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.