TITLE: When is $f(x,y)=|x|^a |y|^b$ on the unit disk integrable?
QUESTION [0 upvotes]: When is $f(x,y)=|x|^a |y|^b$ (for $x,y \neq 0; 0$ otherwise) on the unit disk  in $\mathbb{R}^2$ integrable?

REPLY [3 votes]: Since you tagged measure theory, I assume you mean Lebesgue integrable. If $a,b>-1$ then $f$ is integrable because
\begin{align}
\int_Df\, dm_2 \leq \int_{[-1,1]^2}f\, dm_2 = 4\int_{[0,1]^2}f\, dm_2 = 4 \left(\int_0^1x^a\, dx\right)\left(\int_0^1y^b\, dy\right)<\infty.
\end{align}
Conversely, if $a\leq -1$ or $b\leq -1$, then $f$ is not integrable, because the square $[-\epsilon,\epsilon]^2$ is contained in the unit disk $D$ if $\epsilon>0$ is small enough, so
\begin{align}
\int_Df\,dm_2 \geq \int_{[-\epsilon,\epsilon]^2}f\, dm_2 = 4 \left(\int_0^{\epsilon}x^a\,dx\right)\left(\int_0^{\epsilon}y^b\,dy\right) = \infty,
\end{align}
where the last line is because $a\leq -1$ or $b\leq -1$ implies one of the bracketed terms is infinite. In other words, $f$ is integrable on $D$ if and only if $a,b>-1$.
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