Can someone explain me the meaning of this integral: $I(\gamma)=\frac {1}{2\pi} \int_\gamma\frac {fdg-gdf}{f^2+g^2}$ where $\gamma$ is a closed path in $\mathbb R^2$ and $f:\mathbb R^2 \to \mathbb R$, $g:\mathbb R^2\to \mathbb R$ are continuous functions.
I am studying dynamical system and on the book is written that $I(\gamma)\in \mathbb Z$ tell us how many laps the vectorial field $E(x,y)=(f(x,y),g(x,y))$ is doing along the path $\gamma$. But i am not understanding it.
I appreciate some hints.
$fdg -gdf$ most likely means $(f\nabla g - g\nabla f)\cdot d\mathbf{r}$ (which is easy to see if you think of the gradient as being $\frac{df}{d\mathbf{r}}$ in the first place).
In fact, it's not hard to see that the line integral is the integral of the gradient of $\tan^{-1}\left(\frac{g}{f}\right)$. Imagining $g$ as $y$ and $f$ as $x$, what we're doing is we're counting up how many times the vector makes a full rotation by imagining $\tan^{-1}\left(\frac{g}{f}\right)=\theta$ and $\frac{1}{2\pi}\int_\gamma \frac{fdg -gdf}{f^2+g^2} =$ "$\frac{1}{2\pi}\int_\gamma d\theta$"
If the vector field is conservative, then why isn't the line integral always $0$? Well, if the vector makes a full rotation, it has to encircle the point where $f=g=0$, which is a singularity for the gradient; and when it does, it is no longer conservative. In fact it will give us integer multiples of $2\pi$ for every full rotation it does.