On the Summation of Hypergeometric series

Since, $_2F_1(a,b,c,z)=1+\frac{ab}{c}z+\frac{a(a+1)b(b+1)}{c(c+1)}\frac{z^2}{2!}+\frac{a(a+1)(a+2)b(b+1)(b+2)}{c(c+1)(c+2)}\frac{z^3}{3!}+...$ Then using the following result by Sir Leonard Euler, $\Beta(b,c-b)_2F_1(a,b,c,z)=\int_0^1x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a}dx$ We will convert it as follows by saying$a=-a$ , dividing $-1$ and $a$ to both sides, $\frac{-\Beta(b,c-b)_2F_1(-a,b,c,z)}{a}=\frac{(-1)^{a-1}}{a}\int_0^1x^{b-1}(1-x)^{c-b-1}(zx-1)^{a}dx$ $-\Beta(b,c-b)\sum_{a=1}^{\infty}\frac{_2F_1(-a,b,c,z)}{a}=\int_0^1x^{b-1}(1-x)^{c-b-1}\sum_{a=1}^{\infty}\frac{(-1)^{a-1}(zx-1)^a}{a}dx$ Applying the Taylor series of the logarithm,we get, $-\Beta(b,c-b)\sum_{a=1}^{\infty}\frac{_2F_1(-a,b,c,z)}{a}=\int_0^1x^{b-1}(1-x)^{c-b-1}\ln (zx)dx$ Splitting the integral part we yield, $-\Beta(b,c-b)\sum_{a=1}^{\infty}\frac{_2F_1(-a,b,c,z)}{a}=\int_0^1x^{b-1}(1-x)^{c-b-1}\ln (x)dx+\ln (z)\int_0^1x^{b-1}(1-x)^{c-b-1}dx$ Then realising that the integral with $\ln (z)$ is simply $\Beta(b,c-b)$ but, $\int_0^1x^{b-1}(1-x)^{c-b-1}\ln (x)dx=\Beta(b,c-b)(\psi(b)-\psi(c))$ I'll show this result in the next discussion ,right now let's take it for granted.Inputting it to the integral and dividing both sides by $-\Beta(b,c-b)$ we get this surprising result , $\sum_{a=1}^{\infty}\frac{_2F_1(-a,b,c,z)}{a}=\psi(c)-\psi(b)+\ln (1/z)$ Corollary c=1-b and z=1 we get the reflection equation of diagamma function i.e.$\psi(1-s)-\psi(s)=\pi\cot(\pi s)$ $\sum_{a=1}^{\infty}\frac{_2F_1(-a,b,1-b,1)}{a}=\pi\cot(\pi b)$ So it also provides an alternative representation of the Cotangent function and as always another wonderful equation is born.(Please report any problem in this discussion)