I encountered the following "claim" in a probability exercise solution (find the distribution of $Y=X^2$ where $X\sim N(0,2)$
$$2\int_{0}^{\sqrt{t}}\frac{1}{\sqrt{8\pi}}e^{-\frac{x^2}{8}}dx=\int_{0}^{t}\frac{1}{\sqrt{8\pi}}e^{-\frac{y}{8}}\sqrt{y}dy$$
The change of variable itself was not specified.
According to my understanding, the possible change of variable they did here is $y=x^2$ but then $dx=\frac{dy}{2x}=\frac{dy}{2\sqrt{y}}$, that also makes sense with the limit because when $x\rightarrow0$ then $y\rightarrow 0$, and when $x \rightarrow \sqrt{t}$ then $y \rightarrow t$, but the integral itself doesn't make sense to me.
Is that a typo in the solution, or am I completly missing something here?
Yes, you are correct, there is a typo. By letting $x=\sqrt{y}$ then $dx=\frac{dy}{2\sqrt{y}}$, and, for $t\geq 0$, $$2\int_{0}^{\sqrt{t}}\frac{1}{\sqrt{8\pi}}e^{-\frac{x^2}{8}}\,dx=2\int_{0}^{t}\frac{1}{\sqrt{8\pi}}e^{-\frac{y}{8}}\,\frac{dy}{2\sqrt{y}}=\int_{0}^{t}\frac{1}{\sqrt{8\pi}}e^{-\frac{y}{8}}\,\frac{dy}{\sqrt{y}}.$$