Are there order statistics for a Gaussian variable raised to a power? Let $X$ be a random variable with a standard normal distribution. Let $Y = |X|^{2p}$. I am trying to find the distribution for $Y_{(n)}$, i.e., the largest value of $Y$ out of $n$ samples. 
I have derived the pdf to be:
$$f_{Y_{(n)}} = n \left(\frac{1}{p\sqrt{2\pi}} y^{\frac{1}{2p} - 1} \exp\left(-\frac{1}{2}y^{1/p} \right)\right) \left(\int_0^y \frac{1}{p\sqrt{2\pi}} t^{\frac{1}{2p} - 1} \exp\left(-\frac{1}{2}t^{1/p}\right) \, dt \right)^{n-1}$$
But Mathematica says $EY_{(n)}$ is infinite. Intuitively, I feel that it should be some finite value in terms of p and n. Any ideas?

In this answer I will try to derive an analytic formula for you. I think my answer is the same as yours. Anyway, you can use it to compare your result and check what might have gone wrong.
Let $\Phi$ denote the cumulative distribution function of the standard normal distribution. That is
$$
\Phi(x) = \int_{-\infty}^{x}\frac{1}{\sqrt{2\pi}}e^{-t^2/2}dt
$$
In the following let the $X_i$ be standard normally distributed and $Y_i = |X_i|^{2p}$. Then
$$
P(Y_{(n)}\le y) = P(Y_1\le y;\cdots;Y_n\le y) = \prod_{i=1}^{n}P(Y_i\le y) =: F_Y(y)^n
$$
and thus
$$
f_{Y_{(n)}}(y) = \frac{d}{dy}F_Y(y)^n = nF_Y(y)^{n-1}f_Y(y).
$$
We conclude that if we can determine $F_Y(y)$, then we have the pdf of $Y_{(n)}$.
$$
F_Y(y) = P(Y\le y) = P(|X|^{2p}\le y) = P(-y^{1/2p}\le X \le y^{1/2p}) = 2\Phi(y^{1/2p})-1
$$
From this we see that
$$
f_Y(y) = \frac{d}{dy}2\Phi(y^{1/2p})-1 = 2f_X(y^{1/2p})\frac{1}{2p}y^{-\frac{2p-1}{2p}}.
$$
Finally we get that
\begin{align}
f_{Y_{(n)}}(y) &= n\left(2\Phi(y^{1/2p})-1\right)^{n-1}2f_X(y^{1/2p})\frac{1}{2p}y^{-\frac{2p-1}{2p}}\\
&= n\left(2\int_{-\infty}^{y^{1/2p}}\frac{1}{\sqrt{2\pi}}e^{-t^2/2}dt-1\right)^{n-1}\left(\frac{1}{\sqrt{2\pi}}e^{-y^{1/p}/2}\right)\frac{1}{p}y^{-\frac{2p-1}{2p}}
\end{align}
