Strange application of Cauchy's Integral Theorem

According to my book, Riemann's Zeta Function, Cauchy's Integral Formula is applicable to the following integral for all negative values of $s$:

$$-\frac{\Pi(-s)}{2\pi i}\int_{|z|=\epsilon}(-2\pi in - z)^{s-1}\frac{z}{e^z - 1}\frac{dz}{z} = -\Pi(-s)(-2\pi in)^{s-1}$$

where $\Pi(-s) := \Gamma(-s+1)$.

Could someone explain to me how exactly this works, I can't seem to figure it out, thanks.

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1 Answer

Notice that for $n\neq 0,$ the factor $(-2\pi in-z)^{s-1}\frac{z}{e^z-1}$ is holomorphic near $z=0.$ To see this, expand $\frac{z}{e^z-1}=\frac{z}{(1+z+z^2/2+\cdots) -1}=\frac{1}{1+z/2+\cdots}.$ Then we can apply Cauchy's integral formula to see that the integral equals $2\pi i \times (-2\pi in - 0)^{s-1}\times\frac{1}{1+0+\cdots},$ which gives the result.

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Ohhh I see it! Thank you for this. –  cactuar Jul 22 '12 at 22:27
You're very welcome. –  Andrew Jul 22 '12 at 22:27