## Extra factors appear when evaluating Euler integrals

13

Note: this is fixed in version 9.

When I perform the double integral in Mathematica,

Integrate[(x (1 - x))^z (y (1 - y))^z, {x, 0, 1}, {y, 0, 1}]


which should give

$$B(z+1,z+1)^2 = \frac{\Gamma(z+1)^4}{\Gamma\left(2(z+1)\right)^2}$$

where $B(x,y)$ is the Beta function and $\Gamma(z)$ is the Gamma function because the integral is a product of two Beta functions, I instead get this ratio of Gamma functions times the extra factor $(-1)^{2z}$. What is going on here? To make matters stranger, if I do the integral instead using two nested calls to Integrate (one to integrate out $x$ and one to integrate out $y$), I get the ratio of Gamma functions without the incorrect extra factor.

After playing around and breaking the problem up, it looks like a legitimate issue with Integrate. Send the example in to support@wolfram.com. – Searke – 2012-07-19T20:12:24.753

Are you sure about the validity of the expression you gave over the entire complex plane? It seems that the $(-1)^{2z}$ factor comes from a resultant of the Gammas. – gpap – 2012-07-19T22:03:26.310

@gpap, that's not the issue: try running Assuming[Re[z]>0 && Im[z]==0, Integrate[(x (1 - x))^z (y (1 - y))^z, {x, 0, 1}, {y, 0, 1}]. $(-1)^{2z}$ should definitely not be there. – Mike – 2012-07-19T22:12:32.400

5Definitely a bug. Hoping the fix creates no trouble of its own. – Daniel Lichtblau – 2012-07-19T23:06:57.633

@Mike Please post that as an answer and accept when the system lets you (so the question doesn't appear unanswered). – Szabolcs – 2012-07-21T15:28:24.323