7 Zeta function estimates

7.1 Definitions

Definition 7.1.1

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$\rho $ is understood to lie in the set $\{ s: \zeta (s)=0\} $, counted with multiplicity. We will often restrict the zeroes $\rho $ to a rectangle $\{ \Re \rho \in I, \Im \rho \in J \} $, for instance through sums of the form $\sum _{\Re \rho \in I, \Im \rho \in J} f(\rho )$.

Definition 7.1.2

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We say that the Riemann hypothesis has been verified up to height $T$ if there are no zeroes in the rectangle $\{ \Re \rho \in (0.5, 1), \Im \rho \in [0,T] \} $.

Definition 7.1.3 Section 1.1, FKS2

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We say that one has a classical zero-free region with parameter $R$ if $zeta(s)$ has no zeroes in the region $Re(s) \geq 1 - 1 / R * \log |\Im s|$ for $\Im (s) {\gt} 3$.

Definition 7.1.4 Zero counting function N(T)

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The number of zeroes of imaginary part between 0 and T, counting multiplicity

Definition 7.1.5 Riemann von Mangoldt estimate

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An estimate of the form $N(T) - \frac{T}{2\pi } \log \frac{T}{2\pi e} + \frac{7}{8}| \leq b_1 \log T + b_2 \log \log T + b_3$ for $T \geq 2$.

Definition 7.1.6 Zero density bound

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An estimate of the form $N(\sigma ,T) \leq c₁ T^p \log ^q T + c₂ \log ^2 T - \frac{T}{2\pi } \log \frac{T}{2\pi e} + \frac{7}{8}| \leq b_1 \log T + b_2 \log \log T + b_3$ for $T \geq 2$.

7.2 The estimates of Kadiri, Lumley, and Ng

In this section we establish the primary results of [ 9 ] .

7.3 The zeta function bounds of Rosser and Schoenfeld

In this section we formalize the zeta function bounds of Rosser and Schoenfeld.

Theorem 7.3.1 Rosser–Schoenfeld Theorem 19

One has a Riemann von Mangoldt estimate with parameters 0.137, 0.443, and 1.588.

Proof ▶

7.4 Approximating the Riemann zeta function

We want a good explicit estimate on

\[ \sum _{n\leq a} \frac{1}{n^s} - \int _0^{a} \frac{du}{u^s}, \]

for $a$ a half-integer. As it turns out, this is the same problem as that of approximating $\zeta (s)$ by a sum $\sum _{n\leq a} n^{-s}$. This is one of the two ¹ main, standard ways of approximating $\zeta (s)$.

The non-explicit version of the result was first proved in [ 8 , Lemmas 1 and 2 ] . The proof there uses first-order Euler-Maclaurin combined with a decomposition of $\lfloor x\rfloor - x +1/2$ that turns out to be equivalent to Poisson summation. The exposition in [ 14 , § 4.7–4.11 ] uses first-order Euler-Maclaurin and van de Corput’s Process B; the main idea of the latter is Poisson summation.

There are already several explicit versions of the result in the literature. In [ 2 ] , [ and [ 12 ] , what we have is successively sharper explicit versions of Hardy and Littlewood’s original proof. The proof in [ 4 , Lemma 2.10 ] proceeds simply by a careful estimation of the terms in high-order Euler-Maclaurin; it does not use Poisson summation. Finally, [ 3 ] is an explicit version of [ 14 , § 4.7–4.11 ] ; it gives a weaker bound than [ 12 ] or [ 4 ] . The strongest of these results is [ 12 ] .

We will give another version here, in part because we wish to relax conditions – we will work with $\left|\Im s\right| {\lt} 2\pi a$ rather than $\left|\Im s\right| \leq a$ – and in part to show that one can prove an asymptotically optimal result easily and concisely. We will use first-order Euler-Maclaurin and Poisson summation. We assume that $a$ is a half-integer; if one inserts the same assumption into [ 4 , Lemma 2.10 ] , one can improve the result there, yielding an error term closer to the one here.

For additional context, see the Zulip discussion at https://leanprover.zulipchat.com/#narrow/channel/423402-PrimeNumberTheorem.2B/ topic/Let.20us.20formalize.20an.20appendix

Definition 7.4.1 e

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We recall that $e(\alpha ) = e^{2\pi i \alpha }$.

7.4.1 The decay of a Fourier transform

Our first objective will be to estimate the Fourier transform of $t^{-s} \mathbb {1}_{[a,b]}$. In particular, we will show that, if $a$ and $b$ are half-integers, the Fourier cosine transform has quadratic decay when evaluated at integers. In general, for real arguments, the Fourier transform of a discontinuous function such as $t^{-s} \mathbb {1}_{[a,b]}$ does not have quadratic decay.

