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I want to compute the following integral $$\oint_{|z|=1}\frac{\exp \left (\frac{1}{z} \right)}{z^2-1}\,dz$$ The integrand has essential singularity at the origin, and $2$-poles at $\pm 1$,which lie on the curve $|z|=1$ so I can't apply residue formula. How can I proceed?

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the integral doesn't exits in the usual sense; if you want the principal value (which is most probable) then you should count the two residues on the circle with $1/2$ (it works for simple poles –  user8268 Nov 28 '12 at 20:44
    
since this is a part of a bigger exercise with parameters ( here a=1 in the denominator) and the text says that for some values of $a$ the integral doesn't exists, i think this is the case. But how can i justify this? simply saying that residue formula doesn't apply? –  bateman Nov 28 '12 at 21:12
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