Why Does the Fundamental Solution of Laplace's Equation Cause Infinity in n=3?

[ak416]

Homework Statement

From the book Evans-PDE, p.24, equation (12),
It is written that

$$C ||D^2f||_{L_{\infty}(R^n)} \int_{B(0,\epsilon)}|\Phi(y)|dy<br /> \\ \leq \begin{cases} C \epsilon^2 |\log{}\epsilon| & (n=2) \\ C \epsilon^2 & (n \geq 3) \end{cases}$$

How is this?

Homework Equations

$$\Phi(y) = \begin{cases} -\frac{1}{2\pi}\log{}|y| & (n=2) \\ \frac{1}{n(n-2)\alpha(n)} \frac{1}{|y|^{n-2}} & (n \geq 3) \end{cases}$$
for $$y \in \mathbb{R}^n-0$$ (the fundamental solution of Laplace's equation).

$$\alpha(n)$$ is the volume of the unit ball in $$\mathbb{R}^n$$.

$$C$$ is a constant.

The Attempt at a Solution

Take n=3. Then,

$$\int_{B(0,\epsilon)}|\Phi(y)|dy = C\int_{0}^{\epsilon}\int_{0}^{2\pi}\int_{0}^{2\pi}\frac{1}{r} d\theta d\phi dr$$

$$= C \int_{0}^{\epsilon}\frac{1}{r} dr$$

$$= C (\log{}(\epsilon)-\log{}(0)) = \infty$$

What am I doing wrong?

 

[hunt_mat]

I think you got your Jacobian transformation wrong. If memory serves:
$$dxdydz=r^{2}\sin\theta drd\theta d\phi$$

 

[ak416]

ya you're right. I didn't account for the change of coordinates. thanks.