Utilizing Smoothing Techniques to Bound $|\zeta(1+it)|$

Andrew Christensen Email address: andrewchristensenj@gmail.com and Kyle Pratt Email address: kyle.pratt@mathematics.byu.edu Address: Department of Mathematics, Brigham Young University, Provo, UT 84602, USA

Abstract.

We demonstrate an improved explicit upper bound of $|\zeta(1+it)|$ for $3\leq t\leq 10^{9}$ using smoothing techniques. Our method sharpens previous bounds relying on the Riemann–Siegel formula and the triangle inequality. In particular, we prove that for $t\geq 3$ ,

\displaystyle|\zeta(1+it)|\leq\frac{1}{2}\log t+1.57

and for $t\geq 10^{8}$ ,

|\zeta(1+it)|\leq\frac{1}{3}\log t+2\log\log t-1.16.

Key words and phrases:

Riemann zeta function, upper bound, 1-line, numerical integration, smoothing

2020 Mathematics Subject Classification

11M06, 11Y35

1. Introduction

Studying the behavior of the Riemann zeta function on the one-line has been of particular interest for a number of years, largely due to its utility in expanding the zero-free regions of $\zeta(s)$ [3]. It is often useful to have quality explicit upper bounds on $|\zeta(1+it)|$ , particularly when probing zero-free regions of the zeta function by computational means. Several such bounds have been proved, including in [9] and [6]. A vast majority of these bounds rely heavily on the Riemann–Siegel formula or similar approximate functional equations of the zeta function where the value of $\zeta(1+it)$ is approximated by

\zeta(1+it)\approx\sum_{n=1}^{N(t)}\frac{1}{n^{1+it}}

for some $N(t)$ .

For large $t$ , previous researchers have applied exponential sum techniques to the Riemann–Siegel formula to achieve quality explicit upper bounds of $|\zeta(1+it)|$ . In [9], Patel proved

|\zeta(1+it)|\leq\min\left(\log t,\frac{1}{2}\log t+1.93,\frac{1}{5}\log t+44.02\right)

(1.1)

for $t\geq 3$ . The bounds $\log t$ and $\frac{1}{2}\log t+1.93$ are essentially trivial bounds arising from applying the triangle inequality to the Riemann–Siegel formula. The bound $\frac{1}{5}\log t+44.02$ arises via exponential sum estimates and is better than the previous bounds only when $t$ is sufficiently large.

Hiary, Leong, and Yang [6] refined the work of Patel and showed

|\zeta(1+it)|\leq 1.731\frac{\log t}{\log\log t}

(1.2)

for $t\geq 3$ . Hiary, Leong, and Yang use further exponential sum arguments to improve (1.1) for large $t$ , but for $t\leq e^{16}\approx 8.88\cdot 10^{6}$ they use Patel’s bound $|\zeta(1+it)|\leq\frac{1}{2}\log t+1.93$ .

Working with smoothed sums can sometimes yield improved quantitative or qualitative results. One noteworthy example of this philosophy is Helfgott’s proof of the ternary Goldbach conjecture (see [4]), in which smoothed sums appear throughout the argument. We show that utilizing smoothing techniques can improve the “trivial” bounds of $|\zeta(1+it)|$ for small $t$ . In particular, we obtain improved upper bounds for $t\leq 10^{9}$ . We essentially work with a smooth representation

\zeta(1+it)\approx\sum_{n=1}^{\infty}{\frac{1}{n^{1+it}}\exp{\left(-\frac{n}{X}\right)}}

(1.3)

for suitable $X$ . The following is our main result.

Theorem 1.1.

If $t\geq 3$ , then

|\zeta(1+it)|\leq\frac{1}{2}\log t+1.57.

(1.4)

Furthermore, if $t\geq 10^{8}$ , then

|\zeta(1+it)|\leq\frac{1}{3}\log t+2\log\log t-1.16.

(1.5)

Theorem 1.1 improves upon (1.1) and (1.2) for $t\leq 10^{9}$ . While both bounds rely heavily on smoothing, the bound

|\zeta(1+it)|\leq\frac{1}{3}\log t+2\log\log t-1.16

also depends in part on exponential sum techniques. However, the bound

|\zeta(1+it)|\leq\frac{1}{2}\log t+1.57

relies only on a trivial bound for $|\zeta(1/2+it)|$ . This bound, without using exponential sums, is an improvement of (1.1) and (1.2) for $t\leq 4\cdot 10^{8}$ .

In this paper, we focus on smoothings of the form seen in (1.3), but it would be interesting to consider the effects of different smoothing functions. It is possible that using exponential sums and smoothing techniques in tandem yields even better results. It would also be worthwhile to see if using smoothings could sharpen bounds on $|\zeta^{-1}(1+it)|$ or $|\frac{\zeta^{\prime}}{\zeta}(1+it)|$ in ranges of interest.

In Section 2 we prove a smoothed representation of $\zeta(1+it)$ of the form seen in (1.3) and show that that smoothed sum can be bounded by $\log X$ plus some small error. In Section 3, we twice bound our main error term, once optimizing for smaller $t$ and once for larger $t$ . Finally, in Section 4 we prove Theorem 1.1.

