0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – user583893 Aug 21 '18 at 9:44
  • $\begingroup$ 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. $\endgroup$ – gammatester Aug 21 '18 at 9:56
  • $\begingroup$ What i need is what works for negative values of b. I already tried in the mathematica, and it gave me the correct answer. $\endgroup$ – user583893 Aug 21 '18 at 10:36
  • $\begingroup$ 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. $\endgroup$ – user583893 Aug 21 '18 at 10:40
  • $\begingroup$ 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). $\endgroup$ – gammatester Aug 21 '18 at 10:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.