I need to calculate the integral

$$\int_0^1 \prod_{i=1}^N dx_i \delta \left( \sum_{j=1}^N x_j-1 \right) (x_1 x_2 \cdots x_N)^\alpha(x_2 x_3 \cdots x_{N} +x_1 x_3 \cdots x_{N} + \text{other terms with one } x_i \text{ absent})^\beta.$$

Here $\alpha$ and $\beta$ are real numbers. For instance, for $N=3$ we would have

$$\int_0^1 dx_1 dx_2 dx_3 \delta \left( x_1+x_2+x_3-1 \right) (x_1 x_2 x_3)^\alpha(x_2 x_3+x_1x_3+x_1x_2)^\beta.$$

Utilising the delta function and writing, say, $x_1$ in terms of $x_2, x_3$ does not help much.

Does anyone have any ideas about how to attack such a problem? If we can calculate the $N=3$ case, then I guess the generalisation will not be that difficult.

  • $\begingroup$ Is there a reason you expect this to have an analytic answer? $\endgroup$ Apr 23, 2020 at 23:49

2 Answers 2


Integrate over this $n-1$-dimensional section of a hyperplane (the set of points $\sum\limits_{i=1}^n x_i = 1$):

enter image description here

The red polygon is the set of points that contribute to the integral (thanks to the $\delta$ function). That is, any points off that polygon make the $\delta$ function vanish.

The integral has no simple analytic form. For the case where the second term is not considered, the solution in the $n=3$ case is:

$$\int\limits_{x_1 = 0}^1 dx_1 \int\limits_{x_2 = 0}^1 dx_2\ (x_1 x_2 (1 - x_1 - x_2))^\alpha = $$

$$\frac{1}{4} \left(\frac{2 (-1)^{\alpha } \, _3F_2(1,-\alpha ,2 \alpha +2;\alpha +2,2 \alpha +3;1)}{(\alpha +1)^2}+\frac{\Gamma (\alpha +1)^2 \left(4 \Gamma (\alpha +1)+\frac{4^{-\alpha } e^{i \pi \alpha } \Gamma \left(-\alpha -\frac{1}{2}\right) \Gamma (2 \alpha +2)}{\sqrt{\pi }}+\frac{4 \pi (-1)^{\alpha } \alpha \csc (2 \pi \alpha )}{\Gamma (1-\alpha )}\right)}{\Gamma (3 \alpha +3)}\right)$$

where $F$ is the generalized hypergeometric function.

Assuming[\[Alpha] > 0 && \[Alpha] \[Element] Reals ,
 Integrate[(x1 x2 (1 - x1 - x2))^\[Alpha] , 
  {x1, 0, 1}, {x2, 0, 1}]
  • $\begingroup$ Thanks! I don't see how that works though..What was the change of variables? Edit:oops, just saw that you were editing it.. $\endgroup$
    – Valentina
    Apr 24, 2020 at 0:15
  • $\begingroup$ Any closed form solution will do! But how did you calculate that. Did you use mathematica? Because I've been unable to get even that $\endgroup$
    – Valentina
    Apr 24, 2020 at 0:34

Here is a method that produces an explicit evaluation of your integral if $\beta$ is a positive integer.

For simplicity, suppose that $N=3$. With $x_3 = 1 - x_1 - x_2$, a simple calculation shows that $$ x_2 x_3 + x_1 x_3 + x_1 x_2 = x_1 + x_2 - x_1 x_2 - x_1^2 - x_2^2. $$ Raising this expression to the power $\beta$ and inserting it in the integrand, your integral becomes $$ \int_\Delta x_1^\alpha x_2^\alpha (1-x_1-x_2)^\alpha (x_1 + x_2 - x_1 x_2 - x_1^2 - x_2^2)^\beta \, {\rm{d}}x_1 \, {\rm{d}}x_2, $$ where the domain of integration is the simplex $\Delta = \{(x_1,x_2): x_1 > 0, x_2 > 0, x_1 + x_2 < 1\}$.

If $\beta$ is a positive integer then the term $(x_1 + x_2 - x_1 x_2 - x_1^2 - x_2^2)^\beta$ can be expanded via the multinomial theorem as a linear combination of monomials $x_1^j x_2^k$, where $j, k$ are nonnegative integers. Then your integral is a linear combination of integrals of the form $$ \int_\Delta x_1^{\alpha+j} x_2^{\alpha+k} (1-x_1-x_2)^\alpha \, {\rm{d}}x_1 \, {\rm{d}}x_2, $$ and each of these integrals is a well-known classical Dirichlet integral.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .