I am trying to find $\mathbb{E}(\text{ln}(X))^2$ where $Y=\text{ln}(X)$ and $Y\sim N(0,\sigma^2)$. $$f_Y(y)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{y^2}{2\sigma^2}}$$
Squaring a normal r.v, I am expecting to arrive at a gamma distribution.
I let $$Z=Y^2=(\text{ln}(X))^2\Rightarrow Y=\sqrt{Z} \ \ \ \ \text{for} \ \ y>0$$ Hence $$f_Z(z)=f_Y(\sqrt{z})\Bigg|\frac{dy}{dz}\Bigg|$$ $$f_Z(z)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{z}{2\sigma^2}}\frac{1}{2\sqrt{z}}$$ $$f_Z(z)=\frac{1}{\sigma 2\sqrt{2} \ \Gamma(\frac{1}{2})}e^{-\frac{z}{2\sigma^2}}z^{-\frac{1}{2}}$$ Now, removing the 2 infront of the $\sqrt{2}$ term, this looks like $$Z\sim\text{Gamma}\Big(\frac{1}{2},2\sigma^2\Big)$$ Hence, $\mathbb{E}(Z)=\sigma^2$, which is exactly what I need to prove a result for the next part of this question.
Have I made a mistake somewhere?