# Solving integral with exponential function

I’m interested in solving the integral $$\int\limits_{0}^{\infty} {\frac{2}{3x^{4/3}} \sqrt{\frac{\lambda}{2\pi}} \exp \left( \frac{-\lambda (x^{2/3}-\mu)^2}{2\mu^2 x^{2/3}} \right) dx}$$ for $\mu>0, \lambda>0$. WolframAlpha does not give me a solution but maybe you know some tricks how to simplify the integral. Thanks!

• have you proved that your integral does converge on the given interval? – Dr. Sonnhard Graubner May 24 '17 at 19:32

First, do the change of variables $x=u^{3/2}$. That will turn your integral into $$I=\sqrt{\frac{\lambda}{2\pi}}\int_0^{+\infty}\frac{1}{u^{3/2}}\exp\biggl(-\frac{\lambda(u-\mu)^2}{2\mu^2u}\biggr)\,du.$$ Next, performing $u\mapsto \mu^2/u$, you will find that $$I=\sqrt{\frac{\lambda}{2\pi}}\int_0^{+\infty}\frac{1}{\mu u^{1/2}}\exp\biggl(-\frac{\lambda(u-\mu)^2}{2\mu^2u}\biggr)\,du.$$ Adding these, you get $$2I=\sqrt{\frac{\lambda}{2\pi}}\int_0^{+\infty}\frac{\mu+u}{\mu u^{3/2}}\exp\biggl(-\frac{\lambda(u-\mu)^2}{2\mu^2u}\biggr)\,du.$$ Now, you are lucky, since $$\frac{d}{du}\frac{2(u-\mu)}{\mu\sqrt{u}}=\frac{\mu+u}{\mu u^{3/2}}.$$ Hence, set $$t=\frac{2(u-\mu)}{\mu\sqrt{u}}.$$ You find that $$I=\frac{1}{2}\sqrt{\frac{\lambda}{2\pi}}\int_{-\infty}^{+\infty}\exp\biggl(-\frac{\lambda t^2}{8}\biggr)\,du.$$ We are back in our (mine!) comfort zone with the gaussians, and since $$\int_{-\infty}^{+\infty}e^{-t^2}\,dt=\sqrt{\pi}$$ we use scaling to conclude that
$$I=1.$$