In the derivation of the beta function the following proof is given:
$\begin{align} \Gamma(x)\Gamma(y) &= \int_{u=0}^\infty\ e^{-u} u^{x-1}\,du \cdot\int_{v=0}^\infty\ e^{-v} v^{y-1}\,dv \\[6pt] &=\int_{v=0}^\infty\int_{u=0}^\infty\ e^{-u-v} u^{x-1}v^{y-1}\,du \,dv. \end{align}$
Changing variables by u=zt and v=z(1-t) shows that this is
$\begin{align} \Gamma(x)\Gamma(y) &= \int_{z=0}^\infty\int_{t=0}^\color{red}{1} e^{-z} (zt)^{x-1}(z(1-t))^{y-1}\big|J(z,t)\big|\,dt \,dz \tag{1}\\[6pt] &= \int_{z=0}^\infty\int_{t=0}^\color{red}{1} e^{-z} (zt)^{x-1}(z(1-t))^{y-1}z\,dt \,dz \\[6pt] &= \int_{z=0}^\infty e^{-z}z^{x+y-1} \,dz\cdot\int_{t=0}^\color{red}{1}t^{x-1}(1-t)^{y-1}\,dt\\ &=\Gamma(x+y)\,\beta(x,y), \end{align}$
How was the upper limit of integration calculated after the change of variables. I thought I understood the Fubini-Tonnelli theorem but I don't understand how the value of 1 was calculated. Why is there no $\frac{1}{z}$ dependence in the upper limit of (1)?
In a single dimension its easy to get the limits 0,1 using u=1/x-1 but in the two dimensional plane I'm confused by this substitution.
The easiest way to see what the limits of integration should be is to write out the inverse transformations. In this case, we have $$ u + v = zt + z(1 - t) = z \quad \Rightarrow \quad z = u + v $$ and $$\frac{v}{u} = \frac{1 - t}{t} = \frac{1}{t} - 1 \quad \Rightarrow \quad t = \frac{1}{v/u + 1}. $$
We then need to look at the range of values these quantities can take on when $u \in (0, \infty)$ and $v \in (0, \infty)$. For $z$, $0 < u < \infty$ and $0 < v < \infty$ imply that $0 < z < \infty$. The values of $t$ are a little more subtle. The ratio $v/u$ also satisfies $0 < v/u < \infty$; in the limit $v/u \to 0$, we have $t \to 1$, while in the limit $v/u \to \infty$ we have $t \to 0$. Since this is an invertible one-to-one coordinate transformation, we can conclude that the limits of integration under this change of variables are $z \in (0, \infty)$ and $t \in (0,1)$.