Lemma 7.4.1 Fourier transform of a truncated power law

# Discussion

Let $s = \sigma + i \tau $, $\sigma \geq 0$, $\tau \in \mathbb {R}$. Let $\nu \in \mathbb {R}\setminus \{ 0\} $, $b{\gt}a{\gt}\frac{|\tau |}{2\pi |\nu |}$. Then

\begin{equation} \label{eq:aachquno}\int _a^b t^{-s} e(\nu t) dt = \left. \frac{t^{-\sigma } e(\varphi _\nu (t))}{2\pi i \varphi _\nu '(t)}\right|_a^b + \sigma \int _a^b \frac{t^{-\sigma -1}}{2\pi i \varphi _\nu '(t)} e(\varphi _\nu (t)) dt + \int _a^b \frac{t^{-\sigma } \varphi _\nu ''(t)}{2\pi i (\varphi _\nu '(t))^2} e(\varphi _\nu (t)) dt, \end{equation}

where $\varphi _\nu (t) = \nu t - \frac{\tau }{2\pi } \log t$.

Proof ▶

We write $t^{-s} e(\nu t) = t^{-\sigma } e(\varphi _\nu (t))$ and integrate by parts with $u = t^{-\sigma }/(2\pi i \varphi _\nu '(t))$, $v = e(\varphi _\nu (t))$. Here $\varphi _\nu '(t) = \nu - \tau /(2\pi t)\ne 0$ for $t\in [a,b]$ because $t\geq a {\gt} |\tau |/(2\pi |\nu |)$ implies $|\nu |{\gt}|\tau |/(2\pi t)$. Clearly

\[ u dv = \frac{ t^{-\sigma }}{2\pi i \varphi _\nu '(t)} \cdot 2\pi i \varphi _\nu '(t) e(\varphi _\nu (t)) dt = t^{-\sigma } e(\varphi _\nu (t)) dt, \]

while

\[ du = \left(\frac{-\sigma t^{-\sigma -1}}{2\pi i \varphi _\nu '(t)} - \frac{t^{-\sigma } \varphi _\nu ''(t)}{2\pi i (\varphi _\nu '(t))^2}\right) dt. \]

Lemma 7.4.2 Total variation of a function with monotone absolute value

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# Discussion

Let $g:[a,b]\to \mathbb {R}$ be continuous, with $|g(t)|$ non-increasing. Then $g$ is monotone, and $\| g\| _{\mathrm{TV}} = |g(a)|-|g(b)|$.

Proof ▶

Suppose $g$ changed sign: $g(a'){\gt}0{\gt}g(b')$ or $g(a') {\lt}0 {\lt} g(b')$ for some $a\leq a'{\lt} b'\leq b$. By IVT, there would be an $r\in [a',b']$ such that $g(r)=0$. Since $|g|$ is non-increasing, $g(b')=0$; contradiction. So, $g$ does not change sign: either $g\leq 0$ or $g\geq 0$.

Thus, there is an $\varepsilon \in \{ -1,1\} $ such that $g(t) = \varepsilon |g(t)|$ for all $t\in [a,b]$. Hence, $g$ is monotone. Then $\| g\| _{\mathrm{TV}} = |g(a)-g(b)|$. Since $|g(a)|\geq |g(b)|$ and $g(a)$, $g(b)$ are either both non-positive or non-negative, $|g(a)-g(b)| = |g(a)|-|g(b)|$.

Lemma 7.4.3 Non-stationary phase estimate

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# Discussion

Let $\varphi :[a,b]\to \mathbb {R}$ be $C^1$ with $\varphi '(t)\ne 0$ for all $t\in [a,b]$. Let $h:[a,b]\to \mathbb {R}$ be such that $g(t) = h(t)/\varphi '(t)$ is continuous and $|g(t)|$ is non-increasing. Then

\[ \left|\int _a^b h(t) e(\varphi (t)) dt\right|\leq \frac{|g(a)|}{\pi }. \]

Proof ▶

Since $\varphi $ is $C^1$, $e(\varphi (t))$ is $C^1$, and $h(t) e(\varphi (t)) = \frac{h(t)}{2\pi i \varphi '(t)} \frac{d}{dt} e(\varphi (t))$ everywhere. By Lemma 7.4.2, $g$ is of bounded variation. Hence, we can integrate by parts:

\[ \int _a^b h(t) e(\varphi (t)) dt = \left. \frac{h(t) e(\varphi (t))}{2\pi i \varphi '(t)}\right|_a^b - \int _a^b e(\varphi (t))\, d\left(\frac{h(t)}{2\pi i \varphi '(t)}\right). \]

