Find $ \int_0^\infty \frac{\sqrt x e^{-x}}{b^2 +x^2} dx $ While trying to solve a certain Laplace transform, this spicy integral developed.
$$ \int_0^\infty \frac{\sqrt x e^{-x}}{b^2 +x^2}  dx  $$
I am not sure how to approach this problem, and would appreciate any help.  One attempt was to expand the exponential into a power series.  This would lead to something like:
$$\sum_{n=0}^\infty \frac{{(-1)}^n}{n!}  \int_0^\infty \frac{x^{n+1/2}}{b^2+x^2}dx$$
This integral looks like it could be approached with some complex analysis for $n < 3/2$  but it doesn't really make sense for all except for the first two terms.  Am I missing something here?
I would appreciate any help solving this integral! Thanks.
 A: According to Maple it is (for $b > 0$)
$$ 2\,\sqrt {\pi} \left( {\it LommelS2} \left( 1,1/2,b \right) -1
 \right) 
$$
where the Lommel S2 function is defined here.
So I don't think you're going to get an elementary answer.
EDIT:
Writing $$ \frac{1}{b^2 + x^2} = \frac{i}{2 b (x + i b)} - \frac{i}{2 b (x - i b)}$$
I get something slightly more elementary:
$$ -{\frac {\pi\,\sqrt {2} \left(  \left( 1+i \right) {{\rm e}^{-ib}
}{\rm erf} \left( \left( 1/2-i/2 \right) \sqrt {2}\sqrt {b}\right)+
 \left( 1-i \right) {{\rm e}^{ib}}{\rm erf} \left( \left( 1/2+i/2
 \right) \sqrt {2}\sqrt {b}\right)- \left( 1+i \right) {{\rm e}^{-ib}}
- \left( 1-i \right) {{\rm e}^{ib}} \right) }{4 \sqrt {b}}}
$$
A: Not sure how helpful this is but, we can turn the problem into a second order linear differential equation:
\begin{equation}
I(a)=\int\limits_{0}^{+\infty} \frac{x^{k}e^{-ax}}{x^{2}+b^{2}}\,dx
\end{equation}
for some positive real $k$. Using Leibniz's rule, the first and second derivatives with respect to $a$ are:
\begin{equation}
I'(a)=\int\limits_{0}^{+\infty} \frac{x^{k}e^{-ax}(-x)}{x^{2}+b^{2}}\,dx
\end{equation}
\begin{equation}
I''(a)=\int\limits_{0}^{+\infty} \frac{x^{k}e^{-ax}x^{2}}{x^{2}+b^{2}}\,dx
\end{equation}
In the second derivative, add and subtract $b^{2}$ in the $x^{2}$ term:
\begin{equation}
I''(a)=\int\limits_{0}^{+\infty} \frac{x^{k}e^{-ax}(x^{2}+b^{2}-b^{2})}{x^{2}+b^{2}}\,dx
\end{equation}
\begin{equation}
I''(a)=\int\limits_{0}^{+\infty} x^{k}e^{-ax}\,dx-b^{2}\underbrace{\int\limits_{0}^{+\infty} \frac{x^{k}e^{-ax}}{x^{2}+b^{2}}\,dx}_{I(a)}
\end{equation}
\begin{equation}
I''(a)+b^{2}I(a)=\int\limits_{0}^{+\infty} x^{k}e^{-ax}\,dx
\end{equation}
With the substitution $u=ax$, one can expressed the last remaining integral in terms of the gamma function:
\begin{equation}
\int\limits_{0}^{+\infty} x^{k}e^{-ax}\,dx = \frac{\Gamma(k+1)}{a^{k+1}} = \frac{k!}{a^{k+1}}
\end{equation}
Then, in order to compute $I(a)$, we need to solve the following ODE:
\begin{equation}
I''(a)+b^{2}I(a)-\frac{k!}{a^{k+1}}=0
\end{equation}
For the integral in the question above, we have the case where $k=1/2$, then we would need to solve the following:
\begin{equation}
I''(a)+b^{2}I(a)-\frac{a^{-\frac{3}{2}}\sqrt{\pi}}{2}=0
\end{equation}
The solution to this ODE given by WolframAlpha is quite nasty: https://www.wolframalpha.com/input/?i=y%27%27%28x%29%2Bcy%28x%29-%28%5Csqrt%28%5Cpi%29%2F2x%5E%28-3%2F2%29%29%3D0.
