- #1

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##I(s) = \frac{1}{2\pi i} \int_{\gamma}\text{d}z\ z^{-s} \frac{\text{d}\ln\mathcal{F(z)}}{\text{d}z}##,

where ##\mathcal{F(z)}## is analytic everywhere on the complex plane except at the zeroes of the function.

For the purpose of integration,

*the branch cut which is implied by ##z^{-s}## is chosen to be on the negative real ##z-##axis*, as follows.

The following link shows the branch cut: http://s24.postimg.org/eoxg0dbo5/Stack_Exchange_Question.jpg

I understand that the function $z^{-s}$ is multivalued, but that function has $s$ complex roots over the complex plane, so should we not choose a branch that corresponds to an angle of $\frac{2\pi}{s}$ of the complex plane?