# What is the value of the regularized incomplete beta function at x=0.5?

What is $I_{0.5}(a,b)$ where I is the regularized incomplete beta function?

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Could you please give us a reference to the definition of $I$? –  Robin Chapman May 6 '10 at 9:25
You can get your answer yourself from mathworld.wolfram.com/IncompleteBetaFunction.html. –  Jacques Carette May 6 '10 at 10:06
There's no $I_{0.5}$ there :-( –  Robin Chapman May 6 '10 at 10:40
@Neil: I'd join Robin and recommend you to decipher your $I_{0.5}(a,b)$ by giving an integral or series expression. Otherwise it sounds like you are not interested in getting an answer. –  Wadim Zudilin May 6 '10 at 12:07
sorry for the late reply. $I$ is defined here: mathworld.wolfram.com/RegularizedBetaFunction.html but there is no expression for $I_0.5$ –  Neil May 7 '10 at 2:10

## 1 Answer

You mean this? http://en.wikipedia.org/wiki/Beta_function $$\frac{\int_{0}^{\frac{1}{2}} t^{a - 1} (1 - t)^{b - 1} d t}{\int_{0}^{1} t^{a - 1} (1 - t)^{b - 1} d t} = \\ \quad{}\quad{}\frac{\mathrm{hypergeom} \Bigl([a,-b + 1],[1 + a],\frac{1}{2}\Bigr) \Gamma (a + b)}{2^{a} a \Gamma (b) \Gamma (a)}$$ Why do you think there is anything simpler in general?

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Thanks, Gerald, for demystefying the piece! Yes, there is no reduction of the $_2F_1$ hypergeometric series for general $a$ and $b$. But there are other ways to write it hypergeometrically. –  Wadim Zudilin May 6 '10 at 13:35
I thought there would be something simpler because when a and b are natural numbers, the function yields a natural number. –  Neil May 7 '10 at 2:11
But, thank you for your answer... –  Neil May 7 '10 at 2:11
(...yields a rational number) –  Neil May 7 '10 at 7:39