I am not sure with the solution of the integral

Question

I am having trouble with this integral. I know that the limits of the integral should be from $0$ to $v$ since we need to find the marginal $fv(v)$ and $0<u<v$ and then we put all the variables that are not $U$ in front of the integral. And from that point I don't know what to do.

Answer 1

You almost done.

$f_V(v)=\frac{e^{-v/2}}{2\pi}\int^v_0\frac{1}{\sqrt{u(v-u)}}du$.

The integral part is

$\int^v_0\frac{1}{\sqrt{u(v-u)}}du=\int^{v/v}_0\frac{1}{\sqrt{\frac{u}{v}(1-\frac{u}{v})}}d(\frac{u}{v})=\int^1_0\frac{1}{\sqrt{t(1-t)}}dt=\pi$.

where $t=\frac{u}{v}$. So, the final result is

$f_V(v)=\frac{e^{-v/2}}{2}$.