Showing that the unsigned area bounded by plane curve $\gamma$ is $-\frac1{4\pi}\oint_\gamma\oint_\gamma\vec{dx}\cdot\vec{dy}\log(\|x-y\|^2)$ Let $\gamma$ be a curve in the plane. I wish to show:
$$A=\frac{-1}{4\pi}\oint_\gamma\oint_\gamma\overrightarrow{dx}\cdot\overrightarrow{dy}\log\left(\|x-y\|^2\right),$$
where here $A$ is the unsigned area bounded by $\gamma$. For instance, the unsigned area of the lemniscate is strictly positive whereas its signed area is zero.



*

*I believe this formula is true because I have checked it for rectangles and ellipses, and numerically for a variety of other curves. Nevertheless I haven't been able to prove it myself or find a proof.

*I believe Stokes' theorem is not helpful because I would guess proofs based on it would lead to signed area.

*The physical motivation for this comes from thinking about electromagnetism in 1+1 dimensions, which might be a helpful starting point.

 A: The statement is true when $\gamma$ is a positively oriented simple closed curve bounding some Jordan domain $\Omega$. 
Let $\mathcal{I}$ be the integral at hand. Identify Euclidean plane $\mathbb{R}^2$ with the complex plane $\mathbb{C}$ and introduce complex coordinates $x,y$:
$$
\begin{cases}
\vec{x} = (x_1,x_2) & \leftrightarrow & x = x_1 + x_2 i\\
\vec{y} = (y_1,y_2) &\leftrightarrow  & y = y_1 + y_2 i
\end{cases}$$
Let $z = y - x$. For any $\vec{\rho} = (\rho_1,\rho_2) \leftrightarrow \rho = \rho_1 + i\rho_2 \in \mathbb{C}$, we will use the notation $\Omega + \rho$ and $\gamma + \rho$ to denote the image of $\Omega$ and $\gamma$ under translation $\vec{\rho}$.
In terms of complex coordinates, we have
$d\vec{x} \cdot d\vec{y} = \frac12 \left(dx d\bar{y} + dy d\bar{x}\right)$. This leads to
$$\begin{align}
\mathcal{I} 
&= -\frac{1}{8\pi} \int_{\gamma \times \gamma} \log(z\bar{z}) (dx d\bar{y} + dyd\bar{x})
= -\frac{1}{4\pi}\Re \left[\int_{\gamma\times\gamma} \log(z\bar{z}) dx d\bar{y}\right]\\
&= -\frac{1}{4\pi}\Re\left[\int_{x \in \gamma} \left(\int_{z \in \gamma - x}\log(z\bar{z}) d\bar{z}\right) dx\right]
\end{align}\tag{*1}
$$
Apply Stoke's theorem (the version for complex coordinates) to the inner integral, we obtain
$$\begin{align}
\mathcal{I}
&= -\frac{1}{4\pi}\Re\left[\int_{x\in\gamma} \left(\int_{z \in \Omega - x}
\left(\frac{dz}{z} + \frac{d\bar{z}}{\bar{z}}\right)\wedge d\bar{z}\right) dx\right]\\
&=  \frac{1}{4\pi}\Re\left[\int_{x\in\gamma} \left(
\int_{z\in \Omega - x}\frac{d\bar{z} \wedge dz}{z}
\right) dx\right]\\
&= \frac{1}{4\pi}
\Re\left[\int_{x\in\gamma} \left(
\int_{y\in \Omega}\frac{d\bar{y} \wedge dy}{y-x}
\right) dx\right]\\
&= \frac{1}{4\pi}
\Re\left[
\int_{y\in\Omega}\left(\int_{x\in\gamma}\frac{dx}{y-x}\right) d\bar{y} \wedge dy
\right]
\end{align}
$$
By Cauchy's integral formula, we have $$\int_{x\in\gamma}\frac{dx}{y-x} = -2\pi i\quad\text{ for } y \in \Omega$$ As a result,
$$\begin{align}\mathcal{I} 
&= \frac{1}{4\pi} \Re\left[ \int_{y \in \Omega} (-2\pi i)(2i dy_1 \wedge dy_2 )\right]
= \Re\left[ \int_{y \in \Omega} dy_1 \wedge dy_2 \right]\\
&= \Re\left[\verb/Area/(\Omega)\right] = \verb/Area/(\Omega)
\end{align}
$$

Update - Generalization to non-simple closed curves.
For non-simple closed curve $\gamma$ and $y \not\in \gamma$, let $W(\gamma;y)$ be the winding number of $\gamma$ around $y$. We have following generalization of Cauchy integral formula:
$$\int_\gamma \frac{dx}{y-x} = -2\pi i W(\gamma;y)\tag{*2}$$
When $\gamma$ satisfies following conditions:


*

*$\gamma$ can be decomposed into finitely many curve segments, the segments either completely coincides or their interior (as a curve segment) are disjoint from each other.

*$\gamma$ divides $\mathcal{C}\setminus \gamma$ into finitely many connected  components, the boundaries of these components are finite combinations of curve segments from step $1$.


We can break any contour integral over $\gamma$ to integral combinations of contour integrals over boundaries of connected components in step $2$.
We can apply Stoke's theorem to the individual components and recombine the results. 
Apply this procedure to inner contour integral in $(*1)$ and with help of $(*2)$, we obtain:
$$
\mathcal{I}
= \frac{1}{4\pi}
\Re\left[\int_{x\in\gamma} \left(
\int_{y\in \mathbb{C}\setminus\gamma}\frac{W(\gamma;y)}{y-x} d\bar{y} \wedge dy
\right) dx\right]
= \int_{y\in \mathbb{C}\setminus\gamma} W(\gamma;y)^2 dy_1 \wedge dy_2$$
Recall winding number is constant over each connected component. If $\Omega_1, \ldots, \Omega_m$ are the connected components of $\mathbb{C}\setminus\gamma$ with non-zero winding numbers and $W_i$ is the winding number for $\Omega_i$,
we can rewrite last expression as
$$\mathcal{I} = \sum_{i=1}^m W_i^2 \verb/Area/(\Omega_i)$$
The integral $\mathcal{I}$ is simply a weighted sum of the areas of connected components and the weight equals to the square of corresponding winding number.
In the special case where all $|W_i| \le 1$, above formula simplifies to
$$\mathcal{I} = \sum_{i=1}^m \verb/Area/(\Omega_i)$$
$\mathcal{I}$ becomes the unsigned area of those connected components whose winding number is non-zero.
