How would you calculate $E[\mid x \mid ^{\alpha }], \alpha \in \Re$? Here $x \sim N(0,1)$.
I realize that the expectation won't be defined for $\alpha$ when the integral goes to infinity. I can't seem to figure out which specific values of $\alpha$ would cause this. 
My intuition is that it would have to be negative values of alpha, as I have already derived the values of 
\begin{align}
E[x^{2n+1}] &= 0  \\[5pt]
E[x^{2n}] &= \frac{2n!}{n!2^n} 
\end{align}
 A: Because the distribution is symmetric about zero, with PDF $\exp(-x^2/2)/\sqrt{2\pi}$, the expectation is just twice the integral over the positive numbers
$$\mathbb{E}(|x|^\alpha) = 2 \frac{1}{\sqrt{2\pi}}\int_0^\infty x^\alpha \exp(-x^2/2)\, \text{d}x.$$
The one-to-one transformation $y = x^2/2$ entails $\text{d}y=x\, \text{d}x$ whence $$\text{d}x=\frac{\text{d}y}{x} = \frac{\text{d}y}{\sqrt{2y}},$$ converting the integral into
$$\frac{2}{\sqrt{2\pi}}\int_0^\infty (2y)^{\alpha/2}e^{-y} \frac{\text{d}y}{\sqrt{2 y}}=\frac{2^{\alpha/2}}{\sqrt{\pi}}\int_0^\infty y^{(1+\alpha)/2-1}e^{-y}\text{d}y=\frac{2^{\alpha/2}}{\sqrt{\pi}}\Gamma\left(\frac{1+\alpha}{2}\right)$$
provided $\mathfrak{R}(\alpha) \gt -1$.  When the real part of $\alpha$ is $-1$ or less, the integrand between $0$ to $1$ is bounded below by $y^{-1}e^{-1/2}$.  This integral diverges logarithmically at $0$, showing the expectation does not exist in such cases.  It is noteworthy that negative real values of $\alpha$ between $-1$ and $0$ still lead to finite expectations.
