Average of irrational flow on the torus

Question

Let $$F(x,y) = \frac{1}{\sqrt{2-\sin(2\pi x) - \sin(2\pi y)}}$$ defined on $\mathbb{T}^2$. Here $\mathbb{T}^2 = \mathbb{R}^2/ \mathbb{Z}^2$ is the 2-torus. How can I show that $$ \lim_{T\longrightarrow \infty} \frac{1}{T}\int_0^T F(x,\sqrt{2}x)\;dx = \iint_{[0,1]^2} F(x,y)\;dxdy?$$ What can we say about the rate of convergence? Say $\frac{1}{T^\alpha}$ for some $\alpha \in (0,1)$?

Answer 1

In the original version of the question $F$ was continuous.

This is true for any continuous $F$. First do it for $$F(x,y)=e^{2\pi i(nx+my)},$$for $n,m\in\Bbb Z$. So it holds for $F$ equal to a trigonometric polynomial. Now show the property is preserved under uniform limits...

Of course that no longer works, at least not so simply, with the current version of $F$. One might approximate $F$ from below by continuous functions and try to estimate the error at the fairly mild singularity.

Edit: I believe it's true. Don't have time to try to write a formal proof, but I can explain sort of why I believe it:

The crucial thing is that $$\lim_{\delta\to0}\int_{-\delta}^\delta|t|^{-1/2}\,dt=0.$$

Notation: Let's write $$\int F=\int_0^1\int_0^1 F(x,y)\,dxdy$$and $$I_F(T)=\frac1T\int_0^TF(x,\sqrt 2x)\,dx.$$ For $\lambda>0$ let $$F_\lambda=\min(F,\lambda).$$Then $$\lim_{T\to\infty}I_{F_\lambda}(T)=\int F_\lambda,$$since $F_\lambda$ is continuous. So if we can show that $$I_{F_\lambda}(T)\to I_F(T)$$uniformly for $T>0$ as $\lambda\to\infty$ we're done. And it seems to me that this convergence is in fact uniform in $T$, because for any positive integer $n$ the integral $$\int_{n/\sqrt2}^{(n+1)/\sqrt2}(F(x,\sqrt 2x)-F_\lambda(x,\sqrt 2x))\,dx$$is zero except for maybe one or two singularities no worse than $\int_{-\delta}^\delta|t|^{-1/2}\,dt$, where $\delta\to0$ as $\lambda\to\infty$.