Consider $\displaystyle \int \int _{\Omega} \frac{\sin(x)\sin(y) dx dy}{(x+y)^{p}}$, where $\Omega = \{(x,y)\in\mathbb{R}^2 : x+y \ge 1\}$.
Now we need to prove that integral is divergent for all $p \in \mathbb{R}$.

My attempt : if it diverges , we need to find an $\{\Omega_{n} \}$ and $\{\Phi_{n}\}$, such as $\displaystyle \lim_{n \to \infty} \int\int_{\Omega_{n}} \frac{\sin(x)\sin(y)}{(x+y)^{p}} = \infty \ne const = \lim_{n \to \infty} \int \int _{\Phi_{n}}\frac{\sin(x)\sin(y)}{(x+y)^{p}}$.

Let's simplify this function : $\sin(x)\sin(y) = \frac{1}{2}(\cos(x-y) -\cos(x+y))$, so let $x-y = u$ and $x+y = v$, so $\mathfrak{J} = \frac{1}{2}$ (Jacobian). So we have (with no respect to Jacobian) : $$\displaystyle \lim_{n \to \infty} \int_{1}^{\infty} \int_{-\infty}^{+\infty} \frac{\cos(u)-\cos(v)}{v^{p}}$$

Now let's $-2\pi n\le u \le 2\pi n$, so we will have $$\displaystyle \lim_{n \to \infty} \int_{-2\pi n}^{2\pi n} \int_{1}^{\infty}f(u,v) = \lim_{n \to \infty} 4\pi n\int_{1}^{\infty}\frac{\cos(v)}{v^{p}}$$

Now there I'm stuck , how can we show two $\Omega_{n}$ and $\Phi_{n}$ such as they have different values.

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    $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 14 '18 at 19:45
  • $\begingroup$ Are you going to tell us anything about $p?$ $\endgroup$ – zhw. Mar 14 '18 at 19:47
  • $\begingroup$ @zhw. we need to show that answer doesn't depend of $p$ $\endgroup$ – openspace Mar 14 '18 at 19:47
  • $\begingroup$ There still are some things to fix: many integrals do not have differentials attached (so it is hard to understand where the variables range) and there is a $\lim_{n\to +\infty}$ of something that does not depend on $n$. Additionally, is $$\Omega=\{(x,y)\color{red}{\in\mathbb{R}^+\times\mathbb{R}^+}:x+y\geq 1\}$$ or something else? $\endgroup$ – Jack D'Aurizio Mar 14 '18 at 19:50
  • $\begingroup$ OK, then tell us about $p$ in the statement please. Also tell us what it means for such a double integral to be divergent. We need a precise definition. $\endgroup$ – zhw. Mar 14 '18 at 19:51

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