Show that $\frac{X}{X+Y}\sim Beta(\alpha,\beta)$ 
Let IG denote Inverse-Gamma distribution
  Inverse-Gamma.
  If $X\sim IG(\alpha,1)$ and $Y\sim IG(\beta,1)$. Show that
  $\frac{X}{X+Y}\sim Beta(\alpha,\beta)$

I tried with jacobian transformation taking
$Z=\frac{X}{X+Y}$ and $W=Y$ then $$X=\frac{WZ}{1-Z};Y=W$$
$$\frac{\partial(z,w)}{\partial(x,y)}=\frac{1}{(1-z)^2}$$
$$f_{z,w}(z,w)=f_x(\frac{wz}{1-z})f_y(w)(1-z)^2$$
$$f_{z,w}(z,w)=\frac{1}{\Gamma{(a)}}(\frac{wz}{1-z})^{-\alpha-1}e^{-\frac{(1-z)}{wz}}\frac{1}{\Gamma{(\beta)}}w^{-\beta-1}e^{-\frac{1}{w}}(1-z)^2$$
but I'm stuck, in some place I read that $\frac{X}{X+Y}$ is a type 3 Beta distribution, but I can't show that.
 A: Here is my solution from first principles:
$$P\left(z' \leq z\ \left|\ z' = \frac{x}{x+y},\ x \sim X,\ y \sim Y\right. \right) = P\left(\frac{x}{x+y} \leq z\right) = $$
$$ \int dP_{X}(x) \int dP_{Y}(y)\ I\left[ 
\frac{x}{x+y} \leq z\right] =
 \iint dx\ dy \ f_{X}(x)\ f_{Y}(y)\ \theta\left(y \frac{z} {1-z} - x\right),$$
where $\theta(x)$ is a Heaviside Step function.
Differentiating $P(z)$ by $z$ one can obtain that
$$f_{Z}(z) = \frac{d}{dz} P(z' < z) =
\int dx\ f_{X}(x) \int dy\ f_{Y}(y)\ \delta\left(y \frac{z} {1-z} - x\right) \frac{y}{(1-z)^2},$$
where $\delta(x)$ is a Dirac Delta function. So integrating over $x$ one could see that
$$f_{Z}(z) = \int dy \ f_{Y}(y)\ f_X\left(y \frac{z}{1-z}\right) \frac{y}{(1-z)^2} =$$
$$\frac{1}{\Gamma(\alpha)\Gamma(\beta)} (1-z)^{-2} \int dy\
y^{-\beta -1}e^{-\frac{1}{y}}\cdot y^{-\alpha - 1} z^{-\alpha -1} (1-z)^{\alpha+1} e^{-\frac{1-z}{yz}},$$
and after simplification and integration by $y$
$$f_{Z}(z) = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)} (1-z)^{\alpha-1} z^{\beta - 1}$$ which is probability density of beta  distribution $Beta(\beta, \alpha).$
P.S. You can avoid using of Step/Delta functions by specifying precise limits of the inner integral, but I find that way less handy.
