How to prove that


$$\int (1-z)^{u} z^{v} dz = \frac{z^{v+1}}{v+1} \, \_2F\_1(-u, v+1; v+2; z)?$$



How to prove that $\int (1-z)^{u} z^{v} dz$ is equal to $\frac{z^{v+1}}{v+1}_2F_1(-u, v+1; v+2; z)$?
The generalized binomial coefficients are
$$
\binom{u}{n} := \frac{(-1)^n(-u)\_n}{n!} ,
$$
here we used the notation $(-u)\_n$ for the rising factorial: $(-u)\_n := \prod\_{k=0}^{n-1}(-u+k)$.


Taylor's expansion of $(1-z)^u$ around $z=0$ is
$$
(1-z)^u = \sum\_{n=0}^\infty \binom{u}{v}(-z)^n = \sum\_{n=0}^\infty \frac{(-u)\_n}{n!} z^n ,\quad |z| < 1.
$$


Thus we have
$$\int (1-z)^{u} z^{v} dz=
\sum\_{n=0}^{\infty}\frac{(-u)\_{n}}{n!}\int z^{n+v}dz = z^{v+1}\sum\_{n=0}^{\infty}\frac{(-u)\_{n}}{n! (n+v+1)}z^{n}.$$


Next, $\frac{1}{n+v+1}$ can be expressed as
$$\frac{1}{n+v+1}=\frac{\Gamma(n+v+1)}{\Gamma(n+v+2)} = \frac{\Gamma(v+1)(v+1)\_n}{\Gamma(v+2)(v+2)\_n} = \frac{(v+1)\_n}{(v+1)(v+2)\_n}.$$


Then
$$\int (1-z)^{u} z^{v} dz=
\frac{z^{v+1}}{v+1}\sum\_{n=0}^{\infty}\frac{(-u)\_{n}(v+1)\_n}{n!(v+2)\_n}z^{n}=\frac{z^{v+1}}{v+1}\,\_2F\_1(-u,v+1;v+2;z).$$


