I have this equality of integrals, and I don't know how to prove this. It seems that a change of variables is necessary, but I don't know how to start. Please, can you help me?
$$\int_{a}^{\infty}z^{N-s-1}(M+z-1)^{s-MN}dz = \int_0^{\frac{1-M}{a}}z^{MN-N}(1-z)^{s-MN+1}dz$$
where $a,z\in\mathbb{R}, s\in\mathbb{C}$, and $M,N\in\mathbb{Z}$. The right part of the above equation is from the Incomplete Beta Function.
A change of variables is how to start. To see which, look to the bounds of the integrals.
You want to map from $(a..\infty)$ to $(0..(1-M)/a)$.
So substitute $B:=(1-M)$ , $z_{\tiny 0}:=Bz^{-1}$ and $\mathrm d z_{\tiny 0}:=-Bz^{-2}\,\mathrm d z$.
$${\quad\int_a^\infty z_{\tiny 0}^{N-s-1}(M+z_{\tiny 0}-1)^{s-MN}\mathrm dz_{\tiny 0} \\= -\int_{B/a}^0 (Bz^{-1})^{N-s-1}(Bz^{-1}-B)^{s-MN}Bz^{-2}\,\mathrm d z\\=B^{N-MN}\int_0^{B/a}z^{s-N-1}(z^{-1}-1)^{s-MN}\,\mathrm d z\\\\=B^{N-MN}\int_0^{B/a}z^{MN-N-1}(1-z)^{s-MN}\,\mathrm d z\\~\\=(1-M)^{N-MN}\int_0^{(1-M)/a}z^{MN-N-1}(1-z)^{s-MN}\,\mathrm d z}$$
Next task: integration by parts.