I am trying to solve the integral (see also here)
$$I(\gamma,b,\beta)=\frac{\beta}{\sqrt{\beta^2+b^2}}\int_{0}^{\infty} \frac{1}{x^2+\gamma^2\,\frac{\beta^2}{b^2+\beta^2}} \, e^{- \sqrt{\beta^2+b^2} \, \sqrt{x^2+\gamma^2}} \, \mathrm{d}x$$
I was wondering if it makes any sense to use the power series of
$$e^{- \sqrt{\beta^2+b^2} \, \sqrt{x^2+\gamma^2}} = \sum_{n=0}^\infty \frac{(- \sqrt{\beta^2+b^2} \, \sqrt{x^2+\gamma^2})^n}{n!}$$
and the indefinite integral (from Mathematica)
$$I_1(x,\gamma,b,\beta,n)=\frac{\beta}{\sqrt{\beta^2+b^2}}\int \frac{(- \sqrt{\beta^2+b^2} \, \sqrt{x^2+\gamma^2})^n}{\left(x^2+\gamma^2\,\frac{\beta^2}{b^2+\beta^2}\right)n!}\, \mathrm{d}x=\frac{(-1)^n}{\beta\,\gamma^2\,n!}\,x\,(\beta^2+b^2)^{\frac{1+n}{2}}\,\left(1+\frac{x^2}{\gamma^2}\right)^{-n/2}\,(x^2+\gamma^2)^{\frac{n}{2}}\,\mathrm{F}_1\left(\tfrac{1}{2};-\tfrac{n}{2},1,\tfrac{3}{2};-\tfrac{x^2}{\gamma^2},-\tfrac{x^2\,(b^2+\beta^2)}{\beta^2\,\gamma^2}\right)$$
with $\mathrm{F}_1$ the Appell Hypergemetric Function.
Since $$\lim_{x\to0} I_1(x,\gamma,b,\beta,n)=0$$ I would guess that $$I(\gamma,b,\beta)= \lim_{x\to\infty}\sum_{n=0}^\infty \frac{(-1)^n}{\beta\,\gamma^2\,n!}\,x\,(\beta^2+b^2)^{\frac{1+n}{2}}\,\left(1+\frac{x^2}{\gamma^2}\right)^{-n/2}\,(x^2+\gamma^2)^{\frac{n}{2}}\,\mathrm{F}_1\left(\tfrac{1}{2};-\tfrac{n}{2},1,\tfrac{3}{2};-\tfrac{x^2}{\gamma^2},-\tfrac{x^2\,(b^2+\beta^2)}{\beta^2\,\gamma^2}\right)$$
If this is even correct, is there a way to get a closed form solution from this approach? I know there is a solution for $b=0$. Thank you.
Edit: I think it can be simplified to $$I(\gamma,b,\beta)= \lim_{x\to\infty}\sum_{n=0}^\infty \frac{(-1)^n}{\beta\,\gamma^{2-n}\,n!}\,x\,(\beta^2+b^2)^{\frac{1+n}{2}}\,\,\mathrm{F}_1\left(\tfrac{1}{2};-\tfrac{n}{2},1,\tfrac{3}{2};-\tfrac{x^2}{\gamma^2},-\tfrac{x^2\,(b^2+\beta^2)}{\beta^2\,\gamma^2}\right)$$