Compute $\oint\limits_{\partial\Sigma} \frac{xdy-ydx}{x^{2}+y^{2}}\,$

Question

Assume $ \partial\Sigma$ is a positively oriented, piecewise smooth, simple closed curve in a plane and let $ \Sigma$ be the region bounded by $ \partial\Sigma$ containing $(0,0)$, then compute $$\oint\limits_{\partial\Sigma} \frac{xdy-ydx}{x^{2}+y^{2}}\,$$

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The functions are not defined at the origin, so we can't use Green's theorem, but:

Let $B_\epsilon(0,0):=\left\{(x,y):x^2+y^2 \le \epsilon^2\right\}$ be a circle centered at the origin, and $ \Sigma_\epsilon=\Sigma \setminus B_\epsilon(0,0)$, so I think this is true to claim that $$\partial \Sigma_\epsilon=\partial \Sigma \cup \partial B_\epsilon(0,0)$$

So:

$$0=\iint \limits_{\Sigma_\epsilon}0 \; dxdy=\oint\limits_{\partial\Sigma_\epsilon} \frac{xdy-ydx}{x^{2}+y^{2}}\,=\oint\limits_{\partial \Sigma} \frac{xdy-ydx}{x^{2}+y^{2}}\,+\oint\limits_{\partial B_\epsilon(0,0)} \frac{xdy-ydx}{x^{2}+y^{2}}\,$$$$=\oint\limits_{\partial \Sigma} \frac{xdy-ydx}{x^{2}+y^{2}}\,+\oint\limits_{x^2+y^2 = \epsilon^2} \frac{xdy-ydx}{x^{2}+y^{2}}\,$$

How to continue?

Answer 1

Let $B\varepsilon(0,0)$ is a disk with radius $\varepsilon$ and centered at the origin. Then, on $\Sigma\setminus B_\varepsilon(0,0)$ the functions $L=-\frac{y}{x^2+y^2}$ and $M=\frac{x}{x^2+y^2}$ are continuously differentiable and Green's Theorem applies. Therefore,

$$\begin{align} \int_{\Sigma\setminus B_\varepsilon(0,0)}\left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)\,dx\,dy =\oint_{\partial \Sigma}(L\,dx+M\,dy) -\oint_{\partial B_\varepsilon(0,0)}(L\,dx+M\,dy)\tag1 \end{align}$$

where the minus sign on the second integral on the right-hand side of $(1)$ is a consequence of the orientation of the line integrals that bound $\Sigma\setminus B_\varepsilon(0,0)$.

It is easy to show that the integrand of the integral on the left-hand side of $(1)$ is identically $0$. Hence, we see that

$$\oint_{\partial \Sigma}(L\,dx+M\,dy) =\oint_{\partial B_\varepsilon(0,0)}(L\,dx+M\,dy)\tag2$$

We can parameterize the integral on the right-hand side of $(2)$ with $x=\varepsilon \cos(\phi)$ and $y=\varepsilon \sin(\phi)$, to find that

$$\begin{align} \oint_{\partial B_\varepsilon(0,0)}(L\,dx+M\,dy)&=\int_0^{2\pi} \left(\frac{\varepsilon \sin(\phi)\varepsilon \sin(\phi)}{\varepsilon^2}+\frac{\varepsilon \cos(\phi) \varepsilon \cos(\phi)}{\varepsilon^2}\right)\,d\phi\\\\ &=2\pi \end{align}$$

from which we conclude that

$$\oint_{\partial \Sigma }\left(\frac{x\,dy-y\,dx}{x^2+y^2}\right)=2\pi$$

Answer 2

Another way is to write this as a complex integral. Take $$\begin{eqnarray} z &=& x + \mathrm i \ y \\ \mathrm d z &=& \mathrm d x + \mathrm i\ \mathrm d y \\ \mathrm d \overline z &=& \mathrm d x - \mathrm i\ \mathrm d y.\end{eqnarray}$$

Then $x^2 + y^2 = z \ \overline z$ and $$ x \ \mathrm d y - y \ \mathrm d x = \frac1{2 \mathrm i}(\overline z \ \mathrm d z - z \ \mathrm d \overline z).$$ Now use Cauchy’s integral formula to evaluate your integral. (Use the substitution $z \leftarrow \overline z$ for the “conjugate” part.) The result is $\pi + \pi$.