Why is the value of this line integral constant Consider the line integral given by $$\int_C \frac{(x+y)\,dx-(x-y)\,dy}{x^2+y^2}$$ where $C$ is any simple closed curve around the origin. Can someone explain, without using complex analysis, why this is always $-2 \pi$?
 A: Do a parametrization of your curve
$$
x(t) = r(t) \cos t \\
y(t) = r(t) \sin t
$$
then 
\begin{align}
dx &= (r'\cos t - r\sin t) dt \\
dy &= (r'\sin t + r\cos t) dt
\end{align}
so
\begin{align}
I &= \int_C \frac {(x+y)dx - (x-y)dy}{x^2+y^2} = \\
&= \int_0^{2\pi} \frac {r(\cos t + \sin t)(r'\cos t - r\sin t) dt - r(\cos t - \sin t)(r'\sin t + r\cos t) dt}{r^2} = \\
&= \int_0^{2\pi} \frac {r'-r}r dt = \int_0^{2\pi} d(\ln r) - \int_0^{2\pi} dt = \left . \ln r \right|_{0}^{2\pi} - 2\pi = r(2\pi) - r(0) - 2\pi = -2\pi
\end{align}
A: The form $$\omega = \frac{(x+y)\,dx - (x-y)\,dy}{x^2+y^2}$$ is closed. Stokes' theorem asserts
$$\int_{\partial V} \omega = \int_V d\omega$$
for every bounded open set $V \subset \mathbb{R}^2 \setminus \{(0,0)\}$ with sufficiently smooth boundary, so
$$\int_C \omega = \int_{x^2+y^2 = \varepsilon} \omega$$
if we choose $\varepsilon$ so small that $C$ is entirely outside the disk $x^2+y^2 \leqslant \varepsilon$, since then the two curves - with proper orientation - make up the boundary of a region that is topologically an annulus.
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{{\cal I}
     \equiv
     \int_{C}{\pars{x + y}\,\dd x - \pars{x - y}\,\dd y \over x^{2} + y^{2}}
      = -2\pi:\ {\large ?}}.\quad$
With
$\ds{r \equiv \root{x^{2} + y^{2}}\,,\quad\nabla\ln\pars{r} = {\vec{r} \over r^{2}}}$

\begin{align}
{\cal I}&=\int_{C}\pars{{x + y \over r^{2}},{y - x \over r^{2}}}\cdot\dd\vec{r}
=\int_{C}\pars{%
\partiald{\ln\pars{r}}{x} + \partiald{\ln\pars{r}}{y},
\partiald{\ln\pars{r}}{y} - \partiald{\ln\pars{r}}{x}}\cdot\dd\vec{r}
\\[3mm]&=\int_{S}\braces{%
\partiald{}{x}\bracks{\partiald{\ln\pars{r}}{y} - \partiald{\ln\pars{r}}{x}}
-
\partiald{}{y}\bracks{\partiald{\ln\pars{r}}{x} + \partiald{\ln\pars{r}}{y}}}\,\dd S
\\[3mm]&=-\,\,\,\,\,\overbrace{\int_{S}\nabla^{2}\ln\pars{r}\,\dd S}^{\ds{=\ 2\pi}}\tag{1}
=\color{#00f}{\Large -2\pi}
\end{align}

In this answer, we show a derivation of $\nabla^{2}\ln\pars{r} = 2\pi\,\delta\pars{\vec{r}}$ which, indeed, is a well known result.
