"fractional" expectation of zero-mean normal distribution

Question

I'm trying to calculate $E[X^{\frac{2}{3} } ] $ of a zero-mean normal distribution. Any help to solve $$ E[X^{\frac{2}{3} } ] = \int\limits_{-\infty}^{\infty} x^{\frac{2}{3} } \frac{1}{\sigma \sqrt{2\,\pi}}\,\mathrm e^{ -\frac{x^2}{2\sigma^2} } \,\mathrm d x $$ will be appreciated.

Answer 1

Write it as $\displaystyle\dfrac{2}{\sigma\sqrt{2\pi}}\int_{0}^{\infty}x^{\tfrac{2}{3}}e^{-\tfrac{x^2}{2\sigma^2}}\,dx$.

Then, substitute $u = \dfrac{x^2}{2\sigma^2}$ and do some simplification to get:

$\left(\dfrac{4\sigma^4}{\pi^3}\right)^{1/6}\displaystyle\int_{0}^{\infty}u^{-1/6}e^{-u}\,du$ $= \left(\dfrac{4\sigma^4}{\pi^3}\right)^{1/6}\Gamma\left(\dfrac{5}{6}\right)$,

where $\Gamma()$ denotes the Gamma Function.

I don't think there is a nicer form for this value.