The first term on the right has absolute value $\leq \frac{|g(a)|+|g(b)|}{2\pi }$. Again by Lemma 7.4.2,

\[ \left| \int _a^b e(\varphi (t))\, d\left(\frac{h(t)}{2\pi i \varphi '(t)}\right) \right|\leq \frac{1}{2\pi } \| g\| _{\mathrm{TV}} = \frac{|g(a)|-|g(b)|}{2\pi }. \]

We are done by $\frac{|g(a)|+|g(b)|}{2\pi } + \frac{|g(a)|-|g(b)|}{2\pi } = \frac{|g(a)|}{\pi }$.

Lemma 7.4.4 A decreasing function

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# Discussion

Let $\sigma \geq 0$, $\tau \in \mathbb {R}$, $\nu \in \mathbb {R}\setminus \{ 0\} $. Let $b{\gt}a{\gt}\frac{|\tau |}{2\pi |\nu |}$. Then, for any $k\geq 1$, $f(t) = t^{-\sigma -k} |2\pi \nu -\tau /t|^{-k-1}$ is decreasing on $[a,b]$.

Proof ▶

Let $a\leq t\leq b$. Since $\left|\frac{\tau }{t \nu }\right| {\lt} 2\pi $, we see that $2\pi -\frac{\tau }{\nu t} {\gt}0$, and so $|2\pi \nu -\tau /t|^{-k-1} = |\nu |^{-k-1} \left(2\pi - \frac{\tau }{t \nu }\right)^{-k-1}$. Now we take logarithmic derivatives:

\[ t (\log f(t))' = - (\sigma +k) - (k+1) \frac{\tau /t}{2\pi \nu - \tau /t} = -\sigma - \frac{2\pi k + \frac{\tau }{t \nu }}{2\pi - \frac{\tau }{t \nu }} {\lt} -\sigma \leq 0, \]

since, again by $\frac{|\tau |}{t |\nu |} {\lt} 2\pi $ and $k\geq 1$, we have $2\pi k + \frac{\tau }{t \nu }{\gt}0$, and, as we said, $2\pi - \frac{\tau }{t \nu }{\gt}0$.

Lemma 7.4.5 Estimating an integral

# Discussion

Let $s = \sigma + i \tau $, $\sigma \geq 0$, $\tau \in \mathbb {R}$. Let $\nu \in \mathbb {R}\setminus \{ 0\} $, $b{\gt}a{\gt}\frac{|\tau |}{2\pi |\nu |}$. Then

\[ \int _a^b t^{-s} e(\nu t) dt = \left. \frac{t^{-\sigma } e(\varphi _\nu (t))}{2\pi i \varphi _\nu '(t)}\right|_a^b + \frac{a^{-\sigma -1}}{2\pi ^2} O^*\left(\frac{\sigma }{(\nu -\vartheta )^2} + \frac{|\vartheta |}{|\nu -\vartheta |^3}\right), \]

where $\varphi _\nu (t) = \nu t - \frac{\tau }{2\pi } \log t$ and $\vartheta = \frac{\tau }{2\pi a}$.

Proof ▶

Apply Lemma 7.4.1. Since $\varphi _\nu '(t) = \nu - \tau /(2\pi t)$, we know by Lemma 7.4.4 (with $k=1$) that $g_1(t) = \frac{t^{-\sigma -1}}{(\varphi _\nu '(t))^2}$ is decreasing on $[a,b]$. We know that $\varphi _\nu '(t)\ne 0$ for $t\geq a$ by $a{\gt}\frac{|\tau |}{2\pi |\nu |}$, and so we also know that $g_1(t)$ is continuous for $t\geq a$. Hence, by Lemma 7.4.3,

\[ \left|\int _a^b \frac{t^{-\sigma -1}}{2\pi i \varphi _\nu '(t)} e(\varphi _\nu (t)) dt \right| \leq \frac{1}{2\pi } \cdot \frac{|g_1(a)|}{\pi } = \frac{1}{2\pi ^2} \frac{a^{-\sigma -1}}{\left|\nu - \vartheta \right|^2}, \]

since $\varphi _\nu '(a) = \nu - \vartheta $. We remember to include the factor of $\sigma $ in front of an integral in 1.

