# Analytic continuation of the incomplete beta function

Is there a rigorous proof for the analytic continuation of the incomplete beta function $$B(x;a,b)$$ for all values of $$a$$ and $$b$$? The incomplete beta function normally restricts the values of $$a,b$$ as $$a>0$$, $$b>0$$. So I would like to extend these values to negative but I cannot find a good and comprehensive reference for this.

There is a given expression in wikipedia for the analytic continuation of beta function but I cannot see one for the incomplete one. By the way, the integral that I am working on is:

$$\rho=\frac{b_0}{1-q}\int_0^{1-(b_0/x)^{1-q}}v^{-1/2}(1-v)^{\frac{1}{q-1}-1}dv=\frac{b_0}{1-q}Beta(1-(b_0/x)^{1-q};\frac{1}{2},\frac{1}{q-1})$$

And I am working on $$-\infty. Thanks for the help.

There is the general complete description with the regularized Gauss hypergeometric function ${{_2}\tilde{F}_{1}}(a,b,c,x)$, see 1 and 2: $$B_x(a,b)= \Gamma(a)\,x^a \;{{_2}\tilde{F}_{1}}(a,1-b,a+1,x), \qquad -a \notin \mathbb{N}$$
When $a \le 0$ or $b \le 0$, the Gauss hypergeometric function ${{_2}F_{1}}$ function can be used: If $a \neq 0$ is no negative integer, the result is (3) $$B_x(a,b)= \frac{x^a}{a}\,{{_2}F_{1}}(a,1-b,a+1,x), \qquad -a \notin \mathbb{N},$$ else if $b \neq 0$ is no negative integer, then (4): $$B_x(a,b)= B(a,b)- \frac{(1-x)^b x^a}{b}\;{{_2}F_{1}}(1,a+b,b+1,1-x), \qquad -b \notin \mathbb{N}.$$
• Hi. Thanks for the answer. I don't actually understand this. If the parameter $b$ takes a value of zero or any negative value, I can use the last equation? Is that right? – user583893 Aug 21 '18 at 9:44
• No, you can use it if $b \ne -1, -2, \dots$ otherwise $c= b+1$ becomes zero, and the the Gauss function is undefined. The regularized function $${{_2}\tilde{F}_{1}}(a,b,c,x) = \frac{1}{\Gamma(c)} {{_2}F_{1}}(a,b,c,x)$$ can handle this because together with $\Gamma(c)$ you get a finite limit if $c$ approaches a negative integer or zero. – gammatester Aug 21 '18 at 9:56
• What i need is an extension of incomplete beta function that works for negative values of the parameter $b$. But i tried feeding up negative values of $b$ to the last equation and it gave me the right answers. Now I am confused since the last equation must only work for positive $b$'s. – user583893 Aug 21 '18 at 10:40
• No it does work for negative non-integer $b$. If $b$ is a negative integer and $a$ is not you can use the second formula, otherwise use the regularized function. But note the hint given at functions.wolfram.com/GammaBetaErf/Beta3/02/02. Using my own implementation I get for example $B_{0.5}(-1.5, -2.5)= 6.4$ (verified on Wolfram function evaluation). – gammatester Aug 21 '18 at 10:56