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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?

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Set $$ I(a,b):=\int^{\infty}_{0}\frac{e^{-t}}{\sqrt{(t+a)^2+b^2}}dt.\tag 0 $$ Write $$ I(a,b)=e^a\int^{\infty}_{a}\frac{e^{-x}}{\sqrt{x^2+b^2}}dx =e^a\int^{0}_{a}\frac{e^{-x}}{\sqrt{x^2+b^2}}dx+e^a\int^{\infty}_{0}\frac{e^{-x}}{\sqrt{x^2+b^2}}dx.\tag 1 $$ But $$ \int^{c}_{0}\frac{x^k}{\sqrt{x^2+b^2}}dx=b^{-1}c\frac{c^k}{k+1}\cdot {}_2F_1\left[\frac{1}{2},\frac{k+1}{2};\frac{k+3}{2};-\frac{c^2}{b^2}\right] $$ and $$ e^a\int^{\infty}_{0}\frac{e^{-x}}{\sqrt{x^2+b^2}}dx=\frac{\pi e^a}{2}\left(-Y_{0}(b)+H_0(b)\right), $$ where $Y_{0}(x)$ is the usual Bessel function and $H_{\nu}(x)$ is the Struve function. For integer $\nu\geq 0$, $H_{\nu}(x)$ satisfies the differential equation $$ x^2y''+xy'+(x^2-\nu^2)y=\frac{2}{\pi}\frac{x^{\nu+1}}{(2\nu-1)!!}. $$ Also $$ \int H_{0}(x)dx=\frac{x^2}{\pi}\cdot {}_2F_3\left[1,1;\frac{3}{2},\frac{3}{2},2;\frac{-x^2}{4}\right] $$ Hence $$ I(a,b)=\frac{e^a\pi}{2}\left(-Y_{0}(b)+H_0(b)\right)-\frac{ae^a}{b}\sum^{\infty}_{k=0}\frac{(-a)^k}{(k+1)!}\cdot {}_2F_1\left[\frac{1}{2},\frac{k+1}{2};\frac{k+3}{2};-\frac{a^2}{b^2}\right]. $$ But for the incomplete beta functionwe have $$ B(z;a_1,b_1)=\frac{z^{a_1}}{a_1}\cdot {}_2F_1\left(1-b_1,a_1;1+a_1;z\right) $$ $$ B(z;a_1,b_1)=z^{a_1}\sum^{\infty}_{n=0}\frac{(1-b_1)_n}{n!(n+a_1)}z^n. $$ Hence $$ \frac{2}{k+1}\left(\frac{-a^2}{b^2}\right)^{\frac{k+1}{2}}\cdot {}_2F_1\left[\frac{1}{2},\frac{k+1}{2};\frac{k+3}{2};-\frac{a^2}{b^2}\right]=B\left(-\frac{a^2}{b^2};\frac{k+1}{2},\frac{1}{2}\right) $$ and $I(a,b)$ becomes $$ I(a,b)=\frac{e^a\pi}{2}\left(-Y_{0}(b)+H_0(b)\right)+\frac{ie^a}{2}\sum^{\infty}_{k=0}\frac{(ib)^k}{k!}\cdot B\left(-\frac{a^2}{b^2};\frac{k+1}{2},\frac{1}{2}\right)= $$ $$ =\frac{e^a\pi}{2}\left(-Y_{0}(b)+H_0(b)\right)+ie^a\sum^{\infty}_{k,n=0}\frac{\left(\frac{1}{2}\right)_n}{n!k!\left(2n+k+1\right)}z^kw^{n+\frac{k+1}{2}}, $$ where $z=ib$ and $w^2=-a^2/b^2$. But $$ \sum^{\infty}_{k=0}\frac{z^k w^{\frac{k+1}{2}}}{k!(2n+k+1)}=-\frac{(-z\sqrt{w})^{-2n}\Gamma\left(2n+1,0,-z\sqrt{w}\right)}{z}, $$ where $$ \Gamma(s,a,b):=\int^{b}_{a}e^{-t}t^{s-1}dt $$ and $-z\sqrt{w}=a$. Hence $$ I(a,b)=\frac{e^a\pi}{2}\left(-Y_{0}(b)+H_0(b)\right)- $$ $$ -ie^az^{-1}\sum^{\infty}_{n=0}\frac{\left(\frac{1}{2}\right)_n}{n!}\Gamma\left(2n+1,0,-z\sqrt{w}\right)\left(-z\sqrt{w}\right)^{-2n}w^n= $$ $$ =\frac{e^a\pi}{2}\left(-Y_{0}(b)+H_0(b)\right)-\frac{e^a}{b}\sum^{\infty}_{n=0}\frac{(-1)^n\left(\frac{1}{2}\right)_n}{n!}\Gamma(2n+1,0,a)b^{-2n}= $$ $$ =\frac{e^a\pi}{2}\left(-Y_{0}(b)+H_0(b)\right)-\frac{e^a}{b}\sum^{\infty}_{n=0}\frac{(-1)^n\left(\frac{1}{2}\right)_n(2n)!}{n!}b^{-2n}\left(1-e^{-a}\sum^{2n}_{k=0}\frac{a^k}{k!}\right)\Rightarrow $$ $$ I(a,b)=\frac{e^a\pi}{2}\left(-Y_{0}(b)+H_0(b)\right)-\sum^{\infty}_{n=0}\frac{(-1)^n\left(\frac{1}{2}\right)_n(2n)!}{n!}C_{2n}(a)b^{-2n-1},\tag 2 $$ where $$ C_{n}(a)=e^a-\sum^{n}_{k=0}\frac{a^k}{k!