# Extra factors appear when evaluating Euler integrals

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.

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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 Jul 19 '12 at 20:12
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 Jul 19 '12 at 22:03
@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 Jul 19 '12 at 22:12
Definitely a bug. Hoping the fix creates no trouble of its own. – Daniel Lichtblau Jul 19 '12 at 23:06
@Mike Please post that as an answer and accept when the system lets you (so the question doesn't appear unanswered). – Szabolcs Jul 21 '12 at 15:28

They always say the same about Integrate. Look e.g. here mathematica.stackexchange.com/questions/8161/… and here mathematica.stackexchange.com/questions/3568/… and read the comments here : stackoverflow.com/questions/8353441/… – Artes Jul 24 '12 at 19:33