Let $X,Y$ be continuous random variables with joint pdf: $$f_{XY}(x,y)=\frac{1}{x^2y^2},x \geq 1, y \geq 1$$
Two random variables $U,V$ are defined as: $$U=XY,V=\frac{X}{Y}$$ Find the marginal pdf's of $U,V$
My try:
I figured out the Jacobian of the transformation as $|J|=\frac{1}{2v}$
Since $$X=\sqrt{UV},Y=\sqrt{\frac{U}{V}}$$
We have the joint pdf of $U,V$ given by: $$f_{UV}(u,v)=|J|f_{XY}(x(u,v),y(u,v))=\frac{1}{2vu^2}$$
The bounds of $u,v$ are: $$y\geq 1 \implies \sqrt{\frac{u}{v}}\geq \implies u \geq v$$
Also: $$x \geq 1 \implies \sqrt{uv}\geq 1\implies v \geq \frac{1}{u}$$
So the joint pdf of $U,V$ is given by: $$f(u,v)=\frac{1}{2vu^2},\:u \geq 1, \frac{1}{u} \leq v \leq u$$
Now the marginal PDF of $U$ is: $$f(u)=\int_{v=\frac{1}{u}}^{u} \frac{1}{2 u^{2}} \frac{1}{v} d v=\frac{\ln u}{u^2}, u \geq 1$$
But i am unable to find PDF of $V$
$V$ takes all positive values. For $0<v<1$, $uv >1 $ automatically gives $u >v$ also and $f_V(v)=\int_{1/v}^{\infty} \frac 1 {2vu^{2}}du=\frac 1 2$. Similarly, for $v >1$ we get $f_V(v)=\int_v^{\infty}\frac 1 {2vu^{2}}du=\frac 1 {2v^{2}}$.