I was going over my notes from a class on analytical number theory and we use a bound for the $\zeta$ function on the $1$ line as $\vert \zeta(1+it) \vert \leq \log(\vert t \vert) + \mathcal{O}(1)$ for $t$ bounded away from $0$, say $\vert t \vert \geq 1$. I don't seem to have a proof for this in my notes.

How does one prove the following bound for $\zeta(s)$ on the one line?

$$\zeta(1+it) \leq \log(|t|) + \mathcal{O}(1) \, \, \forall t \geq 1$$

  • $\begingroup$ Isn't the Zeta function defined for complex numbers $z=x+iy$ with $x>1$? $\endgroup$ – Mercy King Jun 23 '12 at 23:03
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    $\begingroup$ @Mercy Initially yes, but it can be extended to a meromorphic function in the entire plane, with just one pole at $s=1$. $\endgroup$ – user31373 Jun 23 '12 at 23:06

Here is a long answer. The main ingredient to get your bound is to find the analytic continuation to a region which includes the $1$ line.

You can skip the first part if you know how to continue $\zeta(s)$ analytically to the left of $\text{Re}(s) =1$.

First note that for $\text{Re}(s) > 1$, we have $$\zeta(s) = \sum_{n=1}^{\infty} \dfrac1{n^s}$$

As stated earlier, to have a bound on the $1$ line, we need to have $\zeta(s)$ to be defined in a half-plane containing the $1$ line i.e. we need to extend the above definition from $\text{Re}(s) > 1$ to a region $\text{Re}(s) > a$ where $a < 1$.

Analytic continuation of $\zeta(s)$ to the left of $1$ line:

One way to do it is as follows. We know that for $\text{Re}(s) > 1$, $\displaystyle \zeta(s) = \sum_{n=1}^{\infty} \dfrac1{n^s}$. We will extend this analytically to the region $\text{Re}(s) > 0$. In some sense, this is all we need not just for this problem but also for most of the problems on $\zeta$, since most often we are only interested in the behavior of the $\zeta(s)$ inside the critical strip i.e. $\{s \in \mathbb{C}: \text{Re(s)} \in (0,1)\}$. We make use of the powerful Euler–Maclaurin formula to make this analytic extension.

$$\displaystyle \zeta(s) = \sum_{n=1}^{\infty} \dfrac1{n^s} = \sum_{n=1}^{N} \dfrac1{n^s} + \sum_{n=N+1}^{\infty} \dfrac1{n^s}$$ The reason why we split this into two parts will become clear as well proceed. But to give a quick answer, the reason for this splitting into a sum $\leq N$ and a sum $>N$, is to allow us to play around with $N$ and optimize the error term accordingly.

We will now focus our attention on $\displaystyle \sum_{n=N+1}^{\infty} \dfrac1{n^s}$. Note that $$\displaystyle \sum_{n=N+1}^{\infty} \dfrac1{n^s} = \int_{N^+}^{\infty} \dfrac{d \lfloor t\rfloor}{t^s} = \left. \dfrac{\lfloor t \rfloor}{t^s} \right \vert_{N^+}^{\infty} + s \int_{N^+}^{\infty} \dfrac{\lfloor t \rfloor}{t^{s+1}} dt$$ So far, we have $\text{Re}(s) > 1$. Hence, $\left. \dfrac{\lfloor t \rfloor}{t^s} \right \vert_{N^+}^{\infty} = - \dfrac{N}{N^s} = - N^{1-s}$. Further, $$\int_{N^+}^{\infty} \dfrac{\lfloor t \rfloor}{t^{s+1}} dt = \int_{N^+}^{\infty} \dfrac{t - \{ t \}}{t^{s+1}} dt = \int_{N^+}^{\infty} \dfrac{dt}{t^s} - \int_{N^+}^{\infty} \dfrac{\{ t \}}{t^{s+1}} dt$$ Again, since $\text{Re}(s) > 1$, we have $$\int_{N^+}^{\infty} \dfrac{dt}{t^s} = - \dfrac{N^{1-s}}{1-s}$$ Hence, putting all these together, we now have that $$\displaystyle \sum_{n=N+1}^{\infty} \dfrac1{n^s} = - N^{1-s} - \dfrac{s}{1-s} N^{1-s} - s \int_{N^+}^{\infty} \dfrac{\{ t \}}{t^{s+1}} dt = \dfrac{ N^{1-s}}{s-1} - s \int_{N^+}^{\infty} \dfrac{\{ t \}}{t^{s+1}} dt$$ Hence, $$\zeta(s) = \sum_{n=1}^{N} \dfrac1{n^s} + \dfrac{N^{1-s}}{s-1} - s \int_{N^+}^{\infty} \dfrac{\{ t \}}{t^{s+1}} dt$$ However, note the little miracle here. The expression on the right now makes sense for $\text{Re}(s) > 0$, since $\displaystyle \int_{N^+}^{\infty} \dfrac{\{ t \}}{t^{s+1}} dt$ converges for $\text{Re}(s) >0$, and is analytic, and it matches with $\displaystyle \sum_{n=1}^{\infty} \dfrac1{n^s}$ for $\text{Re}(s) > 1$. This means that this is the analytic extension we were after.

