Find $ \int_0^\infty \frac{\sqrt x e^{-x}}{b^2 +x^2} dx $

Question

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.

Answer 1

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: $$ -{\frac {\pi\,\sqrt {2} \left( \left( 1+i \right) {{\rm e}^{-ib} }{\rm erf} \left( \left( 1/2-i/2 \right) \sqrt {2}\sqrt {b}\right)+ \left( 1-i \right) {{\rm e}^{ib}}{\rm erf} \left( \left( 1/2+i/2 \right) \sqrt {2}\sqrt {b}\right)- \left( 1+i \right) {{\rm e}^{-ib}} - \left( 1-i \right) {{\rm e}^{ib}} \right) }{4 \sqrt {b}}} $$

Answer 2

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.