How do we calculate the complex integral with poles at 2npi?

[eljose]

complex integral...

let be the integral \int_{-i\infty}^{i\infty}\frac{1}{exp(s)-1}ds then their poles are 2n\pi my question is How would we calculate this integral? i think that the contribution from the poles is -{\pi}Res(z_0) the main problem i find is when i make the change of variables s=iu so we have the improper integral \int_{-\infty}^{\infty}\frac{1}{exp(iu)-1}ds but it has singularities at 2npi so i don,t think if the first integral will be convergent or not...could someone help?..thanks

 

[fresh_42]

You have to find an integration path around the poles.

 

[mfig]

This looks divergent.

$\int_{-\infty}^\infty \frac{1}{e^{ix}-1}dx = -x-i\cdot log \left( 1- e^{ix}\right)\bigg|_{-\infty} ^\infty$