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.

  • $\begingroup$ After playing around and breaking the problem up, it looks like a legitimate issue with Integrate. Send the example in to support@wolfram.com. $\endgroup$
    – Searke
    Jul 19, 2012 at 20:12
  • $\begingroup$ 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. $\endgroup$
    – gpap
    Jul 19, 2012 at 22:03
  • $\begingroup$ @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. $\endgroup$
    – Mike
    Jul 19, 2012 at 22:12
  • 5
    $\begingroup$ Definitely a bug. Hoping the fix creates no trouble of its own. $\endgroup$ Jul 19, 2012 at 23:06
  • $\begingroup$ @Mike Please post that as an answer and accept when the system lets you (so the question doesn't appear unanswered). $\endgroup$
    – Szabolcs
    Jul 21, 2012 at 15:28

1 Answer 1


Wolfram Research has confirmed that this is a bug. From an email I just received: Integrate does appear to be giving an incorrect result, and I have forwarded the example to our developers.


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