The square of a standard Normal random variable I am having a bit of trouble with this: 
Let $U=Z^2$ where Z is a standard Normal random variable with pdf: 
$$f_z(z) = \frac{1}{\sqrt{2\pi}} e^{\frac{-z^2}{2}}$$
I want to use the inversion method but have thus far only learned to use this when functions are strictly increasing or decreasing. Since a standard normal distribution function is strictly increasing and then strictly decreasing I thought perhaps I could find some way to use this method. 
I have the final answer as $f_u(u)= \frac{1}{\sqrt{2\pi}\sqrt{u}}e^{\frac{-u}{2}}$ for $u>0$
But I am not very comfortable with the process of getting to that answer. I used a bit of a walk through and made some assumptions about what was happening. 
Could anyone help me understand how I should use the above information to reach this answer? 
 A: Here is a way to get the answer using more basic principles.
$$f_Z(z) = \frac{1}{\sqrt{2\pi}} e^{-z^2/2}, \quad -\infty < z < \infty.$$  This much you already know.  Then $$\begin{align*} F_U(u) &= \Pr[U \le u] \\ &= \Pr[Z^2 \le u] \\ &= \Pr[-\sqrt{u} \le Z \le \sqrt{u}] \\ &= \Pr[Z \le \sqrt{u}] - \Pr[Z \le -\sqrt{u}] \\ &= F_Z(\sqrt{u}) - F_Z(-\sqrt{u}), \end{align*}$$  where $F_Z(z)$ is the cumulative distribution function for $Z$.  Now differentiate with respect to $u$, taking care to use the chain rule:  $$f_U(u) = \frac{f_Z(\sqrt{u})}{2\sqrt{u}} - \frac{f_Z(-\sqrt{u})}{-2\sqrt{u}} = \frac{f_Z(\sqrt{u}) + f_Z(-\sqrt{u})}{2\sqrt{u}}.$$  Now substitute back into the density for $Z$: $$f_U(u) = \frac{1}{\sqrt{2\pi}}\frac{2e^{-u/2}}{2\sqrt{u}} = \frac{e^{-u/2}}{\sqrt{2\pi u}}, \quad u > 0. $$  The distribution of $U$ is known as the chi-square distribution with $1$ degree of freedom.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
\int_{-\infty}^{\infty}
\frac{\expo{-z^{2}/2}}{\root{2\pi}}\delta\pars{u - z^{2}}\,\dd z}
=2\Theta\pars{u}\,\frac{\expo{-u/2}}{\root{2\pi}}\int_{0}^{\infty}
\frac{\delta\pars{z - \root{u}}}{2\root{u}}\,\dd z
\\[5mm]&=\color{#66f}{\large\Theta\pars{u}\,\frac{u^{-1/2}\ \expo{-u/2}}{\root{2\pi}}}
\end{align}
