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 $-\infty<q<1$. 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): $$ B_x(a,b)= B(a,b)- \frac{(1-x)^b x^a}{b}\;{{_2}F_{1}}(1,a+b,b+1,1-x), \qquad -b \notin \mathbb{N}. $$