Integrales impropias en el plano

ENUNCIADO. Calcúlese la integral $\displaystyle \int \int_{\mathcal{D}} \dfrac{1}{\sqrt[3]{xy}}\,dx\,dy$, donde $\mathcal{D}$ es el cuadrado de lado igual a la unidad: $[0,1]\times [0,1]$, esto es $$\mathcal{D}=\{P(x,y) \in \mathbb{R}^2: 0\le x \le 1; 0\le y\le 1\}$$

SOLUCIÓN.
$\displaystyle \int \int_{\mathcal{D}} \dfrac{1}{\sqrt[3]{xy}}\,dx\,dy=\lim_{(\mu,\delta)\rightarrow (0,0)}\,\int_{\mu}^{1-\mu} \int_{\delta}^{1-\delta} \dfrac{1}{\sqrt[3]{xy}}\,dx\,dy=$
$\displaystyle =\lim_{(\mu,\delta)\rightarrow (0,0)}\,\int_{\mu}^{1-\mu}\,\dfrac{1}{\sqrt[3]{x}} \left( \int_{\delta}^{1-\delta} \dfrac{1}{\sqrt[3]{y}}\,dy \right) \,dx=$
$\displaystyle=\dfrac{3}{2}\,\lim_{\delta\rightarrow 0}\,\left(\,(1-\delta)^{2/3}-\delta^{2/3} \right) \, \lim_{\mu\rightarrow 0}\, \int_{\mu}^{1-\mu}\,\dfrac{1}{\sqrt[3]{x}} \,dx$
$\displaystyle =\left(\dfrac{3}{2}\right)^{2}\,\lim_{\delta\rightarrow 0}\,\left(\,(1-\delta)^{2/3}-\delta^{2/3} \right) \cdot \lim_{\mu\rightarrow 0}\,\left( \,(1-\mu)^{2/3}-\mu^{2/3} \right)=\dfrac{9}{4}$
$\square$