}\textrm{ and }C'_n(a)=C_{n-1}(a)\tag 3 $$ Hence we can state things like $$ \int^{+\infty}_{0}I\left(a,b\right)e^{-sa}da\approx\frac{\pi}{2(s-1)}\left(-Y_{0}\left(b\right)+H_0\left(b\right)\right)- $$ $$ -\frac{1}{s-1}\sum^{\infty}_{n=0}\frac{(-1)^n\left(\frac{1}{2}\right)_n(2n)!}{n!}(bs)^{-2n-1}\textrm{, }s\rightarrow+\infty $$ and if $j=0,1,2,\ldots$, then $$ \partial^j_a\int^{+\infty}_{0}\frac{e^{-t}}{\sqrt{(t+a)^2+b^2}}dt=\frac{e^a\pi}{2}\left(-Y_{0}(b)+H_0(b)\right)- $$ $$ -\sum^{\infty}_{n=0}\frac{(-1)^n\left(\frac{1}{2}\right)_n(2n)!}{n!}C_{2n-j}(a)b^{-2n-1}, $$ where $C_{n}(a)=e^a$ if $n<0$, $C_0(a)=e^a-1$. Hence observing that $C_{2n-j}(0)=1$ if $j=2n+1,2n+2,2n+3,\ldots$ and 0 else, we get $$ \left(\frac{1}{j!}\partial^j_aI(a,b)\right)_{a=0}= $$ $$ =\frac{\pi}{2}\left(-Y_0(b)+H_0(b)\right)-\sum^{\infty}_{n=0}\frac{(-1)^n\left(\frac{1}{2}\right)_n(2n)!}{n!}\frac{X_{j\geq 2n+1}}{j!}b^{-2n-1}, $$ where $X_{j\geq 2n+1}$ is the characteristic function on $j\geq 2n+1$. Hence for all $j=0,1,2,\ldots$, we have $$ \left(\partial^j_aI(a,b)\right)_{a=0}=\frac{\pi}{2}\left(-Y_0(b)+H_0(b)\right)-\sum^{\left[\frac{j-1}{2}\right]}_{n=0}\frac{(-1)^n\left(\frac{1}{2}\right)_n(2n)!}{n!\cdot b^{2n+1}}.\tag 3 $$ Also $$ I(w,b)=\sum^{\infty}_{n=0}\left(\frac{1}{n!}\partial^n_aI(a,b)\right)_{a=0}w^n. $$ Hence $$ I(a,b)=\frac{\pi e^a}{2}\left(-Y_0(b)+H_0(b)\right) -\sum^{\infty}_{n=0}\frac{a^n}{n!}\sum^{\left[\frac{n-1}{2}\right]}_{k=0}\frac{(-1)^k\left(\frac{1}{2}\right)_k(2k)!}{k!\cdot b^{2k+1}}.\tag 4 $$ Another one is from $$ \partial_a\left(e^{-a}C_{2n}(a)\right)=\frac{a^{2n}e^{-a}}{(2n)!} $$ and relation (2): $$ \partial_a\left(e^{-a}I(a,b)\right)=-e^{-a}\sum^{\infty}_{n=0}\frac{(-1)^n\left(\frac{1}{2}\right)_n}{n!}a^{2n}b^{-2n-1}. $$ Hence $$ \partial_a\left(e^{-a}I(a,b)\right)=-\frac{e^{-a}}{\sqrt{a^2+b^2}}. $$ Hence exist function $f(b)$ such that (this is trivial, see relation (1)): $$ I(a,b)=-e^a\int\frac{e^{-a}}{\sqrt{a^2+b^2}}da+e^af(b)\Rightarrow $$ $$ I(a,b)=-e^a\int^{a}_{0}\frac{e^{-t}}{\sqrt{t^2+b^2}}dt+\frac{\pi e^a}{2}\left(-Y_0(b)+H_0(b)\right) $$ Now I use the Fourier transform pairs $$ e^{-t}X_{[0,a]}(t)\leftrightarrow\frac{1-e^{-a(1+iw)}}{1+iw} $$ and $$ \frac{1}{\sqrt{t^2+b^2}}\leftrightarrow 2K_0(b|w|). $$ Hence $$ \int^{a}_{0}\frac{e^{-t}}{\sqrt{t^2+b^2}}dt=\int^{+\infty}_{-\infty}e^{-t}X_{[0,a]}(t)\frac{1}{\sqrt{t^2+b^2}}dt= $$ $$ =\frac{1}{\pi}\int^{+\infty}_{-\infty}K_0\left(b|w|\right)\frac{1-e^{-a(1+iw)}}{1+iw}dw= $$ $$ =\frac{1}{\pi}\int^{+\infty}_{-\infty}\frac{K_0(b|w|)}{1+iw}dw-\frac{e^{-a}}{\pi}\int^{+\infty}_{-\infty}\frac{K_0(b|w|)}{1+iw}e^{-iwa}dw $$ Set now $$ F(a,b):=\int^{+\infty}_{-\infty}\frac{K_0(b|w|)}{1+iw}e^{-iwa}dw, $$ then we have $$ \int^{a}_{0}\frac{e^{-t}}{\sqrt{t^2+b^2}}dt=\pi^{-1}F(0,b)-\pi^{-1}e^{-a}F(a,b). $$

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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.

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