Closed form for this integral $\int_0^\infty \frac{e^{-x} dx}{\sqrt{(x+a)^2+b^2}}$ This is a Laplace transform, however I couldn't find it in the tables and Wolfram doesn't know the answer either:
$$I(a,b)=\int_0^\infty \frac{e^{-x} dx}{\sqrt{(x+a)^2+b^2}}$$
Some kind of closed form (even in terms of special functions) will help me very much with my problem, as this is the simple part of a double integral which will need to be evaluated many times.
With some simple substitutions we can write:
$$I(a,b)=e^a \int_a^\infty \frac{e^{-y} dy}{\sqrt{y^2+b^2}}=e^a \int_{a/b}^\infty \frac{e^{-b z} dz}{\sqrt{z^2+1}}=e^a \int_{\sinh^{-1} a/b}^\infty e^{-b \sinh u} du$$
Looks like some kind of incomplete Bessel function to me, but is there a standart closed form expression?
 A: I've been trying to find a possible way to deal with this integral, an here's one promising attempt:
$$I(a,b)=\int_0^\infty \frac{e^{-x} dx}{\sqrt{x^2+2ax+a^2+b^2}}$$
Let's consider two separate cases:
$$1) \qquad x<\sqrt{a^2+b^2}$$
$$r_1=\frac{x}{\sqrt{a^2+b^2}} <1 \\ q= -\frac{a}{\sqrt{a^2+b^2}}, \quad |q| <1$$
$$x^2+2ax+a^2+b^2=(a^2+b^2) (r_1^2-2 q r_1+1)$$
$$2) \qquad x > \sqrt{a^2+b^2}$$
$$r_2=\frac{\sqrt{a^2+b^2}}{x} <1$$
$$x^2+2ax+a^2+b^2=x^2 (r_2^2-2 q r_2+1)$$
Which means:
$$I(a,b)=\int_0^\sqrt{a^2+b^2} \frac{e^{-x} dx}{\sqrt{a^2+b^2}} \sum_{n=0}^\infty P_n (q) r_1^n+\int_\sqrt{a^2+b^2}^\infty  \frac{e^{-x} dx}{x} \sum_{n=0}^\infty P_n (q) r_2^n$$
Where $P_n$ are Legendre polynomials.
Let's rename $$c=\sqrt{a^2+b^2}, \qquad r=r_1$$
$$I(a,b)= \sum_{n=0}^\infty P_n (q) \left( \int_0^1 r^n e^{-c r} dr+ \int_1^\infty  \frac{e^{-c r} dr}{r^{n+1}} \right)=$$
$$=\sum_{n=0}^\infty P_n (q) \left( \frac{n!-\Gamma(n+1,c)}{c^{n+1}}+ \operatorname{Ei}_{n+1} (c) \right)$$
This is rather ugly, so if anyone has a better option, please share.
