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

[rocket]

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

\oint (1/r^2) dr

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.

 

[e(ho0n3]

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.

 

[Hash]

Since 1/\tau^2 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.

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

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