\begin{align} \frac{\partial}{\partial x} \ln | \vec{r} | &= \frac{1}{2} \frac{\partial}{\partial x} \ln \left( x^2+y^2 \right) = \frac{x}{x^2+y^2} = \frac{x}{r^2} \\ \frac{\partial}{\partial y} \ln | \vec{r} | &= \frac{1}{2} \frac{\partial}{\partial y} \ln \left( x^2+y^2 \right) = \frac{y}{x^2+y^2} = \frac{y}{r^2} \end{align} なので、 \begin{align} \vec{\nabla} \ln | \vec{r} | = \left( \frac{x}{r^2}, \frac{y}{r^2} \right) = \frac{\vec{r}}{r^2} \end{align}
$ x = r \cos \theta, y = r \sin \theta, r = \sqrt{x^2+y^2}, \tan \theta = y/x $ より、 \begin{align} \frac{\partial r}{\partial x} &= \frac{x}{r} = \cos \theta \\ \frac{\partial r}{\partial y} &= \frac{y}{r} = \sin \theta \\ \frac{\partial \theta}{\partial x} &= - \frac{y}{x^2} \cos^2 \theta = - \frac{\sin \theta}{r} \\ \frac{\partial \theta}{\partial y} &= \frac{1}{x} \cos^2 \theta = \frac{\cos \theta}{r} \end{align} であるから、 \begin{align} \frac{\partial}{\partial x} &= \frac{\partial r}{\partial x} \frac{\partial}{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial}{\partial \theta} = \cos \theta \frac{\partial}{\partial r} - \frac{\sin \theta}{r} \frac{\partial}{\partial \theta} \\ \frac{\partial}{\partial y} &= \frac{\partial r}{\partial y} \frac{\partial}{\partial r} + \frac{\partial \theta}{\partial y} \frac{\partial}{\partial \theta} = \sin \theta \frac{\partial}{\partial r} + \frac{\cos \theta}{r} \frac{\partial}{\partial \theta} \end{align} であり、 \begin{align} \frac{\partial^2}{\partial x^2} &= \left( \cos \theta \frac{\partial}{\partial r} - \frac{\sin \theta}{r} \frac{\partial}{\partial \theta} \right) \left( \cos \theta \frac{\partial}{\partial r} - \frac{\sin \theta}{r} \frac{\partial}{\partial \theta} \right) \\ &= \cos^2 \theta \frac{\partial^2}{\partial r^2} + \frac{\sin \theta \cos \theta}{r^2} \frac{\partial}{\partial \theta} - \frac{\sin \theta \cos \theta}{r} \frac{\partial^2}{\partial r \partial \theta} + \frac{\sin^2 \theta}{r} \frac{\partial}{\partial r} - \frac{\sin \theta \cos \theta}{r} \frac{\partial^2}{\partial \theta \partial r} \\ &\ \ \ \ + \frac{\sin \theta \cos \theta}{r^2} \frac{\partial}{\partial \theta} + \frac{\sin^2 \theta}{r^2} \frac{\partial^2}{\partial \theta^2} \\ &= \cos^2 \theta \frac{\partial^2}{\partial r^2} +2 \frac{\sin \theta \cos \theta}{r^2} \frac{\partial}{\partial \theta} -2 \frac{\sin \theta \cos \theta}{r} \frac{\partial^2}{\partial r \partial \theta} + \frac{\sin^2 \theta}{r} \frac{\partial}{\partial r} + \frac{\sin^2 \theta}{r^2} \frac{\partial^2}{\partial \theta^2} \\ \frac{\partial^2}{\partial y^2} &= \left( \sin \theta \frac{\partial}{\partial r} + \frac{\cos \theta}{r} \frac{\partial}{\partial \theta} \right) \left( \sin \theta \frac{\partial}{\partial r} + \frac{\cos \theta}{r} \frac{\partial}{\partial \theta} \right) \\ &= \sin^2 \theta \frac{\partial^2}{\partial r^2} - \frac{\sin \theta \cos \theta}{r^2} \frac{\partial}{\partial \theta} + \frac{\sin \theta \cos \theta}{r} \frac{\partial^2}{\partial r \partial \theta} + \frac{\cos^2 \theta}{r} \frac{\partial}{\partial r} + \frac{\sin \theta \cos \theta}{r} \frac{\partial^2}{\partial \theta \partial r} \\ &\ \ \ \ - \frac{\sin \theta \cos \theta}{r^2} \frac{\partial}{\partial \theta} + \frac{\cos^2 \theta}{r^2} \frac{\partial^2}{\partial \theta^2} \\ &= \sin^2 \theta \frac{\partial^2}{\partial r^2} -2 \frac{\sin \theta \cos \theta}{r^2} \frac{\partial}{\partial \theta} +2 \frac{\sin \theta \cos \theta}{r} \frac{\partial^2}{\partial r \partial \theta} + \frac{\cos^2 \theta}{r} \frac{\partial}{\partial r} + \frac{\cos^2 \theta}{r^2} \frac{\partial^2}{\partial \theta^2} \end{align} したがって、 \begin{align} \Delta &= \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \\ &= \frac{\partial^2}{\partial r^2} + \frac{1}{r} \frac{\partial}{\partial r} + \frac{1}{r^2} \frac{\partial^2}{\partial \theta^2} \\ &= \frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial}{\partial r} + \frac{1}{r^2} \frac{\partial^2}{\partial \theta^2} \end{align} を得る。
$ | \vec{r} | \neq 0 $ のとき、 (2) で得た式をつかって、 \begin{align} \Delta \ln | \vec{r} | = \frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial}{\partial r} \ln r = \frac{1}{r} \frac{\partial}{\partial r} r \frac{1}{r} = 0 \end{align} を得る。
領域 $ | \vec{r} | \leq a $ を $S$ , 閉曲線 $ | \vec{r} | = a $ を $C$ で表して、 次のように計算できる： \begin{align} I &= \iint_S \Delta \ln | \vec{r} | dx dy \\ &= \iint_S \vec{\nabla} \cdot \frac{\vec{r}}{r^2} \ dx dy \\ &= \iint_S \left( \frac{\partial}{\partial x} \frac{x}{r^2} + \frac{\partial}{\partial y} \frac{y}{r^2} \right) dx dy \\ &= \oint_C \left( - \frac{y}{r^2} dx + \frac{x}{r^2} dy \right) \\ &= \int_0^{2 \pi} \left( - \frac{a \sin \theta}{a^2} \cdot (-a \sin \theta) + \frac{a \cos \theta}{a^2} \cdot (a \cos \theta) \right) d \theta \\ &= \int_0^{2 \pi} d \theta \\ &= 2 \pi \end{align}
(3), (4) より、 \begin{align} \Delta \ln | \vec{r} | = 2 \pi \delta^2 ( \vec{r} ) \end{align} がわかる。 ただし、 $\delta^2 ( \cdot )$ は、 2次元のディラックのデルタ関数である。