Since $\varphi _\nu '(t)$ is as above and $\varphi _\nu ''(t) = \tau /(2\pi t^2)$, we know by Lemma 7.4.4 (with $k=2$) that $g_2(t) = \frac{t^{-\sigma } |\varphi _\nu ''(t)|}{|\varphi _\nu '(t)|^3} = \frac{|\tau |}{2\pi } \frac{t^{-\sigma -2}}{|\varphi _\nu '(t)|^3}$ is decreasing on $[a,b]$ we also know, as before, that $g_2(t)$ is continuous. Hence, again by Lemma 7.4.3,

\[ \left|\int _a^b \frac{t^{-\sigma } \varphi _\nu ''(t)}{2\pi i (\varphi _\nu '(t))^2} e(\varphi _\nu (t)) dt\right|\leq \frac{1}{2\pi } \frac{|g_2(a)|}{\pi } = \frac{1}{2\pi ^2} \frac{a^{-\sigma -1} |\vartheta |}{\left|\nu - \vartheta \right|^3}. \]

Lemma 7.4.6 Estimating an sum

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# Discussion

Let $s = \sigma + i \tau $, $\sigma ,\tau \in \mathbb {R}$. Let $n\in \mathbb {Z}_{{\gt}0}$. Let $a,b\in \mathbb {Z} + \frac{1}{2}$, $b{\gt}a{\gt}\frac{|\tau |}{2\pi n}$. Write $\varphi _\nu (t) = \nu t - \frac{\tau }{2\pi } \log t$. Then

\[ \frac{1}{2} \sum _{\nu = \pm n} \left. \frac{t^{-\sigma } e(\varphi _\nu (t))}{2\pi i \varphi _\nu '(t)}\right|_a^b = \left. \frac{(-1)^n t^{-s} \cdot \frac{\tau }{2\pi t}}{2\pi i \left(n^2 - \left(\frac{\tau }{2\pi t}\right)^2\right)}\right|_a^b. \]

Proof ▶

Since $e(\varphi _\nu (t)) = e(\nu t) t^{-i \tau } = (-1)^{\nu } t^{-i \tau }$ for any half-integer $t$ and any integer $\nu $,

\[ \left. \frac{t^{-\sigma } e(\varphi _\nu (t))}{2\pi i \varphi _\nu '(t)}\right|_a^b = \left. \frac{(-1)^{\nu } t^{-s}}{2\pi i \varphi _\nu '(t)}\right|_a^b \]

for $\nu = \pm n$. Clearly $(-1)^{\nu } = (-1)^n$. Since $\varphi _\nu '(t) = \nu - \alpha $ for $\alpha = \frac{\tau }{2\pi t}$,

\[ \frac{1}{2} \sum _{\nu = \pm n} \frac{1}{\varphi _\nu '(t)} = \frac{1/2}{n - \alpha } + \frac{1/2}{- n - \alpha } = \frac{-\alpha }{\alpha ^2-n^2} = \frac{\alpha }{n^2-\alpha ^2}. \]

It is this easy step that gives us quadratic decay on $n$. It is just as in the proof of van der Corput’s Process B in, say, [ 13 , I.6.3, Thm. 4 ] .

Proposition 7.4.1 Estimating a Fourier cosine integral

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# Discussion

Let $s = \sigma + i \tau $, $\sigma \geq 0$, $\tau \in \mathbb {R}$. Let $a,b\in \mathbb {Z} + \frac{1}{2}$, $b{\gt}a{\gt}\frac{|\tau |}{2\pi }$. Write $\vartheta = \frac{\tau }{2\pi a}$. Then, for any integer $n\geq 1$,

\[ \begin{aligned} \int _a^b t^{-s} \cos 2\pi n t\, dt & = \left. \left(\frac{(-1)^n t^{-s}}{2\pi i} \cdot \frac{\frac{\tau }{2\pi t}}{n^2 - \left(\frac{\tau }{2\pi t}\right)^2}\right)\right|_a^b \\ & \quad + \frac{a^{-\sigma -1}}{4\pi ^2}\, O^*\left(\frac{\sigma }{(n-\vartheta )^2} + \frac{\sigma }{(n+\vartheta )^2} + \frac{|\vartheta |}{|n-\vartheta |^3} + \frac{|\vartheta |}{|n+\vartheta |^3}\right).\end{aligned} \]

Proof ▶

Write $\cos 2\pi n t = \frac{1}{2} (e(n t) + e(-n t))$. Since $n\geq 1$ and $a{\gt}\frac{|\tau |}{2\pi }$, we know that $a{\gt}\frac{|\tau |}{2 \pi n}$, and so we can apply Lemma 7.4.5 with $\nu = \pm n$. We then apply Lemma 7.4.6 to combine the boundary contributions $\left. \right|_a^b$ for $\nu =\pm n$.

