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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.

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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".

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  • $\begingroup$ Firstly, that the RHS has an $n$ whilst the LHS doesn't, doesn't in itself mean the inequality doesn't make sense. It only implies, by the arbitrariness of $n$, that the LHS must be equal to $0$, which is actually my main motivation for asking this question since i find that highly implausible. Anyway, what then is the bound just above Titchmarsh's Theorem 4.15 for? $\endgroup$
    – Q_p
    Commented Jun 29, 2023 at 18:16
  • $\begingroup$ By "does not make sense" I meant that your (1) is false, and Titchmarsh never claimed (1). The first half of p.81 gives an alternative proof of Theorem 4.11, which provides a simple approximation to $\zeta(s)$ in the critical strip. The remarks below this alternative proof, including the bound right above Theorem 4.15, concern an alternative proof of Theorem 4.15 (which strengthens Theorem 4.13). This is made clear by Titchmarsh: "we prefer another proof, which will be more useful for other developments". This "other proof", i.e. the proof of Theorem 4.15, starts at the bottom of p.81. $\endgroup$
    – GH from MO
    Commented Jun 30, 2023 at 5:40
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    $\begingroup$ got it. Thanks. $\endgroup$
    – Q_p
    Commented Jun 30, 2023 at 11:35

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