I have recently had occasion to investigate the Fourier series of the function $f(x,y)=\log({2+\cos 2\pi x} +\cos{2\pi y})$. Accordingly, define

$I(m,n)=\int_{0,0}^{1,1}f(x,y)\cos{2\pi mx}\cos{2\pi ny}dxdy$

which is the $(m,n)$th Fourier coefficient. If there is some easy change of variables or flick of the wrist to compute $I$ in closed terms, I haven't been able to find it. One may compute $I(0,0)=\frac{4G}{\pi}-\log 2$, where $G$ is Catalan's constant. As for other values, Mathematica's integration routine returns the values $I(1,1)=\frac{1}{2}+\frac{2}{\pi}$ and $I(2,1)=1-\frac{10}{3\pi}$ (and takes about 10 minutes per computation). I have tried and failed to affirm the following conjecture, which seems reasonable to present here.

Conjecture. If $(m,n)\neq(0,0)$, then $I(m,n)=a+\pi^{-1}b$ with $a,b\in \mathbb{Q}$.

Of course, if you could prove this, it would be great to give a closed formula for $a,b$.

Full disclosure: My interest in this stems from a formula of Kasteleyn for the partition function of dimer coverings of squares; in fact the evaluation of $I(0,0)$ given above can be extracted from his original paper.

  • $\begingroup$ Did you try computing I(1,1) numerically? I am doing this and getting something like -0.14. Did I make a slip? Aah! Conjecture: you wrote "I(1,1)=1/2+2/pi" but mean 1/2-2/pi. Am I right? $\endgroup$ – Kevin Buzzard Feb 19 '10 at 17:04
  • $\begingroup$ Kevin, Mathematica really did give $1/2+2/Pi$ for I(1,1). The integrand has a nasty singularity at (1/2,1/2) - what routine are you using to integrate? $\endgroup$ – David Hansen Feb 20 '10 at 14:03
  • $\begingroup$ Mathematica is wrong, and Kevin is right. $\endgroup$ – Bjorn Poonen Feb 20 '10 at 19:44
  • $\begingroup$ @David: "what routine are you using to integrate?" I used pari-gp, and I also simply summed the integrand over 1,000,000 points in the region of integration and divided by 1,000,000 for a check. $\endgroup$ – Kevin Buzzard Feb 21 '10 at 11:02

Here is a proof of the conjecture, a proof that also shows how to compute the integrals explicitly.

The proof is somewhat similar to David Speyer's approach, but instead of using multivariable residues, I will just shift a one-variable contour. Without loss of generality, $m>0$. Eliminating the trigonometric functions yields $I(m,n)=(-1/4\pi^2) J(m,n)$, where $$ J(m,n) := \int_{|w|=1} \int_{|z|=1} \log(4+z+z^{-1}+w+w^{-1}) \, z^{m-1} w^{n-1} \, dz \, dw.$$ For fixed $w \ne -1$ on the unit circle, the values of $z$ such that $4+z+z^{-1}+w+w^{-1}$ are a negative real number and its inverse; let $g(w)$ be the value in $(-1,0)$. The function $\log(4+z+z^{-1}+w+w^{-1})$ of $z$ extends to the complement of the closed interval $[g(w),0]$ in a disk $|z|<1+\epsilon$. Shrink the contour $|z|=1$ like a rubber band so that it hugs the closed interval. The upper and lower parts nearly cancel, leaving $$ J(m,n) = \int_{|w|=1} \int_{z=0}^{g(w)} -2\pi i \, z^{m-1} w^{n-1} \,dz \,dw,$$ with the $-2\pi i$ coming from the discrepancy in the values of $\log$ on either side of the closed interval. Thus $$ J(m,n) = -\frac{2 \pi i}{m} \int_{|w|=1} g(w)^m w^{n-1} \, dw.$$ Next use the rational parametrization $(g(w),w) = \left( (1-u)/(u+u^2), (u^2-u)/(1+u) \right)$ that David Speyer wrote out. There is a path $\gamma$ from $-i$ to $i$ in the right half of the $u$-plane that maps to $|w|=1$ (counterclockwise from $-1$ to itself) and gives the correct $g(w)$. Integrating the resulting rational function of $u$ gives $$ J(m,n) = -\frac{2 \pi i}{m} \left.(f(u) + r \log u + s \log(1+u))\right|_\gamma = -\frac{2 \pi i}{m} (f(i)-f(-i) + r \pi i + s \pi i/2),$$ for some $f \in \mathbf{Q}(u)$ and $r,s \in \mathbf{Q}$. This shows that $I(m,n)=a + b/\pi$ for some $a,b \in \mathbf{Q}$.

Examples: Mathematica got the answers wrong, presumably because it doesn't understand homotopy very well! Here are some actual values, computed using the approach above: $$I(1,1) = \frac{1}{2} - \frac{2}{\pi}$$ $$I(2,1) = -1 + \frac{10}{3\pi}$$ $$I(20,10) = -\frac{14826977006}{5} + \frac{56979906453888224582}{6116306625 \pi}.$$


I know how I want to answer this question. I'll write up the easy parts here, and leave the hard part for you :).