Some of our work relies on computer calculation. All the code used in this paper is available at our “Utilizing Smoothing Techniques to Bound $\zeta(1+it)$ ” GitHub repository [1].

2. A Smoothed Representation of $\zeta(1+it)$

We begin by proving a smoothed approximation of $\zeta(1+it)$ of the form given in (1.3).

Lemma 2.1.

For all $t\in\mathbb{R}\setminus\{0\}$ and $X>0$ ,

	$\displaystyle\|\zeta(1+it)\|\leq$	$\displaystyle\sum_{n=1}^{\infty}{\frac{1}{n}\exp{\left(-\frac{n}{X}\right)}}+\left\|\Gamma(-it)\right\|$		(2.1)
		$\displaystyle+\frac{X^{-1/2}}{2\pi}\int_{-\infty}^{\infty}\left\|\Gamma(-1/2+iu)\right\|\cdot\left\|\zeta(1/2+i(t+u))\right\|du.$

Proof.

Since the inverse Mellin transform of $\Gamma(s)$ for $\operatorname{Re}(s)>0$ is $e^{-y}$ [8, 2.5.1], we can write

\frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}y^{s}\Gamma(s)\ ds=\exp\left(-\frac{1}{y}\right).

Taking $y=X/n$ and summing over $n$ , we can write

	$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{1+it}}\exp\left(-\frac{n}{X}\right)$	$\displaystyle=\sum_{n=1}^{\infty}\frac{1}{n^{1+it}}\frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}\left(\frac{X}{n}\right)^{s}\Gamma(s)\ ds$
		$\displaystyle=\frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}X^{s}\Gamma(s)\sum_{n=1}^{\infty}\frac{1}{n^{1+it+s}}ds$
		$\displaystyle=\frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}X^{s}\Gamma(s)\zeta(1+it+s)\ ds,$

where interchanging the order of summation and integration is justified by absolute convergence.

We shift the line of integration to $\text{Re}(s)=-\frac{1}{2}$ , picking up contributions from poles at $s=0$ and $s=-it$ . Hence

\displaystyle\sum_{n=1}^{\infty}

\displaystyle\frac{1}{n^{1+it}}\exp\left(-\frac{n}{X}\right)-\frac{1}{2\pi i}\int_{-1/2-i\infty}^{-1/2+i\infty}X^{s}\Gamma(s)\zeta(1+it+s)\ ds=\zeta(1+it)+X^{-it}\Gamma(-it),

and rearranging gives

\zeta(1+it)=\sum_{n=1}^{\infty}\frac{1}{n^{1+it}}\exp\left(-\frac{n}{X}\right)-X^{-it}\Gamma(-it)-\frac{1}{2\pi i}\int_{-1/2-i\infty}^{-1/2+i\infty}X^{s}\Gamma(s)\zeta(1+it+s)\ ds.

Applying the triangle inequality yields

|\zeta(1+it)|\leq\sum_{n=1}^{\infty}{\frac{1}{n}\exp{\left(-\frac{n}{X}\right)}}+\left|\Gamma(-it)\right|+\frac{1}{2\pi}\int_{-1/2-i\infty}^{-1/2+i\infty}\left|{X^{s}\Gamma(s)}\zeta(1+it+s)\right|ds.

We perform a change of variables in the integral to obtain

	$\displaystyle\frac{1}{2\pi}\int_{-1/2-i\infty}^{-1/2+i\infty}\left\|{X^{s}\Gamma(s)}\zeta(1+it+s)\right\|ds$	$\displaystyle=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left\|{X^{-1/2+iu}\Gamma(-1/2+iu)}\zeta(1/2+i(t+u))\right\|du$
		$\displaystyle=\frac{X^{-1/2}}{2\pi}\int_{-\infty}^{\infty}\left\|\Gamma(-1/2+iu)\right\|\cdot\left\|\zeta(1/2+i(t+u))\right\|du,$

which gives the desired bound. ∎

We now proceed by bounding the terms on the right-hand side of Lemma 2.1 individually. We begin with the summation.

Lemma 2.2.

If $X>0$ , $\xi\coloneqq e-2.5$ , and

c_{0}\coloneqq\frac{12\xi-1}{24},\ \ c_{1}\coloneqq\frac{60\xi^{2}-1}{120},

then

\sum_{n=1}^{\infty}\frac{1}{n}\exp\left(-\frac{n}{X}\right)\leq\log X+\frac{1}{2X}-\frac{1}{24X^{2}}+\frac{c_{0}}{X^{3}}+\frac{c_{1}}{X^{4}}.

(2.2)

Proof.

Since $X>0$ , it follows that $|\exp(-1/X)|<1$ . Therefore,

\displaystyle\sum_{n=1}^{\infty}{\frac{1}{n}\exp{\left(-\frac{n}{X}\right)}}

\displaystyle=-\log\left(1-e^{-1/X}\right)=\log X-\log\left(X\left(e^{1/X}-1\right)\right)+\frac{1}{X}.

