$E(x^k)$ under a Gaussian What would be the expectation of $|x|^k$, where $x\sim\mathcal{N}(0,1)$, $k>0$ and $k$ is not an integer?
 A: $E(x^k)$ can be worked out directly from the law of the unconscious statistician
$E(x^k) = \int_{-\infty}^\infty x^k \phi(x) dx$ where $\phi$ is the standard normal pdf
You may be able to make progress with a simple substitution.
See also:
http://mathworld.wolfram.com/NormalDistribution.html (this gives the numeric answers)
and
http://mathworld.wolfram.com/GaussianIntegral.html

Responding to the updated question: we just follow my suggestion above (simple substitution).
\begin{eqnarray}
E(|x|^k) &=& \int_{-\infty}^\infty |x|^k \frac{1}{\sqrt{2\pi}} e^{-x^2/2}dx\\
         &=& 2\frac{1}{\sqrt{2\pi}} \int_{0}^\infty x^k e^{-x^2/2}dx
\end{eqnarray}
Let $u = x^2/2;\,$ so $ du  = x\,dx;\,x=(2u)^{1/2}$
\begin{eqnarray}
   &=& 2\frac{1}{\sqrt{2\pi}} \int_{0}^\infty (2u)^{\frac{k-1}{2}} e^{-u}du\\
   &=& \frac{2^{k/2}}{\sqrt{\pi}} \int_{0}^\infty u^{\frac{k-1}{2}} e^{-u}du\\
   &=& \frac{2^{k/2}}{\sqrt{\pi}} \Gamma\left(\frac{k+1}{2}\right)
\end{eqnarray}
In fact the second link above shows you how to it for the non-integer case - see eqns (9)-(12); there's nothing there requiring the power to be an integer.
