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Let $C$ be the path determined by the square with vertices $(1,1),(-1,1), (-1,-1), (1,-1)$, in the counterclockwise direction. How would one go about finding the following integral?

$\int_C \frac{x}{x^2+y^2}dx+\frac{-y}{x^2+y^2}dy$

Perhaps using the Residue Theorem?

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    Hmm, I supposed (wrongly) that complex analysis would make it easier, but apparently not. I'm confused as to how to interpret the $\int \dots dx+\dots dy$ notation. I'm also unsure as to whether I should parametrize the contour or not. My intuition tells me you need only consider for distinct "easy" integrals (one for each side), because of what @Didier said.2012-05-04

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By definition, for every parametrisation $(t_0,t_1)\mapsto\mathbb R^2$, $t\mapsto (x(t),y(t))$ of the curve $C$, $ \int_Cu(x,y)\mathrm dx+v(x,y)\mathrm dy=\int_{t_0}^{t_1}\left[u(x(t),y(t))x'(t)+v(x(t),y(t))y'(t)\right]\mathrm dt. $ Call $N$, $W$, $S$ and $E$ the contributions of the corresponding sides of the square to the integral under consideration, oriented clockwise. For example, $(t_0,t_1)=(-1,1)$, $(x(t),y(t))=(t,1)$ is a parametrization of $N$, hence $x'=+1$ on $N$, and $ N=\int_{-1}^1\frac{t}{t^2+1^2}\mathrm dt=0. $ By symmetry, $S=0$. The same argument shows that $E=W=0$.

Finally, the integral under consideration is $0$ and this is a consequence of the fact that one is considering odd functions $x\mapsto u(x,1)$, $x\mapsto u(x,-1)$, $y\mapsto v(1,y)$ and $y\mapsto v(-1,y)$ and a path $C$ symmetric with respect to the origin.