I am reading the PDE book of Evans and I am stuck at understanding the following inequality which is used:
$$\int_{B(0,\epsilon)} |\Phi(y)| dy \leq \begin{cases} C \epsilon^2 |\log(\epsilon)| \quad (n=2)\\ C \epsilon^2 \quad (n \geq 3)\end{cases}$$
Note that $B(0,\epsilon)$ denotes the closed ball centered at zero with radius $\epsilon$. Note further that $\Phi$ denotes the fundamental solution of Laplace's equation, i.e.
$$\Phi(x):= \begin{cases} -\frac{1}{2\pi}\log(|x|) \quad n=2 \\ \frac{1}{n(n-2)\alpha(n)}\frac{1}{|x|^{n-2}} \quad n\geq 3\end{cases}$$
Any advice on how to arrive at this inequality would be very helpful! Thank you in advance!
In the case $n=2$, we have in polar coordinates where $ dA= r dr d\theta$
$\int \int \ln (r) dA= 2\pi \int_{r=0}^{r=\epsilon} \ln( r ) r \ dr=C \int_{u=0}^{u=\epsilon^2} \ln u \ du$ where $u=r^2$. This simplifies to $u\ln u- u ]_{u=0}^{u=\epsilon^2} =\epsilon^2 \ln \epsilon^2 -\epsilon^2$. The last term $\epsilon^2$ is negligible in comparison to the dominant term $\epsilon^2 \ln \epsilon^2 $