evans book pde estimate question I'm reading Evans' book on PDE and I'm having troubles understanding one estimate.
He defines the fundamental solution to Laplace' equation as 
$$
\Phi(x) = \begin{cases} -\frac{1}{2\pi} \, \log(|x|), \, & n=2, \\ \frac{1}{n \, (n-2) \, \omega_n} \, \frac{1}{|x|^{n-2}}, \, & n\geq 3, \end{cases} $$
where $\omega_n$ is the volume of the $n$-ball.
For the solution of Poisson's equation $ -\Delta u = f$ he computes the Laplace acting on the convolution of $f$ and $\Phi$, involving this estimate:
$$ \bigg|\int_{B(0,\varepsilon)} \Phi(y) \, \Delta_y f(x-y) \, dy \bigg| \leq C \, \lVert D^2f \rVert_{L^\infty} \int_{B(0,\varepsilon)} |\Phi(y)| \, dy \leq \begin{cases} C \, \varepsilon^2 \, |\log(\varepsilon)|, & n=2, \\ C \, \varepsilon^2, & n\geq 3. \end{cases} $$
How do you obtain the last inequality?
 A: Hint: use polar coordinates to integrate the green's function on the $\epsilon$ ball. 
Edit: Observe for $n\geq 3$ we have
\begin{align}
\int_{B(0, \epsilon)} \frac{dx}{|x|^{n-2}} = \int^\epsilon_0 \int_{|x|=r} \frac{dS(x)}{|x|^{n-2}}\ dr = C\int^\epsilon_0 \frac{r^{n-1}}{r^{n-2}}\ dr = C'\epsilon^2.
\end{align}
A: following jacky chong's hint, i came up with something, and please correct me if I'm wrong.
assuming $\varepsilon < 1$, then $\big|\log |x|\big| = - \log|x|, \, \forall \, |x| \in (0,\varepsilon)$. hence integrating (by parts) we get
$$\int \big|\log|x|\big| \, dx 
= - C' \int \log(r) \, r \, dr 
= C' \, \Big( -\frac{\varepsilon^2}{2} \, \log (\varepsilon) + \frac{\varepsilon^2}{4} \big)
% = C' \, \Big( \varepsilon^2 \, \big|\log (\varepsilon)\big| + \frac{\varepsilon^2}{4} \Big) 
\leq C \, \varepsilon^2 \, \big|\log (\varepsilon)\big| $$
since you can choose a constant $C$ such that $C \, \varepsilon^2 \, \big|\log (\varepsilon)\big| \geq \varepsilon^2$ for all $\varepsilon < 1$.
but for $n \geq 3$ i still have no clue
