Does the following Gauss hypergeometric function increase with $x$

Question

Given a function $f(x)={_2F_1}(-a,1;1-a;-x)$, where $0<a<1$ is a constant. By numerical computation, I found that $f(x)$ increase with $x>0$. But I do not know how to prove it. Could you please help me to prove it or give me some hints? Many Thanks for your help!

Answer 1

By definition, in a neighbourhood of the origin $$\phantom{}_2 F_1\left(-a,1;1-a;-x\right)=\sum_{n\geq 0}\frac{(-a)_n (1)_n}{(1-a)_n}\cdot\frac{(-x)^n}{n!}=\sum_{n\geq 0}\frac{\Gamma(n-a)\Gamma(1-a)}{\Gamma(-a)\Gamma(n+1-a)}(-x)^n $$ or $$\phantom{}_2 F_1\left(-a,1;1-a;-x\right)=\sum_{n\geq 0}\frac{-a}{(n-a)}(-x)^n = -a\sum_{n\geq 0}\int_{0}^{+\infty}e^{-(n-a)z}(-x)^n\,dz $$ or (notice that the integral provides an analytic continuation over $\mathbb{R}^+$) $$ \phantom{}_2 F_1\left(-a,1;1-a;-x\right)=\color{red}{-a\int_{0}^{+\infty}\frac{e^{(1+a)z}}{e^{z}+x}\,dz }.$$ If $x_2>x_1>0$, for any $z\in\mathbb{R}^+$ we have $\frac{e^{(1+a)z}}{e^{z}+x_2}<\frac{e^{(1+a)z}}{e^{z}+x_1}$, hence $$ \phantom{}_2 F_1\left(-a,1;1-a;-x_2\right)>\phantom{}_2 F_1\left(-a,1;1-a;-x_1\right)$$ as wanted.