What is $I_{0.5}(a,b)$ where I is the regularized incomplete beta function?
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$\begingroup$ Could you please give us a reference to the definition of $I$? $\endgroup$– Robin ChapmanCommented May 6, 2010 at 9:25
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$\begingroup$ You can get your answer yourself from mathworld.wolfram.com/IncompleteBetaFunction.html. $\endgroup$– Jacques CaretteCommented May 6, 2010 at 10:06
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$\begingroup$ There's no $I_{0.5}$ there :-( $\endgroup$– Robin ChapmanCommented May 6, 2010 at 10:40
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$\begingroup$ @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. $\endgroup$– Wadim ZudilinCommented May 6, 2010 at 12:07
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$\begingroup$ sorry for the late reply. $I$ is defined here: mathworld.wolfram.com/RegularizedBetaFunction.html but there is no expression for $I_0.5$ $\endgroup$– NeilCommented May 7, 2010 at 2:10
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1 Answer
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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?
answered May 6, 2010 at 12:19
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$\begingroup$ 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. $\endgroup$ Commented May 6, 2010 at 13:35
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$\begingroup$ I thought there would be something simpler because when a and b are natural numbers, the function yields a natural number. $\endgroup$– NeilCommented May 7, 2010 at 2:11
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