7.4.2 Approximating zeta(s)

We start with an application of Euler-Maclaurin.

Lemma 7.4.7 Identity for a partial sum of zeta(s) for integer b

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# Discussion

Let $b{\gt}0$, $b\in \mathbb {Z}$. Then, for all $s\in \mathbb {C}\setminus \{ 1\} $ with $\Re s {\gt} 0$,

\begin{equation} \label{eq:abak1'} \sum _{n \leq b} \frac{1}{n^s} = \zeta (s) + \frac{b^{1-s}}{1-s} + \frac{b^{-s}}{2} + s \int _b^\infty \left(\{ y\} -\frac{1}{2}\right) \frac{dy}{y^{s+1}}. \end{equation}

Proof ▶

Assume first that $\Re s {\gt} 1$. By first-order Euler-Maclaurin,

\[ \sum _{n {\gt} b}\frac{1}{n^s} = \int _b^\infty \frac{dy}{y^s} + \int _b^\infty \left(\{ y\} -\frac{1}{2}\right) d\left(\frac{1}{y^s}\right). \]

Here $\int _b^\infty \frac{dy}{y^s} = -\frac{b^{1-s}}{1-s}$ and $d\left(\frac{1}{y^s}\right) = - \frac{s}{y^{s+1}} dy$. Hence, by $\sum _{n\leq b} \frac{1}{n^s} = \zeta (s) - \sum _{n{\gt}b} \frac{1}{n^s}$ for $\Re s {\gt} 1$,

\[ \sum _{n\leq b} \frac{1}{n^s} = \zeta (s) + \frac{b^{1-s}}{1-s} + \int _b^\infty \left(\{ y\} -\frac{1}{2}\right) \frac{s}{y^{s+1}} dy. \]

Since the integral converges absolutely for $\Re s {\gt} 0$, both sides extend holomorphically to $\{ s\in \mathbb {C}: \Re s{\gt}0, s\ne 1\} $; thus, the equation holds throughout that region.

Lemma 7.4.8 Identity for a partial sum of zeta(s)

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# Discussion

Let $b{\gt}0$, $b\in \mathbb {Z} + \frac{1}{2}$. Then, for all $s\in \mathbb {C}\setminus \{ 1\} $ with $\Re s {\gt} 0$,

\begin{equation} \label{eq:abak1} \sum _{n\leq b} \frac{1}{n^s} = \zeta (s) + \frac{b^{1-s}}{1-s} + s \int _b^\infty \left(\{ y\} -\frac{1}{2}\right) \frac{dy}{y^{s+1}}. \end{equation}

Proof ▶

Assume first that $\Re s {\gt} 1$. By first-order Euler-Maclaurin and $b\in \mathbb {Z}+\frac{1}{2}$,

\[ \sum _{n{\gt}b}\frac{1}{n^s} = \int _b^\infty \frac{dy}{y^s} + \int _b^\infty \left(\{ y\} -\frac{1}{2}\right) d\left(\frac{1}{y^s}\right). \]

\[ \sum _{n\leq b} \frac{1}{n^s} = \zeta (s) + \frac{b^{1-s}}{1-s} + \int _b^\infty \left(\{ y\} -\frac{1}{2}\right) \frac{s}{y^{s+1}} dy. \]

Since the integral converges absolutely for $\Re s {\gt} 0$, both sides extend holomorphically to $\{ s\in \mathbb {C}: \Re s{\gt}0, s\ne 1\} $; thus, the equation holds throughout that region.

Lemma 7.4.9 Estimate for a partial sum of $\zeta (s)$

✓

# Discussion

Let $b{\gt}a{\gt}0$, $b\in \mathbb {Z} + \frac{1}{2}$. Then, for all $s\in \mathbb {C}\setminus \{ 1\} $ with $\sigma = \Re s {\gt} 0$,

\[ \sum _{n\leq a} \frac{1}{n^s} = -\sum _{a {\lt} n\leq b} \frac{1}{n^s} + \zeta (s) + \frac{b^{1-s}}{1-s} + O^*\left(\frac{|s|}{2 \sigma b^\sigma }\right). \]

Proof ▶

By Lemma 7.4.8, $\sum _{n\leq a} = \sum _{n\leq b} - \sum _{a {\lt} n\leq b}$, $\left|\{ y\} -\frac{1}{2}\right| \leq \frac{1}{2}$ and $\int _b^\infty \frac{dy}{|y^{s+1}|} = \frac{1}{\sigma b^\sigma }$.

