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Is this a known special function:

$$\int\nolimits_0^1 a^p(1-a)^{1-p}\\,b^{1-p}\\,(1-b)^p dp\qquad ?$$

I am really only interested in maximizing this over $(a,b)$ in $[0,1] \times [0,1]$, so a pointer to a nice numerical evaluation is appreciated as much or more so than an unstable exact formula.
Thanks for any help

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title: "known" :) – futurebird Nov 9 '10 at 3:50
What's the range of your $p$? – J. M. Nov 9 '10 at 8:43

You can get a closed-form answer.

$$\int_0^1 a^p (1-a)^{1-p} b^{1-p} (1-b)^p dp = b(1-a) \int_0^1 \left(\frac{a(1-b)}{b(1-a)}\right)^p dp = \left. \frac{b(1-a)}{\ln \frac{a(1-b)}{b(1-a)}} \left(\frac{a(1-b)}{b(1-a)}\right)^p \right|_0^1 $$ $$= \frac{a(1-b) - b(1-a)}{\ln a + \ln (1-b) - \ln b - \ln (1-a)} = \frac{a-b}{\ln a + \ln (1-b) - \ln b - \ln (1-a)}.$$

This holds if $a \neq b$ and if neither of $a$ or $b$ is 0 or 1. If $a = b$, then instead we have $$a(1-a) \int_0^1 dp = a - a^{2}.$$ And, of course, if $a$ or $b$ is 0 or 1 then the value of the integral is 0.

So, as far as maximizing, you can use the usual approach of finding where both partial derivatives are 0. I haven't worked through the calculations, but I strongly suspect that because of the symmetry in $a$ and $b$ that the maximum value will occur at $a = b$.

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I think the denominator should be $\ln a - \ln(1-a) - \ln b + \ln(1-b)$ instead. Note that the integrand is identical for $a = b = t$ and for $a = b = 1-t$, but your result is not. – Rahul Nov 9 '10 at 4:29
@Rahul: Thanks. Fixed. – Mike Spivey Nov 9 '10 at 4:32
Shouldn't the numerator be $\left(\frac{a(1-b)}{b(1-a)}\right)-1$? After all $\int x^p dp=x^p/\ln{x}$ so make the big fraction x. – Ross Millikan Nov 9 '10 at 5:09
@Ross: I think I've accounted for that. Remember that the expression you give is being multiplied by $b(1-a)$ to obtain the numerator in my expression. – Mike Spivey Nov 9 '10 at 5:28
You're right, I missed the $b(1-a)$ in front of the integral sign. – Ross Millikan Nov 9 '10 at 5:36

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