Ratio of Gamma random variables If $X_i$, $i=1,2$ are independent gamma$(\alpha_i,1)$ random variables, find the distribution of $\frac{X_1}{X_1+X_2}$ and $\frac{X_2}{X_1+X_2}$.
Attempt:
Let $Y_1 = \frac{X_1}{X_1+X_2}$ and $Y_2=X_1+X_2$.
Then $X_1 = Y_1Y_2$ and $X_2 = Y_2-Y_1Y_2$.
The Jacobian matrix is :
$\hspace{15mm}J = \left| \begin{array}{ccc}
y_2 & y_1 \\
-y_2& 1-y_1 \\
 \end{array} \right| = y_2$
I'm having trouble plugging the values back in to the gamma equation. Since we want to know the distribution of $Y_1$, do we first find the joint distribution of $f_{y_1,y_2}(y_1,y_2)$ and then integrate $y_2$ out?

We have that $X_1$ and $X_2$ are independent $\Gamma(\alpha_i,1)$ random variables. Note that $X_1$ and $X_2$ are non-negative with values in $[0,\infty).$
Given the transformation, the joint density of $Y_1$ and $Y_2$ is:
$$ f_{Y_1Y_2}(y_1,y_2)=f_{X_1X_2}[X_1(y_1,y_2),X_2(y_1,y_2)]|J|= f_{X_1}(y_1y_2)f_{X_2}(y_2-y_1y_2)y_2= \frac1{\Gamma(\alpha_1)}[y_1y_2]^{\alpha_1-1}e^{-y_1y_2}\frac1{\Gamma(\alpha_2)}[y_2-y_1y_2]^{\alpha_2-1}e^{-y_2+y_1y_2}y_2=\frac1{\Gamma(\alpha_1)\Gamma(\alpha_2)}y_1^{\alpha_1-1}(1-y_1)^{\alpha_2-1}y_2^{\alpha_1+\alpha_2-1}e^{-y_2}$$
The marginal density of $Y_1$ is 
$$f_{Y_1}(y_1) = \frac{y_1^{\alpha_1-1}(1-y_1)^{\alpha_2-1}}{\Gamma(\alpha_1)\Gamma(\alpha_2)}\int_{0}^{\infty}y_2^{\alpha_1+\alpha_2-1}e^{-y_2}dy_2=\frac{\Gamma(\alpha_1+\alpha_2)}{\Gamma(\alpha_1)\Gamma(\alpha_2)}y_1^{\alpha_1-1}(1-y_1)^{\alpha_2-1}$$