(2.3)

We write $X\left(e^{1/X}-1\right)$ as a Taylor series to obtain

\displaystyle-\log\left(X\left(e^{1/X}-1\right)\right)

\displaystyle=-\log\left(1+\frac{1}{2X}+\frac{1}{6X^{2}}+\ldots\right).

For ease of notation, we define

	$\displaystyle B$	$\displaystyle\coloneqq\sum_{k=1}^{\infty}\frac{1}{(k+1)!X^{k}}=\frac{1}{2X}+\frac{1}{6X^{2}}+\frac{1}{24X^{3}}+\ldots,$
	$\displaystyle C$	$\displaystyle\coloneqq\sum_{k=2}^{\infty}\frac{1}{(k+1)!X^{k}}=\frac{1}{6X^{2}}+\frac{1}{24X^{3}}+\frac{1}{120X^{4}}+\ldots,$

so that

-\log\left(X\left(e^{1/X}-1\right)\right)=-\log\left(1+B\right)=-\log\left(1+\frac{1}{2X}+C\right).

Furthermore, note that we can write

B=\frac{1}{2X}+C,\ \ \ B^{2}=\frac{1}{4X^{2}}+\frac{C}{X}+C^{2}.

Expanding as a Taylor series gives

	$\displaystyle-\log\left(X\left(e^{1/X}-1\right)\right)$	$\displaystyle=-\log\left(1+B\right)=-B+\frac{B^{2}}{2}-\frac{B^{3}}{3}+\ldots$
		$\displaystyle\leq-B+\frac{B^{2}}{2}=-B+\frac{1}{8X^{2}}+\frac{C}{2X}+\frac{C^{2}}{2}.$

We can bound $C$ by

C=\sum_{k=2}^{\infty}\frac{1}{(k+1)!X^{k}}\leq\frac{1}{X^{2}}\sum_{k=3}^{\infty}\frac{1}{k!}=\frac{e-2.5}{X^{2}}=\frac{\xi}{X^{2}}.

Applying this bound, we find

	$\displaystyle-\log\left(X\left(e^{1/X}-1\right)\right)$	$\displaystyle\leq-\frac{1}{2X}-\frac{1}{6X^{2}}-\frac{1}{24X^{3}}-\frac{1}{120X^{4}}-\ldots+\frac{1}{8X^{2}}+\frac{\xi}{2X^{3}}+\frac{\xi^{2}}{2X^{4}}$
		$\displaystyle\leq-\frac{1}{2X}-\frac{1}{24X^{2}}+\frac{12\xi-1}{24X^{3}}+\frac{60\xi^{2}-1}{120X^{4}}.$

Using this, (2.3) becomes

	$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}\exp\left(-\frac{n}{X}\right)$	$\displaystyle\leq\log X-\frac{1}{2X}-\frac{1}{24X^{2}}+\frac{c_{0}}{X^{3}}+\frac{c_{1}}{X^{4}}+\frac{1}{X}$
		$\displaystyle=\log X+\frac{1}{2X}-\frac{1}{24X^{2}}+\frac{c_{0}}{X^{3}}+\frac{c_{1}}{X^{4}}.\qed$

3. Bounding the Integral

To aid in bounding the integral in (2.1), we make use of two previously proven bounds on the Riemann zeta function.

Lemma 3.1.

If $t\geq 200$ , then

|\zeta(1/2+it)|\leq\frac{4t^{1/4}}{(2\pi)^{1/4}}-2.08.

(3.1)

Furthermore, for $t\geq 3$ , we have

|\zeta(1/2+it)|\leq 0.618t^{1/6}\log t.

(3.2)

Proof.

The first bound (3.1), proved in [5], is an improved version of the Riemann–Siegel–Lehman bound. The bound (3.2) is proved in [7], and is based on exponential sum techniques. ∎

While (3.1) is essentially a trivial bound, the bound (3.2) relies on exponential sums. The trivial bound is better for $t\lessapprox 5\cdot 10^{7}$ while (3.2) is better for larger $t$ .

As (3.1) and (3.2) are both of the form $|\zeta(1/2+it)|\leq c_{2}t^{a}(\log t)^{b}-c_{3}$ , we bound the integral in (2.1) using an arbitrary bound on $|\zeta(1/2+it)|$ of this form.

For the duration of this paper, we fix $A_{0}\coloneqq 200$ .

Lemma 3.2.

