx$
$=\int_1^\infty\dfrac{e^{bk}e^{-bkx}}{x}~dx+\int_1^\infty\sum\limits_{n=1}^\infty\sum\limits_{m=1}^{2n}\dfrac{(-1)^{m+n}a^{2n}b^{2n}e^{bk}C_m^{2n}x^{m-1}e^{-bkx}}{4^n(n!)^2}~dx$
$=e^{bk}E_1(bk)+\int_1^\infty\sum\limits_{n=1}^\infty\sum\limits_{m=1}^n\dfrac{(-1)^na^{2n}b^{2n}e^{bk}(2n)!x^{2m-1}e^{-bkx}}{4^n(2m)!(2n-2m)!(n!)^2}~dx-\int_1^\infty\sum\limits_{n=1}^\infty\sum\limits_{m=1}^n\dfrac{(-1)^na^{2n}b^{2n}e^{bk}(2n)!x^{2m-2}e^{-bkx}}{4^n(2m-1)!(2n-2m+1)!(n!)^2}~dx$
$=e^{bk}E_1(bk)+\int_1^\infty\sum\limits_{m=1}^\infty\sum\limits_{n=m}^\infty\dfrac{(-1)^na^{2n}b^{2n}e^{bk}(2n)!x^{2m-1}e^{-bkx}}{4^n(2m)!(2n-2m)!(n!)^2}~dx-\int_1^\infty\sum\limits_{m=1}^\infty\sum\limits_{n=m}^\infty\dfrac{(-1)^na^{2n}b^{2n}e^{bk}(2n)!x^{2m-2}e^{-bkx}}{4^n(2m-1)!(2n-2m+1)!(n!)^2}~dx$
$=e^{bk}E_1(bk)-\int_1^\infty\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^{m+n}a^{2m+2n+2}b^{2m+2n+2}e^{bk}(2m+2n+2)!x^{2m+1}e^{-bkx}}{4^{m+n+1}(2m+2)!(2n)!((m+n+1)!)^2}~dx+\int_1^\infty\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^{m+n}a^{2m+2n+2}b^{2m+2n+2}e^{bk}(2m+2n+2)!x^{2m}e^{-bkx}}{4^{m+n+1}(2m+1)!(2n+1)!((m+n+1)!)^2}~dx$
$=e^{bk}E_1(bk)+\left[\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{p=0}^{2m+1}\dfrac{(-1)^{m+n}a^{2m+2n+2}b^{2m+2n+2}e^{bk}(2m+2n+2)!x^pe^{-bkx}}{4^{m+n+1}(2m+2)(2n)!((m+n+1)!)^2p!b^{2m-p+2}k^{2m-p+2}}\right]_1^\infty-\left[\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{p=0}^{2m}\dfrac{(-1)^{m+n}a^{2m+2n+2}b^{2m+2n+2}e^{bk}(2m+2n+2)!x^pe^{-bkx}}{4^{m+n+1}(2m+1)(2n+1)!((m+n+1)!)^2p!b^{2m-p+1}k^{2m-p+1}}\right]_1^\infty$
(according to https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions#Integrals_of_polynomials)
$=e^{bk}E_1(bk)-\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{p=0}^{2m+1}\dfrac{(-1)^{m+n}a^{2m+2n+2}b^{2n+p}(2m+2n+2)!}{4^{m+n+1}k^{2m-p+2}(2m+2)(2n)!((m+n+1)!)^2p!}+\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{p=0}^{2m}\dfrac{(-1)^{m+n}a^{2m+2n+2}b^{2n+p+1}(2m+2n+2)!}{4^{m+n+1}k^{2m-p+1}(2m+1)(2n+1)!((m+n+1)!)^2p!}$