Problems when deriving Bernhard Riemann's equation for $\zeta(1-s)$

Question

I am using this source: https://www.maths.tcd.ie/pub/HistMath/People/Riemann/Zeta/EZeta.pdf It includes a translated version of Riemann's famous 1859 paper, and I am trying to understand all the steps he takes to arrive at his formula for $\zeta(1-s)$. At one point (Bottom of page 3) he arrives at the conclusion that if $$2\sin(\pi s)\Gamma(s)\zeta(s)=i \int_\infty^\infty \frac{(-x)^{s-1}}{e^x-1}dx$$ Then $2\sin(\pi s)\Gamma(s)\zeta(s)$ must be equal to the sum of the integrals around the poles of the integrand (seen on the right side of the equation above). He figures out that the values of these integrals must be $(-n2\pi i)^{s-1}(-2\pi i)$ and he then rearranges these into $$2\sin(\pi s)\Gamma(s)\zeta(s)= (2\pi)^s\zeta(1-s)((-i)^{s-1}+i^{s-1}).$$ How did he do this? When I try to do it I end up with: $$=(-2\pi i) \left(\sum_{n=1}^{\infty}(-2\pi n i)^{s-1}+\sum_{n=1}^{\infty}(2\pi n i)^{s-1} \right) \\ =(-2\pi i)(2\pi)^{s-1} \left(\sum_{n=1}^{\infty}(- n i)^{s-1}+\sum_{n=1}^{\infty}(n i)^{s-1} \right)\\ =(-2\pi i)(2\pi)^{s-1} \left((-i)^{s-1}+i^{s-1} \right)\sum_{n=1}^{\infty}n^{s-1}\\ =-i(2\pi)^{s} \left((-i)^{s-1}+i^{s-1} \right)\zeta(1-s)$$ What am I doing wrong? I'm close to his derivation but off by a single factor of $i$, so please, if someone could correct me, feel free. Thank you for reading!