If there exist $a,b,c_{2},c_{3}\in\mathbb{R}_{\geq 0}$ with $a\leq 1$ such that

|\zeta(1/2+iw)|\leq c_{2}w^{a}(\log w)^{b}-c_{3}

(3.3)

for all $w\geq 210$ , then for all $t\geq 210$ we have

\displaystyle\int_{-\infty}^{\infty}\left|\Gamma(-1/2+iu)\right|\cdot

\displaystyle\left|\zeta(1/2+i(t+u))\right|du\leq c_{2}\big(t^{a}I_{1}+I_{2}\big)+I_{3}+2.27\cdot 10^{-5},

(3.4)

where

	$\displaystyle I_{1}$	$\displaystyle\coloneqq\int_{A_{0}}^{\infty}\left\|\Gamma(-1/2+iu)\right\|\log^{b}(2u)du,\ \ \ \ \ \ \ \ I_{2}\coloneqq\int_{A_{0}}^{\infty}\left\|\Gamma(-1/2+iu)\right\|u^{a}\log^{b}(2u)du.$
	$\displaystyle I_{3}$	$\displaystyle\coloneqq\int_{A_{0}-t}^{\infty}\left\|\Gamma(-1/2+iu)\right\|\cdot\left\|\zeta(1/2+i(t+u))\right\|du.$

Remark 3.3.

Since our choice of $A_{0}$ is sufficiently large, the variable $u$ in the integrals $I_{1}$ and $I_{2}$ is far enough from the origin that the exponential decay of the gamma function causes their value to be small. As such, the effect of these integrals on our error is minimal. Most of our error will arise from $I_{3}$ , which is large since its range contains the origin. In an attempt to maximize our savings we bound $I_{3}$ twice, using both bounds of $|\zeta(1/2+it)|$ found in Lemma 3.1. The bound in (3.2) will give better error for large $t$ , while (3.1) will be more effective for small values of $t$ .

Proof of Lemma 3.2.

Due to the conjugate symmetry of the zeta function, it follows that the bound $|\zeta(1/2-iw)|\leq c_{2}w^{a}\log^{b}w-c_{3}$ holds for all $w\geq 210$ . As such, we define

	$\displaystyle I_{4}$	$\displaystyle\coloneqq\int_{-\infty}^{-A_{0}-t}\left\|\Gamma(-1/2+iu)\right\|\cdot\left\|\zeta(1/2+i(t+u))\right\|du,$
	$\displaystyle I_{5}$	$\displaystyle\coloneqq\int_{-A_{0}-t}^{A_{0}-t}\left\|\Gamma(-1/2+iu)\right\|\cdot\left\|\zeta(1/2+i(t+u))\right\|du,$

and split the integral

\displaystyle\int_{-\infty}^{\infty}\left|\Gamma(-1/2+iu)\right|\cdot\left|\zeta(1/2+i(t+u))\right|du=

\displaystyle I_{3}+I_{4}+I_{5}.

(3.5)

We bound $I_{4}$ and $I_{5}$ separately.

First, consider $I_{4}$ . Using our bound (3.3), we have

I_{4}\leq I_{6}-I_{7}

where

	$\displaystyle I_{6}$	$\displaystyle\coloneqq c_{2}\int_{-\infty}^{-A_{0}-t}\left\|\Gamma(-1/2+iu)\right\|\cdot\|t+u\|^{a}\log^{b}\|t+u\|du$
	$\displaystyle I_{7}$	$\displaystyle\coloneqq c_{3}\int_{-\infty}^{-A_{0}-t}\left\|\Gamma(-1/2+iu)\right\|du.$

Consider $I_{6}$ . Then $t\leq|u|$ , so we can bound

\log|t+u|\leq\log(t+|u|)\leq\log(2|u|).

Furthermore, since $a\leq 1$ , we have $|t+u|^{a}\leq t^{a}+|u|^{a}$ . Thus we have

	$\displaystyle I_{6}$	$\displaystyle\leq c_{2}\int_{-\infty}^{-A_{0}-t}\left\|\Gamma(-1/2+iu)\right\|\left(t^{a}+\|u\|^{a}\right)\log^{b}(2\|u\|)du$
		$\displaystyle=c_{2}t^{a}\int_{-\infty}^{-A_{0}-t}\left\|\Gamma(-1/2+iu)\right\|\log^{b}(2\|u\|)du+c_{2}\int_{-\infty}^{-A_{0}-t}\left\|\Gamma(-1/2+iu)\right\|\|u\|^{a}\log^{b}(2\|u\|)du.$

To eliminate the $t$ -dependence in the bounds of integration, we can bound $-A_{0}-t\leq-A_{0}$ and write

\displaystyle I_{6}

\displaystyle\leq c_{2}t^{a}\int_{-\infty}^{-A_{0}}\left|\Gamma(-1/2+iu)\right|\log^{b}(2|u|)du+c_{2}\int_{-\infty}^{-A_{0}}\left|\Gamma(-1/2+iu)\right||u|^{a}\log^{b}(2|u|)du.

However, due to the conjugate symmetry of the gamma function, these integrals are exactly $I_{1}$ and $I_{2}$ , so we have

I_{6}\leq c_{2}\big(t^{a}I_{1}+I_{2}\big).

For $I_{7}$ , since its integrand and $c_{3}$ are nonnegative, we can simply bound $I_{7}\geq 0$ . Thus we have

I_{4}\leq I_{6}-I_{7}\leq c_{2}\big(t^{a}I_{1}+I_{2}\big).

(3.6)

Next consider $I_{5}$ . Letting $\omega=t+u$ , it becomes

I_{5}=\int_{-A_{0}}^{A_{0}}|\Gamma(-1/2+i(\omega-t))|\cdot|\zeta(1/2+i\omega)|d\omega.

