I'm reading Titchmarsh's "The theory of the Riemann zeta function", and on p.81 it is claimed that


$$ \int\_{x}^{x+i\infty} (\cot(\pi z)+i)z^{-s} \, \mathrm{d}z \ll \frac{x^{-\sigma}}{2(n+1)\pi-|t|/x} \tag{1}$$


where $\sigma=\Re(s), t=\Im(s), x$ is a half-positive integer and $n$ is an arbitrary positive integer.


Titchmarsh proves (1) by invoking the identity


$$-\cot(\pi z) - i =2i\sum\_{v=1}^{n} e^{2\pi i vz} + \frac{2ie^{2(n+1)\pi i z}}{1-e^{2\pi iz}},$$


but doesn't show the steps. That's what I'm asking for.



On the bound for $\int_{x}^{x+i\infty} (\cot(\pi z)+ i)z^{-s} \, \mathrm{d}z$
Your inequality does not make sense, since the RHS has $n$ in it, while the LHS does not. What Titchmarsh claims is the bound
$$\int\_{x}^{x+i\infty} (\cot(\pi z)+i)z^{-s} \mathrm{d}z \ll \frac{x^{-\sigma}}{2\pi-|t|/x}.$$
And he provides a detailed proof stretching 5 lines: "In the second integral we put $z=x+ir$ etc." The identity you quote is not needed for this proof. It is mentioned after the half-sentence "and the theorem follows".


