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For $m \in \{1,2,\ldots\}$, I have the sets $A_m=\{x \in [0,1]^2: |x|^2 \geq \frac{1}{m^2},x_1^2 \leq x_2 \leq \sqrt{x_1}\}$, $A_\infty=\{x \in [0,1]^2: x_1^2 \leq x_2 \leq \sqrt{x_1}\}$ (I use the convention $x=(x_1,x_2)$)

I try to find $$\lim_{m \to\infty}\int_{A_m}\left(\frac{\sin(x_1 x_2 m)}{m}+x_2\right)x_1^{-1} \mathrm{ d} \lambda^2$$

What I tried so far:

My goal is to show that I can pull the limit inside to get

$$\lim_{m \to\infty}\int_{A_m}\left(\frac{\sin(x_1 x_2 m)}{m}+x_2\right)x_1^{-1} \mathrm{ d} \lambda^2=\int_{A_\infty}\frac{x_2}{x_1} \mathrm{ d} \lambda^2=\int_0^1\int_{x_1^2}^{\sqrt{x_1}}\frac{x_2}{x_1}\mathrm{ d} x_2 \mathrm{ d} x_1=\frac{3}{8}$$

But to do this (via dominated convergence) I need to know that there is some integrable (over $A_\infty$) function that majorizes

$$g_m(x)=\chi_{A_m}(x)\left(\frac{\sin(x_1 x_2 m)}{m}+x_2\right)x_1^{-1}$$

However I think it is not possible to find such a majorant, because if you bound the $\sin(x_1 x_2 m)$ by anything else than $0$ the function is not integrable. (I know I could use that $\sin(x)\leq x$ for $x \in [0,2]$ but then I get a majorant that depends on my $m$ because I need to assure that $x_1 x_2 m \leq 2$).

Therefore I think maybe a different convergence theorem might be useful here, sadly the $g_m$ are not monotone.

Also I had the idea of splitting the integral into its two sums

$$\int_{A_m}\left(\frac{\sin(x_1 x_2 m)}{m}+x_2\right)x_1^{-1} \mathrm{ d} \lambda^2=\int_{A_m}\left(\frac{\sin(x_1 x_2 m)}{m}\right)x_1^{-1}\mathrm{ d} \lambda^2+\int_{A_m}\frac{x_2}{x_1} \mathrm{ d} \lambda^2$$

Then if we could apply something like the Riemann-Lebesgue Lemma to the left side (as $x_1^{-1}$ is integrable over $A_\infty$) to show that

$$\int_{A_m}\left(\frac{\sin(x_1 x_2 m)}{m}\right)x_1^{-1}\mathrm{ d} \lambda^2\to 0$$

And therefore the identity would follow, but the form I know only works with the dot-product so we would need to have something like $x_1 m+x_2 m$ in the $\sin(x_1 x_2 m)$, it doesn't really seem to work out.

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Why is the inequality $|\sin x|\leq x$ true only for $0\leq x\leq 2$? – Davide Giraudo Dec 30 '11 at 16:37
@DavideGiraudo Hmm, stupid me, of course it is true for bigger $x$ too. Let me see how this affects the calculation. – Listing Dec 30 '11 at 16:42
Ok it works, I can use dominated convergence then and the result follows. You can write it as an answer so I can accept it. – Listing Dec 30 '11 at 16:46
up vote 2 down vote accepted

We have $|\sin x|\leq \int_0^x|\cos t|dt\leq x$, so $$|g_m(x)|\leq \chi_{A_{\infty}}(x)x_1^{-1}(x_1x_2+x_2)=\chi_{A_{\infty}}(x)x_2(1+x_1^{-1})\leq \chi_{A_{\infty}}(x)\sqrt{x_1}(1+x_1^{-1})$$ which is integrable. Since for $x\in (0,1]\times [0,1]$ we have $g_m(x)\to x_2x_1^{-1}$, we can conclude by the dominated convergence theorem.

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