Noting the relation [2, 5.4.4]

|\Gamma(1/2+iy)|^{2}=\frac{\pi}{\cosh(\pi y)},

(3.7)

we use the functional equation of the gamma function to find

|\Gamma(-1/2+i(\omega-t))|=\Bigg[\frac{\pi}{\cosh(\pi(\omega-t))}\frac{1}{\frac{1}{4}+(\omega-t)^{2}}\Bigg]^{1/2}.

Therefore we have

	$\displaystyle I_{5}$	$\displaystyle=\int_{-A_{0}}^{A_{0}}\Bigg[\frac{\pi}{\cosh(\pi(\omega-t))}\frac{1}{\frac{1}{4}+(\omega-t)^{2}}\Bigg]^{1/2}\|\zeta(1/2+i\omega)\|d\omega$
		$\displaystyle\leq\Bigg[\frac{\pi}{\cosh(\pi(t-A_{0}))}\frac{1}{\frac{1}{4}+(t-A_{0})^{2}}\Bigg]^{1/2}\int_{-A_{0}}^{A_{0}}\|\zeta(1/2+i\omega)\|d\omega.$

Bounding this trivially (using $t\geq 210$ ) and evaluating the integral numerically in [1] gives

I_{5}\leq 2.27\cdot 10^{-5}.

(3.8)

Combining our results from (3.5), (3.6), and (3.8) we arrive at the desired expression:

\displaystyle\int_{-\infty}^{\infty}\left|\Gamma(-1/2+iu)\right|\cdot\left|\zeta(1/2+i(t+u))\right|du

\displaystyle\leq c_{2}\big(t^{a}I_{1}+I_{2}\big)+I_{3}+2.27\cdot 10^{-5}.\qed

The integrals $I_{1}$ and $I_{2}$ can both be evaluated numerically, with the tail bounded analytically for various choices of $a$ and $b$ . These bounds are somewhat tedious and uninteresting so we do not give all the details. However, we will demonstrate a bound of the tail of $I_{1}$ with $b=0$ and note that we can bound the others quite similarly.

Lemma 3.4.

If $b=0$ , then

I_{1}=\int_{A_{0}}^{\infty}|\Gamma(-1/2+iu)|du\leq 10^{-133}.

Proof.

We can split $I_{1}$ into two regions:

I_{1}=\int_{A_{0}}^{\infty}|\Gamma(-1/2+iu)|du=\int_{A_{0}}^{B_{0}}|\Gamma(-1/2+iu)|du+\int_{B_{0}}^{\infty}|\Gamma(-1/2+iu)|du.

Taking $B_{0}=1000$ , we bound the first integral using complex ball arithmetic in [1] and find

\int_{A_{0}}^{B_{0}}|\Gamma(-1/2+iu)|du\leq 10^{-134}.

For the tail, we can use the functional equation of the gamma function along with (3.7) to write

I_{1}^{\prime}\coloneqq\int_{B_{0}}^{\infty}|\Gamma(-1/2+iu)|du=\int_{B_{0}}^{\infty}\frac{|\Gamma(1/2+iu)|}{|-1/2+iu|}du=\int_{B_{0}}^{\infty}\frac{\pi^{1/2}}{|-1/2+iu|\sqrt{\cosh(\pi u)}}du.

We then quickly bound the resulting integral in as follows:

\displaystyle I_{1}^{\prime}

\displaystyle\leq 2\sqrt{2\pi}\int_{B_{0}}^{\infty}\frac{du}{\left(e^{\pi u}+e^{-\pi u}\right)^{1/2}}\leq 2\sqrt{2\pi}\int_{B_{0}}^{\infty}e^{-\pi u/2}du=\frac{4\sqrt{2}}{\sqrt{\pi}}e^{-B_{0}\pi/2}\leq 10^{-681}.

With this, we find

I_{1}\leq 10^{-134}+10^{-681}\leq 10^{-133}.\qed

We can similarly bound both $I_{1}$ and $I_{2}$ for various values of $a$ and $b$ . However, the integral $I_{3}$ still has some $t$ dependence. Since $I_{3}$ provides the largest contribution to our eventual error term, we treat it carefully. In the following two lemmas, we provide two different bounds on $I_{3}$ , the first using the bound in (3.1) and the second using the bound in (3.2).

Lemma 3.5.

If $t\geq 210$ , then

I_{3}\leq 4.0315c_{2}t^{1/4}+0.4618c_{2}t^{-3/4}-4.0314c_{3},

(3.9)

where $c_{2}=4/(2\pi)^{1/4}$ and $c_{3}=2.08$ .

Proof.

Since $t+u\geq t+(A_{0}-t)=A_{0}=200$ , Lemma 3.1 yields

\left|\zeta(1/2+i(t+u))\right|\leq c_{2}(t+u)^{1/4}-c_{3}.

