Apply DCT to multivariable function

Question

Let $\epsilon>0$. Define $$D(\epsilon)=\{(x,y)\in\mathbb{R}^2\mid0<x^2+y^2\le\epsilon^2\}$$ Determine the following limit $$\lim_{\epsilon\to0}\frac{1}{\epsilon}\int_{D(\epsilon)}\frac{1+\sin(x)+\sin(y)}{\sqrt{x^2+y^2}}d(x,y)$$

My attempt: I think I can simply apply DCT? But I'm not sure because the limit also determines the integration boundaries? However, if I apply DCT, I dominate the integrand by the function $$(x,y)\mapsto\frac{1+x+y}{\sqrt{x^2+y^2}}$$ Performing change of variables to the cylindrical coordinates, we obtain $$\int_0^{2\pi} 1+\cos\theta+ \sin\theta \,d\theta \int_0^\epsilon r\,dr$$ So we have that the integrand can be dominated by an integrable function. Now I only have to determine the limit of the integrand. But now, I'm not sure how this limit can be calculated? Any tips? Thank you!

Answer 1

One doesn't need the DCT to evaluate the limit. Rather a simple rescaling of variables suffices. To see this, let $I_\varepsilon$ be given by the integral

$$I_\varepsilon = \frac1\varepsilon\int_{-\varepsilon}^\varepsilon \int_{-\sqrt{\varepsilon^2-y^2}}^{\sqrt{\varepsilon^2-y^2}}\frac{1+\sin(x)+\sin(y)}{\sqrt{x^2+y^2}}\,dx\,dy\tag1$$

Enforcing the substitutions $x\mapsto \varepsilon x$ and $y\mapsto \varepsilon y$ in $(1)$, we obtain

$$I_\varepsilon = \int_{-1}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{1+\sin(\varepsilon x)+\sin(\varepsilon y)}{\sqrt{x^2+y^2}}\,dx\,dy\tag2$$

Note that since $|\sin(x)|\le x$, we have

$$\left|\int_{-1}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{\sin(\varepsilon x)}{\sqrt{x^2+y^2}}\,dx\,dy\right|\le |\varepsilon|\int_{-1}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{|x|}{\sqrt{x^2+y^2}}\,dx\,dy=2|\varepsilon|$$

Hence, as $\varepsilon\to 0$ we have

$$\lim_{\varepsilon\to0}I_\varepsilon=\int_{-1}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{1}{\sqrt{x^2+y^2}}\,dx\,dy=2\pi$$

And we are done!

Answer 2

Mark's answer is of course correct, but if you must:

Consider $A_n = \{ x | {1 \over n+1} \le \|x\| < {1 \over n} \}$, then ${1 \over \|x\|} \le \sum_n (n+1) m A_n$ and $mA_n = \pi ({1 \over n^2} - {1 \over (n+1)^2})$. It is not hard to show that the sum is indeed summable and so ${1 \over \|x\|}$ is dominated by the integrable function $\sum_n (n+1) 1_{A_n}$.

Hence $x \mapsto {1 \over \|x\|}$ is integrable and so $|{1+\sin x_1 + \sin x_2 \over \|x\| }| \le {3 \over \|x\|}$ is integrable.