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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!

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  • $\begingroup$ have you proved that your integral does converge on the given interval? $\endgroup$ – Dr. Sonnhard Graubner May 24 '17 at 19:32
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I give you some steps:

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

$$I=1.$$

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    $\begingroup$ +1. Pretty fine. It'll be nice the OP 'remove' all of the annoying factors before he/she sets the question !!!. $\endgroup$ – Felix Marin May 24 '17 at 23:24
  • $\begingroup$ Nice solution!! Thanks a lot! And next time I'll remove the factors. $\endgroup$ – Takashi May 25 '17 at 8:43

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