Integral of gaussian distribution divided by square root of x I, maxima and WolframAlpha are struggling to evaluate the following integral:
$$
\int_{0}^{\infty} {x^{-\frac{1}{2}}\exp{\left(-\dfrac{(x-\mu)^2}{2\,\sigma^2}\right)}}dx
$$
There should be a probability distribution with wich one could model this function and use its normalization for integration but I could not find it either.
 A: 
I, Maxima and Wolfram Alpha are struggling to evaluate the following integral.

Wow ! No wonder you're struggling ! Apparently, $$F\Big(\mu~,~\sigma\Big)~=~\sqrt{-\frac\mu2}\cdot\exp\bigg[-\bigg(\frac\mu{2~\sigma}\bigg)^2~\bigg]\cdot K_{\tfrac14}\bigg[\bigg(\frac\mu{2~\sigma}\bigg)^2~\bigg],$$ for $\color{blue}{\mu<0}$, and $$F\Big(\mu~,~\sigma\Big)~=~\frac\pi2~\sqrt\mu\cdot\exp\bigg[-\bigg(\frac\mu{2~\sigma}\bigg)^2~\bigg]\cdot \bigg\{~I_{\tfrac14}\bigg[\bigg(\frac\mu{2~\sigma}\bigg)^2~\bigg]~+~I_{-\tfrac14}\bigg[\bigg(\frac\mu{2~\sigma}\bigg)^2~\bigg]~\bigg\}~,$$ for $\color{blue}{\mu>0}$, where I and K are the Bessel functions. For $\mu=0$ we have $F\big(\sigma\big)~=~\sqrt[\Large^4]{\dfrac{\sigma^2}8}\cdot\Gamma\bigg(\dfrac14\bigg)$.
A: 
Thanks a lot - in the mean time I was able to more or less guess the solution for μ > 0 and could confirm it by numerical integration. How did you manage to do it? 

For $\mu<0$, just set 
$$x=-2\mu\sinh^2\frac{t}{4},$$
with $0\leq t\leq \infty$, and combine with the integral representation of the modified Bessel function of the second kind
$$K_\nu(x)=\int_0^\infty e^{-x\cosh{t}}\cosh{(\nu t)}dt,\quad Re(x)>0.$$
For $\mu>0$ break the integral into three parts
$$F(\mu,\sigma)=I_1+I_2+I_3$$
where
$$I_1=\int_0^\mu x^{-1/2}\exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right]dx$$
$$I_2=\int_\mu^{2\mu} x^{-1/2}\exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right]dx$$
$$I_3=\int_{2\mu}^\infty x^{-1/2}\exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right]dx,$$
and set
$$x=\mu(1-\sin\frac{\theta}{2})$$
with $0\leq\theta\leq\pi$, for $I_1$,
$$x=\mu(1+\sin\frac{\theta}{2})$$
with $0\leq\theta\leq\pi$, for $I_2$,
$$x=2\mu\cosh^2\frac{t}{4}$$
with $0\leq t\leq\infty$, for $I_3$.
For $\mu=0$, set $x=\sqrt{2}\sigma y$ and use the definition for the gamma function.
