help deriving Beta function as a ratio of Gamma functions

Question

It comes directly from the wikipedia page on the Beta function.

$$ \begin{aligned} \Gamma(x) \Gamma(y) &= \int_{u=0}^\infty e^{-u}u^{x-1} du \;\; \cdot \;\; \int_{v=0}^\infty e^{-v}v^{y-1} dv \\ &= \int_{v=0}^\infty \int_{u=0}^\infty e^{-u - v}u^{x-1} v^{y-1} dudv \\ \end{aligned} $$

$u = zt$ and $v = z(1-t)$ means that $du = zdt$ and $dv = (1-t)dz$ I get the following after the substitution

$$ \begin{aligned} &= \int_{z=0}^\infty \int_{t=0}^1 e^{-z}(zt)^{x-1} (z(1-t))^{y-1} z (1 - t) \; dtdz \\ \end{aligned} $$

But the Wikipdedia page has it as...

$$ = \int_{z=0}^\infty\int_{t=0}^1 e^{-z} (zt)^{x-1}(z(1-t))^{y-1}z\,dt \,dz $$

So I must have gone wrong somewhere, but I cannot see my mistake. Can you find it? If I take the Wikipedia page as correct, I can get the rest of the way there, but I do not see why I am getting an extra $(1-t)$ after the substitution.

Answer 1

$u = zt$ and $v = z(1-t)$ means that $du = zdt$ and $dv = (1-t)dz$

This is your error. For multivariable integrations, you need to use the Jacobian to change the variables. $$ \begin{vmatrix} \dfrac{∂u}{∂t} & \dfrac{∂v}{∂t} \\ \dfrac{∂u}{∂z} & \dfrac{∂v}{∂z} \\ \end{vmatrix} = \begin{vmatrix} z & -z\\ t & 1-t\\ \end{vmatrix} =z $$