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.

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 $$F\big(\sigma\big)~=~\sqrt[\Large^4]{\dfrac{\sigma^2}8}\cdot\Gamma\bigg(\dfrac14\bigg)$$.

• 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? Mar 14 '15 at 16:25

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.