It follows that

I_{3}\leq c_{2}\int_{A_{0}-t}^{\infty}|\Gamma(-1/2+iu)|(t+u)^{1/4}du-c_{3}\int_{A_{0}-t}^{\infty}|\Gamma(-1/2+iu)|du.

We define

	$\displaystyle I_{8}$	$\displaystyle\coloneqq c_{2}\int_{A_{0}-t}^{\infty}\|\Gamma(-1/2+iu)\|(t+u)^{1/4}du,$
	$\displaystyle I_{9}$	$\displaystyle\coloneqq c_{3}\int_{A_{0}-t}^{\infty}\|\Gamma(-1/2+iu)\|du,$

so that $I_{3}\leq I_{8}-I_{9}$ .

We first examine $I_{8}$ . Pulling out a factor of $t^{1/4}$ gives

I_{8}=c_{2}t^{1/4}\int_{A_{0}-t}^{\infty}|\Gamma(-1/2+iu)|\left(1+\frac{u}{t}\right)^{1/4}du.

Since $u/t\geq-1$ , we see by concavity that $(1+u/t)^{1/4}\leq 1+u/4t$ . Inserting this bound and expanding the integral gives

\displaystyle I_{8}\leq

\displaystyle c_{2}t^{1/4}\int_{A_{0}-t}^{\infty}|\Gamma(-1/2+iu)|du+\frac{c_{2}}{4t^{3/4}}\int_{A_{0}-t}^{\infty}|\Gamma(-1/2+iu)|udu.

We obtain an upper bound on the integrals, and simultaneously remove their $t$ -dependence, by extending the integrals to the entire real line. Hence

\displaystyle I_{8}\leq

\displaystyle c_{2}t^{1/4}\int_{-\infty}^{\infty}|\Gamma(-1/2+iu)|du+\frac{c_{2}}{4t^{3/4}}\int_{-\infty}^{\infty}|\Gamma(-1/2+iu)||u|du.

Rigorously integrating these numerically using SageMath in [1] and bounding the tail (similarly to the proof of Lemma 3.4), we find

I_{8}\leq 4.0315c_{2}t^{1/4}+0.4618c_{2}t^{-3/4}.

Similarly we get a lower bound $I_{9}\geq 4.0314c_{3}$ , and we arrive at the desired conclusion. ∎

Lemma 3.6.

If $t\geq 210$ and $c_{2}=0.618$ , then

\displaystyle I_{3}\leq c_{2}\Bigg[4.0315t^{1/6}\log t+1.8469\ t^{-5/6}+3.3032\log t+1.7493\ t^{-1}\Bigg].

(3.10)

Proof.

Since $u\geq A_{0}-t$ , Lemma 3.1 gives

|\zeta(1/2+i(t+u))|\leq c_{2}(t+u)^{1/6}\log(t+u).

Therefore, we have

I_{3}\leq c_{2}\int_{A_{0}-t}^{\infty}|\Gamma(-1/2+iu)|(t+u)^{1/6}\log(t+u)du.

Since $u\geq A_{0}-t>-t$ , we have $\frac{u}{t}>-1$ , and therefore

\log(t+u)=\log t+\log\left(1+\frac{u}{t}\right)\leq\log t+\frac{|u|}{t}.

We also have the upper bound

(t+u)^{1/6}\leq t^{1/6}+|u|^{1/6}.

Inserting both of these bounds gives

I_{3}\leq c_{2}\int_{A_{0}-t}^{\infty}|\Gamma(-1/2+iu)|(t^{1/6}+|u|^{1/6})\left(\log t+\frac{|u|}{t}\right)du.

Extending the range of integration to the whole real line and expanding gives

	$\displaystyle I_{3}\leq$	$\displaystyle c_{2}\Bigg[t^{1/6}\log t\int_{-\infty}^{\infty}\|\Gamma(-1/2+iu)\|du+t^{-5/6}\int_{-\infty}^{\infty}\|\Gamma(-1/2+iu)\|\|u\|du$
		$\displaystyle+\log t\int_{-\infty}^{\infty}\|\Gamma(-1/2+iu)\|\|u\|^{1/6}du+t^{-1}\int_{-\infty}^{\infty}\|\Gamma(-1/2+iu)\|\|u\|^{7/6}du\Bigg].$

The first two integrals we can bound the tails and rigorously integrate using SageMath in the usual way. For the later two integrals, since $|u|^{1/6}$ and $|u|^{7/6}$ are not holomorphic at $u=0$ , we must also analytically bound the integral in a very small neighborhood of the origin before performing numerical integration. Performing the computations in [1], we arrive at

I_{3}\leq c_{2}\Bigg[4.0315t^{1/6}\log t+1.8469\ t^{-5/6}+3.3032\log t+1.7493\ t^{-1}\Bigg].\qed

4. Proof of Theorem 1.1

Proof of Theorem 1.1.

