Hypergeometric function integral representation How to prove the following relation?
$$ \, _2{F}_1(K,K;K+1;1-m) = \frac{\Gamma (K+1)}{\Gamma (K)} \int_0^{\infty } \frac{1}{(1+x) (m+x)^K} \, dx $$
where $_2{F}_1(.,.;.;.)$ is the hypergeometric function, $m\in\mathbb{R}^+$ , and $K \in\mathbb{N}$.
 A: Refer to the integral definition of $ _2F_1(a,b;c;z)$ in the general case :
$$ \, _2{F}_1(a,b;c;z) = \frac{\Gamma (c)}{\Gamma (b) \Gamma (c-b)} \int_0^{\infty} t^{-b+c-1} \, (t+1)^{a-c} \, (t-z+1)^{-a} \, dt $$
In the particular case : $a=K$ ; $b=K$ ; $c=K+1$ ; $z=1-m$ ; $t=x$ 
and with $\Gamma(c-b)=\Gamma(1)=1$  leads to : 
$$ \, _2{F}_1(K,K;K+1;1-m) = \frac{\Gamma (K+1)}{\Gamma (K)} \int_0^{\infty } \frac{1}{(1+x) (m+x)^K} \, dx $$
Note that $\frac{\Gamma(K+1)}{\Gamma(K)}=K$
A: Let $u=1-m$. Then:
$$I=\frac{\Gamma(K+1)}{\Gamma(K)}\int_{0}^{+\infty}\frac{1}{(1+x)(1+x-u)^K}\,dx=K\int_{0}^{+\infty}\frac{\left(1-\frac{u}{x+1}\right)^{-K}}{(1+x)^{K+1}}\,dx$$
but since:
$$(1-z)^{-K} = \sum_{n=0}^{+\infty}\binom{n+K-1}{K-1} x^n$$
it follows that:
$$ I = K\sum_{n=0}^{+\infty}u^n\binom{n+K-1}{K-1}\int_{0}^{+\infty}\frac{dx}{(1+x)^{K+n+1}}=\sum_{n=0}^{+\infty}u^n\binom{n+K-1}{K-1}\frac{K}{K+n}.$$
If we set $A_n = \binom{n+K-1}{K-1}\frac{K}{K+n}$ then we have:
$$\frac{A_{n+1}}{A_n}=\frac{(n+K)^2}{(n+1)(n+K+1)}$$
from which it follows that:
$$ I = \phantom{}_2 F_1(K,K;K+1;u) = \phantom{}_2 F_1(K,K;K+1;1-m)$$
as wanted.
A: This is to answer the question which @sky-light asked in the comment section.
Based on the formula from the original question:
$\frac{_2F_1(K+1,K+1;K+2;1-m)}{K+1} = \int_{0}^{\infty} \frac{dx}{(1+x)(m+x)^{K+1}} = \frac{1}{1-m} \int_{0}^{\infty} \frac{(1+x)-(m+x)}{(1+x)(m+x)^{K+1}}dx = \frac{1}{1-m} \big[ \int_{0}^{\infty}  \frac{dx}{(m+x)^{K+1}} - \int_{0}^{\infty} \frac{dx}{(1+x)(m+x)^k} \big] = \frac{1}{1-m} \big[ -\frac{1}{K(m+x)^K}\big|_0^{\infty} - \frac{_2F_1(K,K;K+1;1-m)}{K}\big] = \frac{m^{-K} - _2F_1(K,K;K+1;1-m)}{K(1-m)}$
Then, the two are equal.
P/S: at the time I posted this answer, my privilege was not enough to answer in the comment.
