Is the Integral of 1/r^2 Over a Closed Loop Zero?

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rocket
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i want to prove that this integral along a closed loop:

[tex]\oint (1/r^2) dr[/tex]

is equal to zero. but I'm not sure how to prove it. i was wondering if someone can show me a rigid proof for this. I think I'm missing something here because I'm not really that familiar with loop integrals.
 
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Pick two points, A and B, and two curves, C_1 and C_2 where C_1 goes from A to B and C_2 from B to A.

Now calculate the line integral along the curve C_1 and C_2. They should both be equal, in which case, the loop integral will be 0.
 
Since [tex]1/\tau^2[/tex] is an analytic function with a singularity in 0 (a pole of order 2), the contour your want to use is closed, then you can use the Cauchy's theorem on residues.

[tex]\oint 1/\tau^2 \, d\tau = 2i\pi \cdot 0[/tex]

where 0 is the residue of [tex]1/\tau^2[/tex]