How to show $|\log | x||^{2 \varepsilon} \cdot|x|^{-2}$ is in $L_{\mathrm{loc}}^{r}$ for every $rLet $\Omega=\mathbb R^{n}$. I wanted to show that $|\log | x||^{2 \varepsilon} \cdot|x|^{-2}$ is in $L_{\mathrm{loc}}^{r}$ for every $r<n / 2$.
Only problem is with compact set including point $0$.
I thought to use polar coordinates.
Consider $B=\overline{B(0,R)}$ and $n\alpha(n)=$ surface area of unit sphere.
$$\begin{align}\int_B|\log | x||^{(n-\delta) \varepsilon} \cdot|x|^{-(n-\delta)}dx&=n\alpha(n)\int_0^R|\log \varrho|^{(n-\delta) \varepsilon} \cdot \varrho^{-(n-\delta)}\varrho^{n-1}d\varrho\\
&=n\alpha(n)\int_0^R|\log \varrho|^{(n-\delta) \varepsilon} \cdot \varrho^{\delta-1}d\varrho
\end{align}$$
How to show that last integral is finite for any $\delta$ such that $0\le\delta\le n$?
Any help will be appreciated.
 A: Assuming $R\ge1$, split $[0,R]$ into $[0,1]$ and $(1,R]$. If $0<R<1$ the integral $\int_0^R$ is finite if $\int_0^1$ is finite as the integrand is non-negative.
There is a closed form on $[0,1]$. Letting $u=-\delta\log\varrho$, we can drop the absolute value to obtain $$\int_0^1|\log\varrho|^{(n-\delta)\varepsilon}\varrho^{\delta-1}\,d\varrho=\frac1\delta\int_0^\infty\left(\frac u\delta\right)^{(n-\delta)\varepsilon}e^{-u}\,du=\frac{\Gamma(1+(n-\delta)\varepsilon)}{\delta^{1+(n-\delta)\varepsilon}}<\infty.$$ On $(1,R]$, the integrand becomes $(\log\varrho)^{(n-\delta)\varepsilon}\varrho^{\delta-1}$ with derivative $$(n-\delta)\varepsilon(\log\varrho)^{(n-\delta)\varepsilon-1}\varrho^{\delta-2}+(\delta-1)(\log\varrho)^{(n-\delta)\varepsilon}\varrho^{\delta-2}.$$ Setting to zero yields the argmax $\varrho^*=\exp((n-\delta)\varepsilon/(1-\delta)).$ This is greater than $1$ if $\delta<1$, so if $\delta\ge1$ the global maximum in $(1,R]$ occurs at $\varrho=R$. Thus $$\int_1^R(\log\varrho)^{(n-\delta)\varepsilon}\varrho^{\delta-1}\,d\varrho<(R-1)\max\left\{(\log\varrho^*)^{(n-\delta)\varepsilon}(\varrho^*)^{\delta-1},(\log R)^{(n-\delta)\varepsilon}R^{\delta-1}\right\}<\infty.$$ The original integral in $[0,R]$ is therefore also finite.
