Find the p.d.f of U=X/Y Known the p.d.fs of X,Y respectively are
$$f(x)=\frac{2}{x^3}\hspace{1em},\hspace{1em}1<x<\infty \\
g(y)=\frac{3}{y^4}\hspace{1em},\hspace{1em}1<y<\infty$$
What is the p.d.f of X/Y ?
I tried change-of-variable technique, but I found something wrong in the support.
Let $$U=\frac{X}{Y}\hspace{2em}V=Y$$
then have $$x=uv\hspace{1em},\hspace{1em}y=v$$
the Jacobian is $$J=
\left |
\begin{matrix}
\frac{\partial x(u,v)}{\partial u} & \frac{\partial x(u,v)}{\partial v}\\
\frac{\partial y(u,v)}{\partial u} & \frac{\partial x(u,v)}{\partial v}
\end{matrix}
\right |
=\left |\begin{matrix}v & u\\
0 & 1\end{matrix}\right |
=v
$$
so the joint p.d.f of U,V is $$h(u,v)=v\cdot f(uv)g(v)=\frac{6}{u^3v^6}\hspace{1em},\hspace{1em}0<u<\infty\hspace{1em},\hspace{1em}1<y<\infty$$
marginal p.d.f of U is $$h_1(u)=\int_1^\infty \frac{6}{u^3v^6}dv=\frac{6}{5u^3}\hspace{1em},\hspace{1em}0<u<\infty$$
Till this step, I am confused about the support of U, because of $$P(0<U<\infty)=\int_0^\infty \frac{6}{5u^3}du$$
tends to infinity.
I think I made some mistakes but can't find.

Hint: Assuming independence,  $P(\frac X Y \leq t)=\int P(X\leq yt) f_Y(y)dy$ and $P(X\leq yt)=\int_1^{yt} \frac 2 {x^{3}} dx$ if $yt>1$ and $0$ if $yt <1$. When you integrate over $y$ you have to integrate over $y$ such that $y>1$ and $y >\frac  1t$ You will have to consider the case $t>1$ and $t <1$ separately. I will let you compute the integrals and then differentiate the answer to get the density function of $\frac X Y$.
