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On page 267 of Titchmarsh's book about the Riemann zeta function he claims that for $s= \frac{1}{2}+it$, $T\leq t \leq 2T$ we have \begin{equation}T^{1/4} \left|\Xi(t)\right| \frac{e^{\frac{1}{4}\pi t}}{t^2 + \frac{1}{4}}> A \left| \zeta\left( \frac{1}{2}+it \right) \right| \hspace{2cm}(1)\end{equation}

for some constant $A >0$.

I feel like I have seen this inequality once but it has been a while since I studied this topic and do not know where it has been shown. I didn't find it in my notes either. Anyone knows where I can find its proof?

Nevertheless, the real problem arises in the next inequality:

He says that from this it follows that for $H\geq 1$

\begin{align*} T^{1/4} \int_t^{t+H} \left|\Xi(u)\right| \frac{e^{\frac{1}{4}\pi u}}{u^2 + \frac{1}{4}}e^{-u/T}\, du & > A \int_{t}^{t+H}\left| \zeta\left(\frac{1}{2}+iu \right) \right|\,du\\& > A \left|\int_{t}^{t+H}\ \zeta\left(\frac{1}{2}+iu \right) \, du\right| \\ &=A \left| \int_{t}^{t+H} \left\{ \sum_{n<AT}\hspace{5pt} \frac{1}{n^{\frac{1}{2}+iu}}+O\left(T^{-\frac{1}{2}}\right) \right\} \, du \right| \end{align*}

$1.$ inequality - is this argument right?

From $(1)$ we get

$$ T^{1/4} \int_t^{t+H} \left|\Xi(u)\right| \frac{e^{\frac{1}{4}\pi u}}{u^2 + \frac{1}{4}}e^{-u/T}\, du > A \int_{t}^{t+H}\left| \zeta\left(\frac{1}{2}+iu \right) \right|\, e^{-u/T}\, du$$

Since $u \leq t+H \leq 2T+H$ it follows that $e^{-u/T}> e^{-\frac{2T+H}{T}}$ for $u\in (t,t+H)$ and $e^{-\frac{2T+H}{T}}$ is constant for fixed $T$.

Thus \begin{align*} \cdots \hspace{1cm} & >\hspace{0.3cm} \underbrace{A\, e^{-\frac{2T+H}{T}}}_{=:A}\int_{t}^{t+H}\left| \zeta\left(\frac{1}{2}+iu \right) \right|\, du \end{align*}

$3.$ inequality

I do not understand this one at all.

As far as I know the big $O$ notation is used when the left hand side is smaller than the right hand side but here we need that it is bigger, so why is he using it there and what does it mean in this context?

Additionally, he seems to approximate $\zeta(1/2+iu)$ with its series expansion (which is valid only for $Re(s)>1$) ; why is that?

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1 Answer 1

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This is the definition $$\Xi(t) = \xi(1/2+it), \qquad\frac{1}{2}\xi(s) = s (s-1) \pi^{-s/2}\Gamma(s/2) \zeta(s)$$ so that $$|\Xi(t)| = |\xi(1/2+it)| = |\Gamma(1/4+it/2)|\,(t^2+1/4)\frac{\pi^{-1/4}}{2} \, |\zeta(1/2+it)|$$ together with an estimate for the decay rate of $|\Gamma(1/4+it/2)|$ using the Stirling approximation. As $|s| \to \infty, Re(s) > c$ : $$\Gamma(s) = \sqrt{\frac{2\pi}{s}}\,{\left(\frac{s}{e}\right)}^s \left(1 + O\left(\frac{1}{s}\right)\right)$$

The substitution of $\zeta(1/2+it)$ by $Re(e^{i \theta t} \sum_{n < AT}n^{-1/2-it})+O(T^{-1/2})$ is the Riemann-siegel formula (approximate functional equation).

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