First some minor changes. It will be convenient to clear out denominators and work with $\log \left( 4 + e^{2 \pi i x} + e^{- 2 \pi i x} + e^{2 \pi i y} + e^{- 2 \pi i y} \right)$. That just changes the constant term of your Fourier series by $\log 2$. Next, it is convenient to focus on $$ \int_0^1 \int_0^1 \log \left( 4 + e^{2 \pi i x} + e^{- 2 \pi i x} + e^{2 \pi i y} + e^{- 2 \pi i y} \right) e^{2 \pi i m x} e^{2 \pi i n y} dx dy.$$ A simple linear transformation goes between this and the cosine formulation. Let $S = \{ (z,w) : |z|=|w|=1 \}$. So we are dealing with $$\frac{1}{(2 \pi i )^2} \int_S \log \left( 4+z+z^{-1} + w +w^{-1} \right) z^{m-1} w^{n-1} dz dw.$$ Dropping out the $4 \pi ^2$, we want to show the integrand is of the form $a \pi + b$.

UPDATE: Thanks to fedja for pointing out that I had oversimplified the next paragraph.

Assuming that $(m,n) \neq (0,0)$, we can integrate by parts with respect to one of the two variables, let's say $z$. Once we do that, we will have a quantity of the form $$ (\mbox{rational number}) \cdot \int_S \frac{(z-z^{-1}) w^k z^{\ell} dw dz} {4+w+w^{-1}+z+z^{-1}}$$

So we'd like to show this quantity is of the form $a+b \pi$.

As fedja points out, we need to be careful here. Without the $z-z^{-1}$ term, the integral diverges like $\int \int ds dt/(s^2 + t^2)$ near $(-1,-1)$.

Whew! Now comes the actual hard part. Let $$E:=\{ (z,w) \in (\mathbb{C}^*)^2 : \ 4+z+z^{-1}+w+w^{-1} =0 \}.$$ This is an elliptic curve with four punctures. As Bjorn points out, this is a nodal cubic and can be parameterized as $$(z,w) = \left( \frac{1-u}{u(1+u)}, \frac{u(u-1)}{1+u} \right).$$ We'll come back to this point later.

The $2$-form $dw dz/(4+z+z^{-1}+w+w^{-1})$ has a simple pole on $E$. Let $\omega$ be the $1$-form on $E$ which is the residue of that $2$-form.

I think there should be a curve $\gamma$ in $E$ such that $S$ is homotopic, in $(\mathbb{C}^*)^2 \setminus E$, to a tubular neighborhood of $\gamma$. So $$\int_S \frac{w^k z^{\ell} (z-z^{-1}) dw dz} {4+w+w^{-1}+z+z^{-1}} = \int_{\gamma} \omega w^k z^{\ell} (z - z^{-1}) .$$

If we substitute in the above parameterization, this will be the integral around a closed loop of some rational function in $\mathbb{Q}(u)$. In particular, we can compute this integral by residues and we will get something of the form $a+b \pi$, as desired.

Actually, it looks to me like we should just get $b \pi$. Maybe the integration by parts doesn't go as well as I hoped?

Obviously, someone should actually work this out explicitly, but I don't think it will be me.

  • 1
    $\begingroup$ Well, your integration by parts already seems very suspicious: the final integral you wrote certainly diverges at $(-1,-1)$. Still I like your general idea :). $\endgroup$ – fedja Feb 18 '10 at 22:13
  • $\begingroup$ Hmmm. You're right that integral z^k w^l/(4+...) diverges at (-1,-1). On the other hand, the particular expressions we get out of integrating by parts look like integral (z-z^{-1}) z^k w^l/(4+...) depending on which variable we integrated on. And that should be convergent. I'll edit to point that out. $\endgroup$ – David E Speyer Feb 18 '10 at 22:28
  • $\begingroup$ As for justifying the integration by parts, the right way to do it is to cut out a small disc D around (-1,-1). There is then a completely valid integration by parts, which winds up with a left over term where we integrate over \partial D. One then has to show that this term drops out when we shrink D. But I'll leave that for someone else. $\endgroup$ – David E Speyer Feb 18 '10 at 22:29
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    $\begingroup$ @David: Don't wimp out yet! Your "elliptic curve" is singular at (-1,-1), so it's just a rational curve, which should make your life much, much easier... $\endgroup$ – Bjorn Poonen Feb 18 '10 at 22:50
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    $\begingroup$ @David: "Dropping out the 4 pi^2, we want to show the integrand is of the form a pi +b.". Either I'm misunderstanding or that should be a.pi + b.pi^2? $\endgroup$ – Kevin Buzzard Feb 19 '10 at 14:18

Mathematica gets the (1,1) case right.

Integrate the x-variable to obtain an answer that involves the absolute value of Cos[Pi y].

Then split the y-integral at y=1/2 to get the correct value.

The answer is 1/2 - 2/Pi


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