Lemma 7.4.10 Poisson summation for a partial sum of $\zeta (s)$

# Discussion

Let $a,b\in \mathbb {R}\setminus \mathbb {Z}$, $b{\gt}a{\gt}0$. Let $s\in \mathbb {C}\setminus \{ 1\} $. Define $f:\mathbb {R}\to \mathbb {C}$ by $f(y) = 1_{[a,b]}(y)/y^s$. Then

\[ \sum _{a {\lt} n\leq b} \frac{1}{n^s} = \frac{b^{1-s} - a^{1-s}}{1-s} + \lim _{N\to \infty } \sum _{n=1}^N (\widehat{f}(n) + \widehat{f}(-n)). \]

Proof ▶

Since $a\notin \mathbb {Z}$, $\sum _{a {\lt} n\leq b} \frac{1}{n^s} = \sum _{n\in \mathbb {Z}} f(n)$. By Poisson summation (as in [ 10 , Thm. D.3 ] )

\[ \sum _{n\in \mathbb {Z}} f(n) = \lim _{N\to \infty } \sum _{n=-N}^N \widehat{f}(n) = \widehat{f}(0) + \lim _{N\to \infty } \sum _{n=1}^N (\widehat{f}(n) + \widehat{f}(-n)), \]

where we use the facts that $f$ is in $L^1$, of bounded variation, and (by $a,b\not\in \mathbb {Z}$) continuous at every integer. Now

\[ \widehat{f}(0) = \int _{\mathbb {R}} f(y) dy = \int _a^b \frac{dy}{y^s} = \frac{b^{1-s}-a^{1-s}}{1-s}. \]

We could prove these equations starting from Euler’s product for $\sin \pi z$.

Lemma 7.4.11 Euler/Mittag-Leffler expansion for cosec

# Discussion

Let $z\in \mathbb {C}$, $z\notin \mathbb {Z}$. Then

\[ \frac{\pi }{\sin \pi z} = \frac{1}{z} + \sum _{n {\gt} 0} (-1)^n\left(\frac{1}{z - n} + \frac{1}{z + n}\right). \]

Proof ▶

Let us start from the Mittag-Leffler expansion $\pi \cot \pi s = \frac{1}{s} + \sum _n \left(\frac{1}{s-n} + \frac{1}{s+n}\right)$.

Applying the trigonometric identity $\cot u - \cot \left(u + \frac{\pi }{2}\right) = \cot u + \tan u = \frac{2}{\sin 2 u}$ with $u=\pi z/2$, and letting $s = z/2$, $s = (z+1)/2$, we see that

\[ \begin{aligned} \frac{\pi }{\sin \pi z} & = \frac{\pi }{2} \cot \frac{\pi z}{2} - \frac{\pi }{2} \cot \frac{\pi (z+1)}{2} \\ & = \frac{1/2}{z/2} + \sum _n \left(\frac{1/2}{\frac{z}{2} -n} + \frac{1/2}{\frac{z}{2} +n}\right) -\frac{1/2}{(z+1)/2} - \sum _n \left(\frac{1/2}{\frac{z+1}{2} -n} + \frac{1/2}{\frac{z+1}{2} +n}\right)\\ & = \frac{1}{z} + \sum _n \left(\frac{1}{z - 2 n} + \frac{1}{z + 2 n}\right) - \sum _n \left(\frac{1}{z - (2 n - 1)} + \frac{1}{z + (2 n - 1)}\right) \end{aligned} \]

after reindexing the second sum. Regrouping terms again, we obtain our equation.

Lemma 7.4.12 Euler/Mittag-Leffler expansion for cosec squared

# Discussion

Let $z\in \mathbb {C}$, $z\notin \mathbb {Z}$. Then

\[ \frac{\pi ^2}{\sin ^2 \pi z} = \sum _{n=-\infty }^\infty \frac{1}{(z-n)^2}. \]

Proof ▶

Differentiate the expansion of $\pi \cot \pi z$ term-by-term because it converges uniformly on compact subsets of $\mathbb {C}\setminus \mathbb {Z}$. By $\left(\pi \cot \pi z\right)' = - \frac{\pi ^2}{\sin ^2 \pi z}$ and $\left(\frac{1}{z\pm n}\right)' = -\frac{1}{(z\pm n)^2}$, we are done.

