How can the double integral $$ \int_{0}^{\infty}\int_{0}^{\infty}e^ {{-\beta(x^2+y^2+xy)}}\cos(2\pi Mx)\cos(2\pi Ny)\,\text{d}x \,\text{d}y, $$ where $\beta>0$ and $M$, $N\in\mathbb{Z}$, be evaluated? The exponent can be separated by completing the square, which seems a natural first step, but I can't seem to make much progress from there.

For those who are interested, the integral arises when applying the Poisson summation formula to the partition function $$ Z(\beta) = \sum_{m,n=1}^{\infty}\exp(-\beta E_{m,n}). $$ Here $E_{m,n}=m^{2}+n^{2}+mn$ gives the eigenvalue spectrum for the Helmholtz equation solved inside a planar domain shaped like an equilateral triangle.


1 Answer 1


Write the cosines in complex form, that will cast the integral into the exponential of a quadratic form in $x,y$ with complex coefficients, then complete the square and the integral becomes a Gaussian integral in two dimensions, which can be evaluated directly.

  • $\begingroup$ Thanks, I tried something along these lines, but couldn't find an expression for multi-dimensional Gaussian integrals when the limits are from 0 to $\infty$. Do you have a reference, or am doing something wrong? $\endgroup$
    – Alex W
    Aug 23, 2015 at 10:06
  • 1
    $\begingroup$ Every multidimensional Gaussian integral factors as a product of one-dimensional Gaussian integrals, once you have completed the squares (i.e. it might be necessary to change coordinates). Once you have done this, each one dimensional Gaussian integral might run over $[0, \infty)$, in such case observe that $e^{-a x^2}$ is an even function, so the integral is half of the complete standard integral running over $(-\infty, \infty)$. However if it runs over $[r, \infty)$, $r>0$, no closed form exists, and the "result" is written using the so called "Error function", $Erf(r)$. $\endgroup$ Aug 23, 2015 at 14:14

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