Back to the problem

Now setting $s = 1 +it$, we get that \begin{align*} \zeta(1+it) & = \sum_{n \leq N} \frac1{n^{1+it}} - \frac{N^{1-(1+it)}}{1-(1+it)} - (1+it) \int_{N^+}^{\infty} \frac{\{y\}}{y^{2+it}} dy\\ & = \sum_{n \leq N} \frac1{n^{1+it}} + \frac{N^{-it}}{it} - (1+it) \int_{N^+}^{\infty} \frac{\{y\}}{y^{2+it}} dy \end{align*} We will now get a bound for $E(N,t) = \displaystyle \frac{N^{-it}}{it} - (1+it) \int_{N^+}^{\infty} \frac{\{y\}}{y^{2+it}} dy$. \begin{align*} \left \lvert E(N,t) \right \rvert & \leq \left \lvert \frac{N^{-it}}{it} \right \rvert + \displaystyle \left \lvert 1+it \right \rvert \left \lvert \int_{N^+}^{\infty} \frac{\{y\}}{y^{2+it}} dy \right \rvert\\ & \leq \frac1{\lvert t \rvert} + \displaystyle \left \lvert 1+it \right \rvert \left \lvert \int_{N^+}^{\infty} \frac1{y^{2+it}} dy \right \rvert\\ & \leq \frac1{\lvert t \rvert} + \displaystyle \left \lvert 1+it \right \rvert \int_{N^+}^{\infty} \left \lvert \frac1{y^{2+it}} \right \rvert dy\\ & \leq \frac1{\lvert t \rvert} + \displaystyle \left \lvert 1+it \right \rvert \int_{N^+}^{\infty} \frac1{y^{2}} dy\\ & = \frac1{\lvert t \rvert} + \frac{\left \lvert 1+it \right \rvert}{N} \end{align*} Now we need to optimally choose $N$ to get a good error bound. This is why we split the original summation into two parts.

Note that since $\lvert t \rvert \geq 1$, we get that $\left \lvert 1 + it \right \rvert \leq \lvert t \rvert \sqrt{2}$. If we choose $N = \lfloor \lvert t \rvert \rfloor$, we get that $$\left \lvert E(N,t) \right \rvert \leq \frac1{\lvert t \rvert} + \frac{\lvert t \rvert \sqrt{2}}{\lfloor \lvert t \rvert \rfloor}.$$ Since, $\displaystyle |t| \geq 1$, we get that $\displaystyle \frac1{\lvert t \rvert} \leq 1$ and also that $\displaystyle \frac{\lvert t \rvert}{\lfloor\lvert t \rvert \rfloor} \leq 2$. Hence, we get that $$\displaystyle \left \lvert E(N,t) \right \rvert \leq 1 + 2 \sqrt{2}$$ i.e. $E(N,t) = \mathcal{O}(1)$. Hence, we have that $$\zeta(1+it) = \sum_{n \leq N} \frac1{n^{1+it}} + \frac{N^{-it}}{it} - (1+it) \int_{N^+}^{\infty} \frac{\{y\}}{y^{2+it}} dy\\ = \sum_{n \leq \lvert t \rvert} \frac1{n^{1+it}} + E(N,t) = \sum_{n \leq \lvert t \rvert} \frac1{n^{1+it}} + \mathcal{O}(1).$$

Finishing steps

Now we are almost done. Taking the absolute value we get that \begin{align*} \lvert \zeta(1+it) \rvert & = \left \lvert \sum_{n \leq \lvert t \rvert} \frac1{n^{1+it}} + E(N,t) \right \rvert\\ & \leq \sum_{n \leq \lvert t \rvert} \left \lvert \frac1{n^{1+it}} \right \rvert + \left \lvert E(N,t) \right \rvert\\ & = \sum_{n \leq \lvert t \rvert} \frac1{n} + \left \lvert E(N,t) \right \rvert\\ & \leq \log \left( \left \lvert t \right \rvert \right) + \mathcal{O}(1) + \mathcal{O}(1) \text{ (Since $\displaystyle \sum_{n \leq |t|} \dfrac1n$ goes as $\log(|t|) + \gamma + \mathcal{O}(1/|t|)$)}\\ & = \log \left( \left \lvert t \right \rvert \right) + \mathcal{O}(1). \end{align*} Hence, we get what we want, viz, $$\lvert \zeta(1+it) \rvert \leq \log \left( \left \lvert t \right \rvert \right) + \mathcal{O}(1).$$