Lemma 7.4.13 Estimate for an inverse cubic series

# Discussion

For $\vartheta \in \mathbb {R}$ with $0\leq |\vartheta |{\lt} 1$,

\[ \sum _n\left(\frac{1}{(n-\vartheta )^3} + \frac{1}{(n+\vartheta )^3}\right) \leq \frac{1}{(1-|\vartheta |)^3} + 2\zeta (3)-1. \]

Proof ▶

Since $\frac{1}{(n-\vartheta )^3} + \frac{1}{(n+\vartheta )^3}$ is even, we may replace $\vartheta $ by $|\vartheta |$. Then we rearrange the sum:

\[ \sum _{n=1}^\infty \left(\frac{1}{(n-|\vartheta |)^3} + \frac{1}{(n+|\vartheta |)^3}\right) = \frac{1}{(1-|\vartheta |)^3} + \sum _{n=1}^\infty \left(\frac{1}{\left(n+1-|\vartheta |\right)^3} + \frac{1}{\left(n+|\vartheta |\right)^3}\right). \]

We may write $(n+1-|\vartheta |)^3$, $(n+|\vartheta |)^3$ as $(n+\frac{1}{2}-t)^3$, $(n+\frac{1}{2} + t)^3$ for $t = |\vartheta |-1/2$. Since $1/u^3$ is convex, $\frac{1}{(n+1/2-t)^3} + \frac{1}{(n+1/2+t)^3}$ reaches its maximum on $[-1/2,1/2]$ at the endpoints. Hence

\[ \sum _{n=1}^\infty \left(\frac{1}{\left(n+1-|\vartheta |\right)^3} + \frac{1}{\left(n+|\vartheta |\right)^3}\right) \leq \sum _{n=1}^\infty \left(\frac{1}{n^3} + \frac{1}{(n+1)^3}\right) = 2 \zeta (3)-1. \]

Lemma 7.4.14 Estimate for a Fourier sum

# Discussion

Let $s = \sigma + i \tau $, $\sigma \geq 0$, $\tau \in \mathbb {R}$, with $s\ne 1$. Let $b{\gt}a{\gt}0$, $a, b\in \mathbb {Z} + \frac{1}{2}$, with $a{\gt}\frac{|\tau |}{2\pi }$. Define $f:\mathbb {R}\to \mathbb {C}$ by $f(y) = 1_{[a,b]}(y)/y^s$. Write $\vartheta = \frac{\tau }{2\pi a}$, $\vartheta _- = \frac{\tau }{2\pi b}$. Then

\[ \begin{aligned} \sum _n (\widehat{f}(n) + \widehat{f}(-n)) & = \frac{a^{-s} g(\vartheta )}{2 i} - \frac{b^{-s} g(\vartheta _-)}{2 i} + O^*\left(\frac{C_{\sigma ,\vartheta }}{a^{\sigma +1}}\right)\end{aligned} \]

with absolute convergence, where $g(t) = \frac{1}{\sin \pi t} - \frac{1}{\pi t}$ for $t\ne 0$, $g(0)=0$, and

\begin{equation} \label{eq:defcsigth}C_{\sigma ,\vartheta }= \begin{cases} \frac{\sigma }{2} \left(\frac{1}{\sin ^2\pi \vartheta } - \frac{1}{(\pi \vartheta )^2}\right) + \frac{|\vartheta |}{2\pi ^2} \left(\frac{1}{(1-|\vartheta |)^3} + 2\zeta (3)-1\right) & \text{for $\vartheta \ne 0$,}\\ \sigma /6 & \text{for $\vartheta = 0$.}\end{cases}\end{equation}

Proof ▶

By Proposition 7.4.1, multiplying by $2$ (since $e(-n t)+e(n t) = 2 \cos 2\pi n t$),

\begin{align} \widehat{f}(n) + \widehat{f}(-n) & = \notag \frac{a^{-s}}{2\pi i} \frac{(-1)^{n+1} 2\vartheta }{n^2 - \vartheta ^2} - \frac{b^{-s}}{2\pi i} \frac{(-1)^{n+1} 2\vartheta _-}{n^2 - \vartheta _-^2} \\ & + \frac{a^{-\sigma -1}}{2\pi ^2} O^*\left(\frac{\sigma }{(n-\vartheta )^2} + \frac{\sigma }{(n+\vartheta )^2} + \frac{|\vartheta |}{(n-\vartheta )^3} + \frac{|\vartheta |}{(n+\vartheta )^3}\right),\label{eq:abaderrcontrib}\end{align}

where $\vartheta _- = \tau /(2\pi b)$. Note $|\vartheta _-|\leq |\vartheta |{\lt}1$. By the Lemma ??,

\[ \sum _n \frac{(-1)^{n+1} 2 z}{n^2 - z^2} = \frac{\pi }{\sin \pi z} - \frac{1}{z} \]

