Cotangent in integral form

Question

Could you help me to establish the following relation, without going through the Eulerian development $\displaystyle \pi \operatorname{cotan}(\pi z) = \frac{1}{z} + \sum_{n=1}^{+\infty} \left( \frac{1}{z+n} + \frac{1}{z-n} \right)$ : $$\pi \operatorname{cotan}(\pi z) = \int_0^{+\infty} \frac{e^{(1-z)t} - e^{tz}}{e^t-1} dt. \quad 0<\Re(z)<1$$

Answer 1

Hoping that you do enjoy hypergeometric functions $$I=\int\frac{e^{(1-z)t} - e^{tz}}{e^t-1} \,dt$$ $$ I=\frac {e^{-t z} \, _2F_1\left(1,-z;1-z;e^t\right)+e^{t z} \, _2F_1\left(1,z;z+1;e^t\right)-e^{-t z} } z$$

$$J=\int_0^\infty\frac{e^{(1-z)t} - e^{tz}}{e^t-1} \,dt=\psi (-z)-\psi (z)-\frac{1}{z}=\pi \cot (\pi z)$$