Constructing functions such that integral along any closed curve is non-zero Consider smooth maps $f: \mathbb R^2 \setminus \{(0,0)\} \to \mathbb R$. 
How can I construct such an $f$ with the property that
$$ \oint_C f \neq 0$$
for any closed curve $C$ around the origin?
Note that I am aware that $\frac{xdy-ydx}{x^2+y^2}$ has this property but knowing the answer is unhelpful to me as it does not tell me how to find such a function. 

Ideally, I would like to see an answer with $$ f \neq
 \frac{xdy-ydx}{x^2+y^2}$$

 A: As JMoravitz said, if you want to understand what the integral represents (and your notation is not quite right.. $dx,dy$ cannot be part of a definition of f), you need to understand complex analysis..
Specifically consider the integral of $\int_C \frac{1}{z}\ dz$ where $C$ is a closed rectifiable curve in the complex plane. This integral is non-zero if and only if $C$ has non-zero winding number around the origin. More precisely, the integral is $2 \pi i k$ where $k$ is the (anti-clockwise) winding number of $C$ around the origin.
Note that though some people define winding number via this integral, that is not the most natural. Instead winding number can be defined for any continuous closed curve that does not pass through the origin, by choosing a fine enough partition such that each piece is within a disk that does not contain the origin. This is possible because a continuous closed curve is bounded and uniformly continuous with non-zero distance from the origin, since $[0,1]$ is compact. Then the winding angle can be defined as the sum of the angles subtended by each piece with respect to the origin. It is clear that all sufficiently fine partitions give the same winding angle, and that it is an integer multiple of $2\pi$, so we define the winding number as the winding angle divided by $2\pi$.
When that integral is translated to $x,y$-coordinates by parametrizing the curve by its real and imaginary components, it would give $2\pi i k = \int_C \frac{1}{x+yi} (dx+i\ dy) = \int_C \frac{x-yi}{x^2+y^2} (dx+i\ dy) = \int_C \frac{x\ dx+y\ dy}{x^2+y^2} + i \int_C \frac{x\ dy-y\ dx}{x^2+y^2}$, and we can see that in the final sum the two integrals are real, and hence the first integral is zero and the second is a multiple of $2\pi$ that is non-zero if and only if $C$ has non-zero winding number around the origin.
Note that $dx,dy$ make no sense without the context of $(x,y)$ being along a continuous curve parametrized in some way, in this case $C$ as a function on $[0,1]$. So technically we can't ask for a function defined in terms of $dx,dy$, though we get what you meant.
Both sides of the if and only if statement are needed for it to be interesting behaviour, otherwise as sranthrop said any strictly positive continuous function will have positive integral along any rectifiable path of non-zero length.
In general, to get such behaviour you can take any meromorphic function on some domain with only one pole at the origin with a non-zero residue there, that is, its power series expansion has a non-zero coefficient for $z^{-1}$. Then $\frac{1}{2\pi} \int_C f(z)\ dz$ will be the winding number of $C$ around $0$ for any rectifiable closed curve $C$. The reason is that $z \mapsto z^n$ has an anti-derivative on $\mathbb{C} \backslash 0$ for every integer $n$ except $-1$, and power series can be integrated term-wise, so integrating along $C$ will cause all terms to vanish except the $z^{-1}$ term.
