# How to prove that $_2F_1(1,-k;1-k;-z) = 1 + \frac{zk}{1-k}~{_2F_1}(1,1-k;2-k,-z).$

Prove that $$_2F_1(1,-k;1-k;-t) = 1 + \frac{tk}{1-k}~{_2F_1}(1,1-k;2-k,-t), \quad(1)$$ where $k \in (0, 1)$, $t > 0$, and $_2F_1(a,b;c;z)$ is the Gauss hypergeometric function given by $$_2F_1(a,b;c;z) \triangleq \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_{0}^{1}x^{b-1}(1-x)^{c-b-1}(1-xt)^{-a}\mathrm{d}x.$$

I tried using the change of variables, but the addition 1 on the right-hand side of (1) creates the problem.

• Why don't you just write down both terms of $(1)$ in their power series form? There is no need to invoke the integral representation for $\phantom{}_2 F_1$. May 9, 2017 at 13:00
• But is it not that the power series definition of the hypergeometric function is valid for $|z| <1$? In my case it is possible that $|z| \geq 1$. Hence I did not consider it.
– user389066
May 9, 2017 at 14:19
• You should be aware of the principle of analytic continuation. As an alternative, you may prove that both sides are solutions of the same differential equation with the same constraints. Still no need for an explicit integral representation. May 9, 2017 at 14:37

The particular case of the hypergeometric function in the left hand side of (1) has $a=1$, $b=-k$ and $c=1-k$ which can be represented as \begin{aligned} {}_2F_1(1, -k; 1-k; \color{red}{z}) &= \frac{\Gamma(1-k)}{\Gamma(-k)\Gamma(1)}\int_0^1 \frac{dx}{x^{k+1}(1-x\color{red}{z})}\\ &=\frac{\Gamma(1-k)}{\Gamma(-k)\Gamma(1)}\int_0^1 \frac{1-xz + xz}{x^{k}x(1-xz)}dx\\ &= \frac{\Gamma(1-k)}{\Gamma(-k)\Gamma(1)} \int_0^1\left[\frac{1}{x^{k+1}} + \frac{z}{x^{k}(1-xz)} \right]dx. \end{aligned} Given the recursive property of Gamma function is $\Gamma(1)=1$ and $\Gamma(x+1)=x\Gamma(x)$, the factor in front of the integral is simplified to $-k$.
You can obtain the exact value of the first term inside the integral. You also need to rewrite the second term inside the integral to represent it as a hypergeometric function. After doing those, just replace $z=-t$ to get your desirable proof.