Say X ~ $\mathcal{N}(0,1)$ what would be the CDF of Y=X^a (where a is an integer). I have seen how to get the PDF which would theoretically give the CDF but I'm not sure how to get a closed form solution, say using erf() like the standard normal cdf form.
Reference: Distribution of higher powers than 2 of a gaussian distribution
If a is a positive odd integer, then $Y=X^a$ has a symmetric distribution. Thus, for the complementary CDF (CCDF) of $Y$, we obtain: $$\overline F_Y(x) = P(X^a>x) = \overline \Phi(x^{1/a}),$$ for all $x>0$, where $\overline \Phi(z) = \frac{1}{\sqrt{2\pi}}\int_z^\infty e^{-u^2/2}du$ is the CCDF of the standard Normal distribution. Or, for the CDF: $$ F_Y(x) = \left\{ \begin{array}{ll} \Phi(x^{1/a})&, x>0\\ 1-\Phi(|x|^{1/a})&, x\le 0 \end{array}\right., $$ Where $\Phi(x) = 1-\overline\Phi(x)$ is the standard Normal CDF.
This solves half of the problem. For $a$ even, $Y$ is positive and for all $x>0$, we have $$ \overline F_Y(x) = P(|X|> x^{1/a}) = 2\overline\Phi(x^{1/a}). $$ Which gives $$ F_Y(x) = 2 \Phi(x^{1/a}) - 1,\ \ x\ge 0. $$