for $z\ne 0$, while $\sum _n \frac{(-1)^{n+1} 2 z}{n^2 - z^2} = \sum _n 0 = 0$ for $z=0$. Moreover, by Lemmas 7.4.12 and 7.4.13, for $\vartheta \ne 0$,

\[ \sum _n \left(\frac{\sigma }{(n-\vartheta )^2} + \frac{\sigma }{(n+\vartheta )^2}\right)\leq \sigma \cdot \left(\frac{\pi ^2}{\sin ^2 \pi \vartheta } - \frac{1}{\vartheta ^2}\right), \]

\[ \sum _n \left(\frac{|\vartheta |}{(n-\vartheta )^3} + \frac{|\vartheta |}{(n+\vartheta )^3}\right) \leq |\vartheta |\cdot \left(\frac{1}{(1-|\vartheta |)^3} + 2\zeta (3)-1\right). \]

If $\vartheta =0$, then $\sum _n \left(\frac{\sigma }{(n-\vartheta )^2} + \frac{\sigma }{(n+\vartheta )^2}\right) = 2 \sigma \sum _{n=1}^\infty \frac{1}{n^2} = \sigma \frac{\pi ^2}{3}$.

Proposition 7.4.2 Approximation of zeta(s) by a partial sum

# Discussion

Let $s = \sigma + i \tau $, $\sigma \geq 0$, $\tau \in \mathbb {R}$, with $s\ne 1$. Let $a\in \mathbb {Z} + \frac{1}{2}$ with $a{\gt}\frac{|\tau |}{2\pi }$. Then

\begin{equation} \label{eq:abadlondie} \zeta (s) = \sum _{n\leq a} \frac{1}{n^s} - \frac{a^{1-s}}{1-s} + c_\vartheta a^{-s} + O^*\left(\frac{C_{\sigma ,\vartheta }}{a^{\sigma +1}}\right), \end{equation}

where $\vartheta = \frac{\tau }{2\pi a}$, $c_\vartheta = \frac{i}{2} \left(\frac{1}{\sin \pi \vartheta } - \frac{1}{\pi \vartheta }\right)$ for $\vartheta \ne 0$, $c_0 =0$, and $C_{\sigma ,\vartheta }$ is as in 4.

Proof ▶

Assume first that $\sigma {\gt}0$. Let $b\in \mathbb {Z}+\frac{1}{2}$ with $b{\gt}a$, and define $f(y) = \frac{1_{[a,b]}(y)}{y^s}$. By Lemma 7.4.9 and Lemma 7.4.10,

\[ \sum _{n\leq a} \frac{1}{n^s} = \zeta (s) + \frac{a^{1-s}}{1-s} - \lim _{N\to \infty } \sum _{n=1}^N (\widehat{f}(n) + \widehat{f}(-n)) + O^*\left(\frac{2 |s|}{\sigma b^\sigma }\right). \]

We apply Lemma 7.4.14 to estimate $\lim _{N\to \infty } \sum _{n=1}^N (\widehat{f}(n) + \widehat{f}(-n))$. We obtain

\[ \sum _{n\leq a} \frac{1}{n^s} = \zeta (s) + \frac{a^{1-s}}{1-s} - \frac{a^{-s} g(\vartheta )}{2 i} + O^*\left(\frac{C_{\sigma ,\vartheta }}{a^{\sigma +1}}\right) + \frac{b^{-s} g(\vartheta _-)}{2 i} + O^*\left(\frac{2 |s|}{\sigma b^\sigma }\right), \]

where $\vartheta _- = \frac{\tau }{2\pi b}$ and $g(t)$ is as in Lemma 7.4.14, and so $-\frac{g(\vartheta )}{2 i} = c_\vartheta $. We let $b\to \infty $ through the half-integers, and obtain 8, since $b^{-\sigma }\to 0$, $\vartheta _-\to 0$ and $g(\vartheta _-)\to g(0) = 0$ as $b\to \infty $.

Finally, the case $\sigma =0$ follows since all terms in 8 extend continuously to $\sigma =0$.

Remark 7.4.1

The term $c_\vartheta a^{-s}$ in 8 does not seem to have been worked out before in the literature; the factor of $i$ in $c_\vartheta $ was a surprise. For the sake of comparison, let us note that, if $a\geq x$, then $|\vartheta |\leq 1/2\pi $, and so $|c_\vartheta |\leq |c_{\pm 1/2\pi }| = 0.04291\dotsc $ and $|C_{\sigma ,\vartheta }|\leq |C_{\sigma ,\pm 1/2\pi }|\leq 0.176\sigma +0.246$. While $c_\vartheta $ is optimal, $C_{\sigma ,\vartheta }$ need not be – but then that is irrelevant for most applications.