I want to prove the following formula: $$\frac{e^{-tE}}{1+e^{-\beta E}} = \frac{1}{\beta}\sum_{k \in \frac{2\pi}{\beta}\mathbb{Z}}\frac{e^{-ikt}}{-ik+E},$$ for $\beta > t > 0$. I know the trick is to use the Residue Theorem for a contour integration. Concretely, the formula must follow from the integral: $$I = \frac{1}{2\pi}\oint dz \frac{e^{-izt}}{(-iz+E)(1+e^{-i\beta z})}.$$ I am a bit confused about dealing the contours to obtain the desired formula.
For example, if I take a curve on the complex plane which goes from $-\infty$ to $+\infty$ right below the real axis, without enclosing the $-iE$ pole and then comes back from $+\infty$ to $-\infty$ just above the real axis, the integral $I$ is supposed to be: $$I = i\sum_{k \in \frac{2\pi}{\beta}\mathbb{Z}}\frac{e^{-ikt}}{-ik+E}$$ It seems that there is a $\beta^{-1}$ missing. Moreover, if a take a contour which only encloses the $-iE$ pole, for instance a semi-circle in the lower half-plane and a semi-circle in the upper half-plane I would get: $$I = -i\frac{e^{-tE}}{1+e^{-\beta E}}$$ the minus sign coming from the clockwise orientation. However, I can't put these two formulas together to obtain the result. First, why are they supposed to be the same (so the desired equality holds) if I took completely different paths with different poles? Also, why I am not getting the $\beta^{-1}$ factor on the first contour?