With bounds on $I_{3}$ , we can now directly bound the integral term in (2.1). We apply Lemma 3.2 to (3.1) and (3.2) to arrive at two distinct bounds of $|\zeta(1+it)|$ . We begin with the bound in (3.1). That is, setting

a=\frac{1}{4},\ \ b=0,\ \ c_{2}=\frac{4}{(2\pi)^{1/4}},\ \ c_{3}=2.08,

Lemma 3.2 gives

\displaystyle I

\displaystyle\coloneqq\int_{-\infty}^{\infty}\left|\Gamma(-1/2+iu)\right|\cdot\left|\zeta(1/2+i(t+u))\right|du\leq c_{2}\big(t^{1/4}I_{1}+I_{2}\big)+I_{3}+2.27\cdot 10^{-5}.

With these parameters, we numerically bound $I_{1}$ and $I_{2}$ in [1] and find

I_{1}\leq 10^{-134},\ \ \ I_{2}\leq 10^{-131}.

Furthermore, Lemma 3.5 gives

I_{3}\leq 4.0315c_{2}t^{1/4}+0.4618c_{2}t^{-3/4}-4.0314c_{3}.

Hence defining

c_{4}\coloneqq 10.18549,\ \ \ c_{5}\coloneqq 1.16673,\ \ \ c_{6}\coloneqq 8.38527,

we have

\displaystyle I

\displaystyle\leq 4.031501\ c_{2}t^{1/4}+0.4618c_{2}t^{-3/4}+c_{2}\cdot 10^{-131}-4.0314c_{3}+2.27\cdot 10^{-5}\leq c_{4}t^{1/4}+c_{5}t^{-3/4}-c_{6}.

This along with Lemmas 2.1 and 2.2 allows us to write

	$\displaystyle\|\zeta(1+it)\|$	$\displaystyle\leq\log X+\frac{1}{2X}-\frac{1}{24X^{2}}+\frac{c_{0}}{X^{3}}+\frac{c_{1}}{X^{4}}+\|\Gamma(-it)\|+\frac{X^{-1/2}}{2\pi}\left(c_{4}t^{1/4}+c_{5}t^{-3/4}-c_{6}\right)$
		$\displaystyle\coloneqq\log X+\mathcal{R}_{1}(X).$

The two largest terms in this bound are $\log X$ and $\frac{c_{4}}{2\pi}\frac{t^{1/4}}{X^{1/2}}$ . As such, we minimize the sum of these terms by choosing

X\coloneqq\frac{c_{4}^{2}t^{1/2}}{(4\pi)^{2}}.

Thus we have

\displaystyle|\zeta(1+it)|

\displaystyle\leq\frac{1}{2}\log t+\log\left(\frac{c_{4}^{2}}{(4\pi)^{2}}\right)+\mathcal{R}_{1}\left(\frac{c_{4}^{2}t^{1/2}}{(4\pi)^{2}}\right),

say. Since

\log\left(\frac{c_{4}^{2}}{(4\pi)^{2}}\right)+\mathcal{R}_{1}\left(\frac{c_{4}^{2}t^{1/2}}{(4\pi)^{2}}\right)

is increasing on $t\geq 10$ , we bound the expression trivially for $210\leq t\leq 5\cdot 10^{8}$ in [1] to find

|\zeta(1+it)|\leq\frac{1}{2}\log t+1.57.

(4.1)

Similarly, we can bound the integral term using the bound given in (3.2). Here we have

a=\frac{1}{6},\ \ \ b=1,\ \ \ c_{2}=0.618,\ \ \ c_{3}=0.

With these parameters, we numerically bound $I_{1}$ and $I_{2}$ in [1] as follows:

I_{1}\leq 10^{-133},\ \ \ I_{2}\leq 10^{-130}.

Furthermore, Lemma 3.6 gives

I_{3}\leq c_{2}\Bigg[4.0315t^{1/6}\log t+1.8469\ t^{-5/6}+3.3032\log t+1.7493\ t^{-1}\Bigg].

Thus if we define

	$\displaystyle c_{7}$	$\displaystyle\coloneqq 2.49147,$	$\displaystyle c_{8}$	$\displaystyle\coloneqq 2.04138,$	$\displaystyle c_{9}$	$\displaystyle\coloneqq 10^{-133},$
	$\displaystyle c_{10}$	$\displaystyle\coloneqq 1.14139,$	$\displaystyle c_{11}$	$\displaystyle\coloneqq 1.08107,$	$\displaystyle c_{12}$	$\displaystyle\coloneqq 2.271\cdot 10^{-5},$

we have

	$\displaystyle I$	$\displaystyle\leq 4.0315c_{2}t^{1/6}\log t+3.3032c_{2}\log t+c_{2}10^{-133}t^{1/6}$
		$\displaystyle\ \ \ \ +1.8469\ c_{2}t^{-5/6}+1.7493\ c_{2}t^{-1}+c_{2}10^{-130}+2.27\cdot 10^{-5}$
		$\displaystyle\leq c_{7}t^{1/6}\log t+c_{8}\log t+c_{9}t^{1/6}+c_{10}t^{-5/6}+c_{11}t^{-1}+c_{12}.$

This along with Lemmas 2.1 and 2.2 allows us to write

	$\displaystyle\|\zeta(1+it)\|$	$\displaystyle\leq\log X+\frac{1}{2X}-\frac{1}{24X^{2}}+\frac{c_{0}}{X^{3}}+\frac{c_{1}}{X^{4}}+\|\Gamma(-it)\|$
		$\displaystyle\ \ \ \ +\frac{X^{-1/2}}{2\pi}\left(c_{7}t^{1/6}\log t+c_{8}\log t+c_{9}t^{1/6}+c_{10}t^{-5/6}+c_{11}t^{-1}+c_{12}\right).$

The two largest terms in this bound are $\log X$ and $\frac{c_{7}}{2\pi X^{1/2}}t^{1/6}\log t$ . As such, we minimize the sum of these terms by choosing

X\coloneqq\frac{c_{7}^{2}\ t^{1/3}\log^{2}t}{(4\pi)^{2}}.

