I've strumble solving this integral that I obtained after a transformation. Consider $f_{X,Y}(x,y)=e^{-(x+y)}, x,y>0$. Let $V=X^2$ and $R=\frac{X}{X+Y}$. I want to get $f_{V,R}$. So $x=\sqrt{v}$ and $y=\frac{\sqrt(v)}{r}-\sqrt{v}$. The Jacobian $J=|-\frac{1}{2r^2}|=\frac{1}{2r^2}$:
$f_{V,R}(v,r)=\frac{1}{2r^2}\cdot e^{-\frac{\sqrt v}{r}}, 0<r<1, v>0$.
Now I want to calculate the integral $\int\limits_0^\infty \frac{1}{2r^2}\cdot e^{-\frac{\sqrt v}{r}} dv$, however, I can't seem to find the answer on this question.. does anyone know whether this integral is improper or not?
$$ \frac{1}{2r^2}\int_0^{\infty} \mathrm{e}^{-\frac{\sqrt{v}}{r}}dv $$ let $v = u^2$ we find $$ \frac{1}{r^2}\int_0^{\infty}u \mathrm{e}^{-\frac{u}{r}}du = \frac{1}{r^2}(-r)\frac{\partial}{\partial \alpha}\int_0^{\infty} \mathrm{e}^{-\alpha\frac{u}{r}}du =\lim_{\alpha\to 1}-\frac{1}{r}\frac{\partial}{\partial \alpha}\int_0^{\infty} \mathrm{e}^{-\alpha\frac{u}{r}}du $$ thus $$ \int_0^{\infty} \mathrm{e}^{-\alpha\frac{u}{r}}du = -\frac{r}{\alpha}\left[0-1\right]=\frac{r}{\alpha} $$ so the integral becomes $$ \lim_{\alpha\to 1}-\frac{1}{r}\frac{\partial}{\partial \alpha}\frac{r}{\alpha}=-\lim_{\alpha\to 1}\frac{-1}{\alpha^2} = 1 $$ which makes sense if it is a pdf.