I found the following integral in $\textit{Integrals and Series Volume 4: Direct Laplace Transform}$ (Prudnikov and Brychkov) p.312 (3.14.2.1)
$$F_1(p)=\int_{0}^{\infty} \mathrm{H}_0^{(1)}\left(a \, \sqrt{x^2-b^2}\right)\, \mathrm{e}^{-p\,x}\,\mathrm{d}x = \frac{\mathrm{e}^{-b\,\sqrt{p^2+a^2}}}{\sqrt{p^2+a^2}}\,\left(1-\frac{2\,\mathrm{i}}{\pi}\,\mathrm{ln}\left(\frac{p+\sqrt{p^2+a^2}}{a}\right)\right)$$ with the given restrictions $b>0$ and Re$(p)>|$Im$(a)|$ and $\mathrm{H}_0^{(1)}()$ the Hankel function of the first kind.
Can anybody tell me how this result is derivated? And if the restrictions can be relaxed?
The integral on p.357 (3.16.5.4) would also be really interesting $$F_2(p)=\int_{0}^{\infty} \frac{1}{\sqrt{x^2-b^2}}\mathrm{K}_\nu\left(a \, \sqrt{x^2-b^2}\right)\, \mathrm{e}^{-p\,x}\,\mathrm{d}x = \\\frac{1}{2}\mathrm{sec}\left(\frac{\nu\,\pi}{2}\right)\,\mathrm{K}_{\nu/2}\left(\frac{b\,(p-\sqrt{p^2-a^2})}{2}\right)\,\mathrm{K}_{\nu/2}\left(\frac{b\,(p+\sqrt{p^2-a^2})}{2}\right)$$ with the restrictions $|$Re$(\nu)<1|$, $b>0$ and Re$(p+a)>0$. I am wondering if this integral becomes divergent if $\nu \to 1$. Especially since I think that $F_2(p)$ can be computed with $\frac{\partial\,F_1(p)}{\partial\,b}$ and the relation between $\mathrm{H}_1^{(1)}()$ and $\mathrm{K}_1()$.
I am interested, because this could maybe solve this problem
Thank you.