When is $|x|^a|\log{|x|}|^b$ integrable over $B(0,\frac{1}{2})$

Question

I am preparing for my final exam and I got stuck at one of the problems on integration in polar coordinates from Folland:

For which values of $a$ and $b$ is $f(x)=|x|^a|\log{|x|}|^b$ integrable over the open ball $B(0,\frac{1}{2})$?

Attempt: By changing to polar coordinates this is equivalent of asking when is $$\int_0^{\frac{1}{2}}r^a|\log{r}|^br^{n-1}dr < \infty$$

Making the substitution $u=-\log{r}$, this is equivalent of asking when is $$\int_{\log{2}}^{\infty}t^be^{-t(a+n)} du < \infty$$

Since exponential decays faster than any polynomial, my guess would be this happens if and only if $a+n>0$ or $a+n=0;b<-1$.

But how do I show this rigourously? I'd really appreciate any insights.

Answer 1

• If $a+n<0$ then there is $T>0$ large enough so that $t^be^{-t(a+n)}>1$. Then $f(t)=t^be^{-t(a+b)}>1$ for all $t\geq T$ sand so $f$ is not integrable over $[T,\infty)$.

• If $a+n=0$, we have the classical case of $\int^\infty_c t^b$, $c>0$, which converges only if $b<-1$.

• If $a+b>0$, then there is $T>0$ large enough so that $t^be^{-\frac{t}{2}(a+n)t}<1$ for all $t\geq T$. Hence $f(t)\leq e^{-\frac{t}{2}(a+n)t}$and so, $f$ is integrable.