3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$ Mar 14 '15 at 16:25
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.