How do you change of variables in integrals (i.e. dxdy into dudv)?

Question

I was looking around the internet for a simple(r) proof of the Gamma Reflection Formula. I found this: Detailed explanation of the Γ reflection formula understandable by an AP Calculus student, and did not understand how they got from the dxdy integral to a dudv integral:

To quote: "$$\Gamma(z)\Gamma(1-z) = \int_{0}^{\infty}\int_{0}^{\infty}s^{-z}t^{z-1}e^{-(s+t)}\,ds\, dt \qquad (0<\Re z <1)$$

With the new variables $u = s + t, v = t/s$ this gives $$\Gamma(z)\Gamma(1-z) = \int_{0}^{\infty}\int_{0}^{\infty}\frac{v^{z-1}}{1+v}e^{-u}\,du\, dv $$"

Can anyone help me? Explanations understandable by an AP Calculus student would be great! If this needs stuff that AP Calculus doesn't cover, explanations of how those work and why or references to other sources would be excellent.

Answer 1

\begin{eqnarray*} dudv= det \left[ \begin{array}{cc} \frac{\partial u}{\partial x} & \frac{\partial v}{\partial x} \\ \frac{\partial u}{\partial y} & \frac{\partial v}{\partial y} \\ \end{array} \right]dxdy \end{eqnarray*}

Answer 2

The multiplication factor is the Jacobian

\begin{eqnarray*} du\, dv = J (x,y,u,v) =Det \left[ \begin{array}{cc} \dfrac{\partial u}{\partial x} & \dfrac{\partial v}{\partial x} \\ \dfrac{\partial u}{\partial y} & \dfrac{\partial v}{\partial y} \\ \end{array} \right]dx\,dy \end{eqnarray*}

For example from metric $ ds^2 = E\, du^2 + 2 F \,du \,dv + G\, dv^2 $ differential area transforms by a factor $ \sqrt {EG-F^2. }$