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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 $K \in\mathbb{N}$.

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3 Answers 3

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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 $\frac{\Gamma(K+1)}{\Gamma(K)}=K$

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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.

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  • $\begingroup$ Could you please tell me what will be the solution for the following case? $$ \int_0^{\infty } \frac{1}{(1+x) (m+x)^{K+1}} \, dx $$ $\endgroup$
    – sky-light
    Commented Nov 11, 2014 at 12:57
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    $\begingroup$ @barznjy: from straightforward manipulations, $$\frac{1}{K+1}\phantom{}_2 F_1(K+1,K+1;K+2;1-m).$$ $\endgroup$ Commented Nov 11, 2014 at 13:02
  • $\begingroup$ Could you please let me know how these two forms are equal? both of them give the same result, however, I don't know how they are equal. $$\frac{m^{-K}-\, _2F_1(K,K;K+1;1-m)}{K(1-m)} = \frac{\, _2F_1(K+1,K+1;K+2;1-m)}{K+1}$$ $\endgroup$
    – sky-light
    Commented Nov 12, 2014 at 20:42
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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.

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