Thus we have

|\zeta(1+it)|\leq\frac{1}{3}\log t+2\log\log t+\log\left(\frac{c_{7}^{2}}{(4\pi)^{2}}\right)+\mathcal{R}_{2}\left(\frac{c_{7}^{2}\ t^{1/3}\log^{2}t}{(4\pi)^{2}}\right),

say. Since

\log\left(\frac{c_{7}^{2}}{(4\pi)^{2}}\right)+\mathcal{R}_{2}\left(\frac{c_{7}^{2}\ t^{1/3}\log^{2}t}{(4\pi)^{2}}\right)

is decreasing on $t\geq 10$ , for $t\geq 10^{8}$ we trivially bound this expression in [1] to arrive at

|\zeta(1+it)|\leq\frac{1}{3}\log t+2\log\log t-1.16.

(4.2)

Noting that the bound in (4.1) is greater than the bound in (4.2) for all $t\geq 2\cdot 10^{8}$ , it follows that for all $t\geq 210$

|\zeta(1+it)|\leq\frac{1}{2}\log t+1.57.

Furthermore, we prove computationally in [1] that this bound holds for all $3\leq t\leq 210$ . Hence we have the desired bound. ∎

5. Acknowledgments

Both authors are partially supported by the National Science Foundation (DMS-2418328). The second author is also partially supported by the Simons Foundation (MPS-TSM-00007959).

References

[1] Andrew Christensen and Kyle Pratt. Utilizing Smoothing Techniques to Bound zeta-1 $+$ it, June 2026. https://github.com/AndrewChristensenJ/Utilizing_Smoothing_Techniques_to_Bound_zeta-1_it.
[2] NIST Digital Library of Mathematical Functions. https://dlmf.nist.gov/, Release 1.2.6 of 2026-03-15. F. W. J. Olver, A. B. Olde Daalhuis, D. W. Lozier, B. I. Schneider, R. F. Boisvert, C. W. Clark, B. R. Miller, B. V. Saunders, H. S. Cohl, and M. A. McClain, eds.
[3] Kevin Ford. Vinogradov’s integral and bounds for the Riemann zeta function. Proc. London Math. Soc. (3), 85(3):565–633, 2002.
[4] Harald Andres Helfgott. The ternary Goldbach problem, 2015. https://arxiv.org/abs/1501.05438.
[5] Ghaith A. Hiary. An explicit van der Corput estimate for $\zeta(1/2+it)$ . Indag. Math. (N.S.), 27(2):524–533, 2016.
[6] Ghaith A. Hiary, Nicol Leong, and Andrew Yang. Explicit bounds for the Riemann zeta-function on the 1-line. Funct. Approx. Comment. Math., 73(1):53–89, 2025.
[7] Ghaith A. Hiary, Dhir Patel, and Andrew Yang. An improved explicit estimate for $\zeta(1/2+it)$ . J. Number Theory, 256:195–217, 2024.
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	$\displaystyle\|\zeta(1+it)\|\leq$	$\displaystyle\sum_{n=1}^{\infty}{\frac{1}{n}\exp{\left(-\frac{n}{X}\right)}}+\left\|\Gamma(-it)\right\|$		(2.1)
		$\displaystyle+\frac{X^{-1/2}}{2\pi}\int_{-\infty}^{\infty}\left\|\Gamma(-1/2+iu)\right\|\cdot\left\|\zeta(1/2+i(t+u))\right\|du.$

Utilizing Smoothing Techniques to Bound |ζ⁡(1+i​t)||\zeta(1+it)|

Abstract.

Key words and phrases:

2020 Mathematics Subject Classification

1. Introduction

Theorem 1.1.

2. A Smoothed Representation of ζ⁡(1+i​t)\zeta(1+it)

Lemma 2.1.

Proof.

Lemma 2.2.

Proof.

3. Bounding the Integral

Lemma 3.1.

Proof.

Lemma 3.2.

Remark 3.3.

Proof of Lemma 3.2.

Lemma 3.4.

Proof.

Lemma 3.5.

Proof.

Lemma 3.6.

Proof.

4. Proof of Theorem 1.1

Proof of Theorem 1.1.

5. Acknowledgments

References

Utilizing Smoothing Techniques to Bound $|\zeta(1+it)|$

2. A Smoothed Representation of $\zeta(1+it)$