Jacobian Transformations of Two Random Variables ( probability density function) Let ${Y}_{1}$ and ${Y}_{2}$ be independent,each with density $\frac{1}{{y}^{2}}$ , $y>1$. Consider the transformation ${U}_{1}=\frac{{Y}_{2}}{{Y}_{1}+{Y}_{2}}$ and ${U}_{2}={Y}_{1}+{Y}_{2}$. Find probability density function of random variables $\left({U}_{1},{U}_{2}\right)$.
Here is my solution:
Since ${Y}_{1}$ and ${Y}_{2}$ are independent then $f({y}_{1},{y}_{2})=\frac{1}{{\left({y}_{1}{y}_{2}\right)}^{2}}$ , ${y}_{1}>1$ , ${y}_{2}>1$.
${U}_{1}=\frac{{Y}_{2}}{{Y}_{1}+{Y}_{2}}$ and ${U}_{2}={Y}_{1}+{Y}_{2}$ then ${Y}_{2}={U}_{1}{U}_{2}$ , ${Y}_{1}={U}_{2}-{U}_{1}{U}_{2}$
From here Jacobian is $J=\left|\begin{array}{cc}{u}_{2}& {u}_{1}\\ -{u}_{2}& 1-{u}_{1}\end{array}\right|={u}_{2}\Rightarrow \left|J\right|={u}_{2}$
Then I stuck here because I found the density function $f({u}_{1},{u}_{2})=\frac{{u}_{2}}{{\left({u}_{2}-{u}_{1}{u}_{2}\right)}^{2}{\left({u}_{1}{u}_{2}\right)}^{2}}$  but when I want to check it this function with integral definition I can not define the interval of integral. Any help will be appreciated.

When defining joint support like this, you need to consider to specify the marginal support of which variable first. If you specify $u_1$ first, since
$$ u_1 = \frac {y_2} {y_1 + y_2} $$
then we see that
when $y_1$ fixed, $y_2 \to +\infty$, $u_1 \to 1$;
when $y_1 \to +\infty$, $y_2$ fixed, $u_1 \to 0$.
And obviously $0 < u_1 < 1$ so the support of $U_1$ is indeed $(0, 1)$
Let say we specify $u_1$ first. Then the "conditional" support of $u_2$ will be in terms of $u_1$. Since
$$ \begin{cases} u_2 - u_1u_2 &= y_1 > 1 \\ u_1u_2 &= y_2 > 1 \end{cases}
\Rightarrow \begin{cases} u_2 &> \displaystyle  \frac {1} {1 - u_1} \\  u_2 &> \displaystyle \frac {1} {u_1} \end{cases} $$
So the joint support can be specified as
$$ 0 < u_1 < 1, u_2 > \max\left\{\frac {1} {1-u_1}, \frac {1} {u_1} \right\}$$
Similarly if we want to specify $u_2$ first, obviously
$$u_2 = y_1 + y_2 > 1 + 1 = 2$$
so the marginal support of $U_2$ is $(2, +\infty)$. And
$$ \begin{cases} u_2 - u_1u_2 &= y_1 > 1 \\ u_1u_2 &= y_2 > 1 \end{cases}
\Rightarrow \begin{cases} u_1 &< \displaystyle 1 - \frac {1} {u_2} \\ u_1 &> \displaystyle \frac {1} {u_2} \end{cases} $$
So the joint support can be equivalently specified as
$$ u_2 > 2, \frac {1} {u_2} < u_1 < 1 - \frac {1} {u_2} $$
Verification:
Note
$$ \int \frac {1} {u^2(1-u)^2} du = \frac {1} {1-u} - \frac {1} {u} 
- 2\ln (1-u) + 2\ln u + C$$
Therefore
$$ \begin{align} 
&~ \int_{1/u_2}^{1-1/u_2} \frac {1} {u_1^2(1-u_1)^2u_2^3} du_1 \\
=&~ \frac {1} {1-1+1/u_2} - \frac {1} {1-1/u_2} 
- 2\ln \left(1-1+\frac {1} {u_2}\right) 
+ 2\ln\left(1- \frac {1} {u_2} \right) \\
&~ - \frac {1} {1-1/u_2} + \frac {1} {1/u_2} 
+ 2\ln \left(1-\frac {1} {u_2}\right) 
- 2\ln\left(\frac {1} {u_2} \right) \\
=&~ 2u_2 - \frac {2u_2} {u_2-1} + 4\ln(u_2-1)\end{align} $$
As a result,
$$ \begin{align} 
&~ \int_2^{+\infty} \int_{1/u_2}^{1-1/u_2} \frac {1} {u_1^2(1-u_1)^2u_2^3} du_1du_2 \\
=&~ \int_2^{+\infty} \left[\frac {2} {u_2^2} - \frac {2} {u_2^2(u_2-1)} + \frac {4\ln(u_2-1)} {u_2^3} \right] du_2 \\ 
=&~ \left. - \frac {2\ln(u_2-1)} {u_2^2} - \frac {2} {u_2}  \right|_2^{+\infty} \\
=&~ 1 \end{align} $$
