3
$\begingroup$

If I start with the definition of the beta function $$B(a,b) = \int_0^1 t^{a-1} (1-t)^{b-1} \operatorname{d}t = \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)}$$ that is valid for $\mathcal{R}(a) >0$ and $\mathcal{R}(b) > 0$, and apply a binomial expansion to the second factor in the argument of the integral, I ultimately get: $$\frac{\Gamma(a)}{\Gamma(a+b)} = \frac{1}{(a)_b} = \sum_{j=0}^\infty \frac{(-1)^j}{j!\, \Gamma(b-j)\, (a+j)},$$ with $(x)_n$ being Pochhammer's symbol.

The starting point of the derivation is only valid for the real part of $a>0$. The use case I have requires it to be valid for $a\in \mathbb{R}$ and $a\notin \{n \in \mathbb{Z}\ |\ n \le 0\}$ (i.e. $a-1$ is not a negative integer). I've looked numerically at various cases of $b$ with $50$ terms in the sum, and it appears to converge to the fraction using gamma functions.

For example, in this graph with $a$ as the $x$-axis I've used $b=2.7$ and plotted the sum as a blue dashed line, and the ratio of gamma functions as the solid red line. The graphs overlap similarly for $b$ as positive and negative integers and non-integers. Pochhammer identity comparison

I'm asking this question because I'm surprised that it appears to be valid for more values than the derivation's starting point allows, and want to know if a more robust derivation is possible.

There's an identity in Wikipedia's beta function article that is identical $$B(x,y) = \sum_{n=0}^\infty \frac{{n-y \choose n}}{x+n}$$ if you use gamma functions to expand the parts of the choose function that involve $y$ and apply the reflection identity to both gamma functions. The only citation for the section is Abramowitz and Stegun, which does not contain the identity from Wikipedia.

The MathWorld page on the incomplete beta function has a formula that reduces to the Wikipedia one when plugging in $z=1$ (equation 5) $$B(z;a,b)= z^a \sum_{n=0}^\infty \frac{(1-b)_n}{n!\, (a+n)} z^n,$$ and the only reference on that page, "Tables of the Incomplete Beta Function" by Karl Pearson (2nd ed), does not contain this formula.

$\endgroup$
  • 1
    $\begingroup$ The key may lie in judicious use of the Pochhammer contour or similar. But it's not unusual for derivations using analytic functions to rely on analytic continuation: functional equations are often proved in an open subset, and extended by analytic continuation, for example. $\endgroup$ – Chappers Sep 14 '17 at 1:41
  • 1
    $\begingroup$ But certainly if the sum works for any $a$, it will work on the set you specify, since the $1/(a+j)$ terms are eventually decreasing no matter what the value of $a$ is. $\endgroup$ – Chappers Sep 14 '17 at 1:44

Your Answer

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

Browse other questions tagged or ask your own question.