You should also look here where another asymptotic for $\zeta(s)$ on the left of the $0$-line is obtained by different argument. You could also follow a similar method as described in the other question i.e. use the functional equation described in that to obtain a bound for $\zeta(s)$ on the $1$-line. You will need bounds for $\zeta(it)$, if you want to make use of the functional equation. But either way you cannot escape some long arguments. And I think what I have written is a bit more straight-forward than trying to make use of the functional equation.

  • $\begingroup$ Thanks for the elaborate answer! $\endgroup$ – abhIta Jun 23 '12 at 23:50
  • $\begingroup$ Will read through and accept it later $\endgroup$ – abhIta Jun 23 '12 at 23:52
  • $\begingroup$ Your analytic continuation of $\zeta$ into the strip $0<\mathrm{Re}(z)<1$ almost duplicates the first step in proving the Euler-Maclaurin Sum Formula. Nice (+1). $\endgroup$ – robjohn Jun 24 '12 at 15:51
  • $\begingroup$ @robjohn I have actually only used the Euler-Maclaurin summation there. I have mentioned in my post as well. (Probably the post was a bit too long that you probably missed reading it.) $\endgroup$ – user17762 Jun 24 '12 at 16:18
  • $\begingroup$ @Marvis: I missed the sentence where you mentioned "the powerful Euler–Maclaurin formula." However, it seems to me that you actually derived the first iteration of the formula for $t^{-s}$ before using it. $\endgroup$ – robjohn Jun 24 '12 at 18:24

Analytic Continuation of $\boldsymbol{\zeta(z)}$ with Euler-Maclaurin

The Euler-Maclaurin Sum formula says that, as a function of $n$, $$ \sum_{k=1}^n\frac{1}{k^z}=\zeta^\ast(z)+\frac{1}{1-z}n^{1-z}+\frac12n^{-z}+O\left(zn^{-1-z}\right)\tag{1} $$ for some $\zeta^\ast(z)$. Note that for $\mathrm{Re}(z)>1$, $\zeta^\ast(z)=\zeta(z)$.

For all $z\in\mathbb{C}\setminus\{1\}$, define $$ \zeta_n(z)=\sum_{k=1}^n\frac{1}{k^z}-\frac{1}{1-z}n^{1-z}\tag{2} $$ Note that each $\zeta_n$ is analytic and equation $(1)$ says that $$ \zeta_n(z)=\zeta^\ast(z)+\frac12n^{-z}+O\left(zn^{-1-z}\right)\tag{3} $$ which says that for $\mathrm{Re}(z)>0$, $$ \lim_{n\to\infty}\zeta_n(z)=\zeta^\ast(z)\tag{4} $$ and the convergence is uniform on compact subsets of $\mathbb{C}\setminus\{1\}$. Thus, the $\zeta^\ast(z)$ defined in $(4)$ is analytic and agrees with $\zeta(z)$ for $\mathrm{Re}(z)>1$. Thus, $\zeta^\ast(z)$ is the analytic continuation of $\zeta(z)$ for $\mathrm{Re}(z)>0$.

Application to $\boldsymbol{|\zeta(1+is)|}$

Therefore, using $(1)$, we get that $$ \zeta(1+it)=\sum_{k=1}^n\frac{1}{k^{1+it}}+\frac{1}{it}n^{-it}-\frac{1}{2n}n^{-it}+O\left(\frac{t}{n^2}\right)\tag{5} $$ So as long as $n\ge|t|$, we get that $$ \zeta(1+it)=\sum_{k=1}^n\frac{1}{k^{1+it}}+O\left(\frac1t\right)\tag{6} $$ Using the Euler-Maclaurin Sum formula again, we get that $$ \sum_{k=1}^n\frac1k=\log(n)+\gamma+O\left(\frac1n\right)\tag{7} $$ Letting $n=\lceil|t|\rceil$, and using $(6)$ and $(7)$, we get for large $|t|$, $$ |\zeta(1+it)|\le\log(|t|)+C\tag{8} $$

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    $\begingroup$ Carrying the Euler-Maclaurin expansion further, we can continue the $\zeta$ function as far to the left of $\mathrm{Re}(z)=1$ as we desire. $\endgroup$ – robjohn Jun 24 '12 at 14:57

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