## Proof of Formula for Volume Integral of Lapacian of Function Divided by r

Theorem
Let
$V$
be a region enclosed by a surface
$S$
.
If
$\mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k}$
, and let
$f=f(x,y,z)$
be continuous with first and second partial derivatives. Then
$\int \int \int_V \frac{\nabla^2 f}{r} dV +A = \in \int_S (\frac{\mathbf{\nabla} f}{r} - f \mathbf{\nabla}(\frac{1}{r})) \cdot \mathbf{n} dS ( \left\{ \begin{array}{cc} 0 & (0,0,0) \notin S \\ 4 \pi & (0,0,0) \in S \end{array} \right.$

where
$A = \left\{ \begin{array}{cc} 0 & (0,0,0) \notin S \\ 4 \pi f(0,0,0) & (0,0,0) \in S \end{array} \right.$

Proof
Apply the Divergence Theorem
$\int \int \int \int_V \mathbf{\nabla} \cdot \mathbf{F} dV = \int \int_S \mathbf{F} \cdot \mathbf{n} dS$
with
$\mathbf{F} = \frac{\mathbf{\nabla} f}{r} - f \mathbf{\nabla}(\frac{1}{r})$
to give
\begin{aligned} \int \int_S (\frac{\mathbf{\nabla} f}{r} - f \mathbf{\nabla}(\frac{1}{r})) \cdot \mathbf{n} dS &= \int \int \int \int_V \mathbf{\nabla} \cdot (\frac{\mathbf{\nabla} f}{r} - f \mathbf{\nabla}(\frac{1}{r})) dV \\ &= \int \int \int \int_V \mathbf{\nabla} \cdot (\frac{\mathbf{\nabla} f}{r}) - \mathbf{\nabla} \cdot (f \mathbf{\nabla}(\frac{1}{r})) dV \\ &= \end{aligned}

Use the identities
$\mathbf{\nabla} (f \mathbf{F}) = \mathbf{F} \cdot (\mathbf{\nabla} f) + f \mathbf{\nabla} \cdot \mathbf{F}$

$\mathbf{\nabla} r^n = n r^{n-2} \mathbf{r}$

The volume integral becomes
\begin{aligned}\int \int \int \int_V \mathbf{\nabla} \cdot \mathbf{F} dV &= \int \int \int_V (\mathbf{\nabla} f) \cdot (\mathbf{\nabla} (\frac{1}{r})) +\frac{1}{r} \nabla^2 f - \cdot (\mathbf{\nabla} (\frac{1}{r})) \cdot (\mathbf{\nabla } f)-f \nabla^2 (\frac{1}{r}) dV \\ &= \int \int \int_V \frac{1}{r} \nabla^2 f 0 f \mathbf{\nabla} \cdot ( \mathbf{\nabla} (\frac{1}{r})) dV \\ &= \int \int \int_V \frac{ \nabla^2 f} {r} dV + \int \int \int_V f \mathbf{\nabla} \cdot \frac{\mathbf{r}}{r^3} dV\end{aligned}

$\mathbf{\nabla} \cdot \frac{\mathbf{r}}{r^3} =0$
everywhere except the origin so the second integral is zero if the origin is outside
$S$
.
If the origin is inside
$S$
, since
$\mathbf{\nabla} \cdot \frac{\mathbf{r}}{r^3} =0$
except the origin we can write
$\int \int \int_V f \mathbf{\nabla} \cdot (\frac{\mathbf{r}}{r^3}) dV = f(0,0,0) \int \int \int_V \mathbf{\nabla} \cdot (\frac{\mathbf{r}}{r^3}) dV = 4 \pi f(0,0,0)$

Using Gauss's Theorem.