# Difference between revisions of "Effective bounds on H t"

## Explicit upper bounds on integrals

We will need effective upper bounds on various integrals that occur as error terms, with explicit constants. Here is a basic tool to do this:

Lemma 1 Let $\phi: [a,b] \to {\bf C}$ be a smooth function on a compact interval $[a,b]$. Let $\psi: [a,b] \to {\bf C}$ be a measurable function. Let $I$ denote the integral $I := \int_a^b e^{\phi(x)} \psi(x)\ dx$.

1. If $\mathrm{Re} \phi(x) \lt 0$ for all $a \leq x \leq b$, then

$|I| \leq e^{\mathrm{Re} \phi(a)} \sup_{a \leq x \leq b} \frac{|\psi(x)|}{|\mathrm{Re} \phi'(x)|}.$

2. If $\mathrm{Re} \phi(x) \gt 0$ for all $a \leq x \leq b$, then

$|I| \leq e^{\mathrm{Re} \phi(b)} \sup_{a \leq x \leq b} \frac{|\psi(x)|}{|\mathrm{Re} \phi'(x)|}.$

3. If there is an point $x_0 \in (a,b)$ such that $\mathrm{Re} \phi'(x)$ is negative for $x \gt x_0$ and positive for $x \lt x_0$ with $\mathrm{Re} \phi''(x_0) \neq 0$ (thus $\mathrm{Re} \phi$ has a non-degenerate maximum at $x_0$), then

$|I| \leq 2\sqrt{\pi} e^{\mathrm{Re} \phi(x_0)} \sup_{a \leq x \leq b: x \neq x_0} \frac{|\psi(x)| \sqrt{\mathrm{Re} \phi(x_0) - \mathrm{Re} \phi(x)}}{|\mathrm{Re} \phi'(x)|}.$

4. With the same hypotheses as part 3, we also have

$|I| \leq \sqrt{\pi} e^{\mathrm{Re} \phi(x_0)} \sup_{a \leq x \leq b: x \neq x_0} \frac{|\psi(x)|}{|\mathrm{Re} \phi'(x)| \sqrt{\mathrm{Re} \phi(x_0) - \mathrm{Re} \phi(x)}}.$

Proof Write $\Phi := \mathrm{Re} \phi$. To prove part 1, we may normalise $\Phi(a)=0$ and the supremum to be $1$, then $\Phi$ is decreasing with $\Phi(b) \lt \Phi(a)=0$. By the triangle inequality and change of variables we then have

$|I| \leq -\int_a^b e^{\Phi(x)} \Phi'(x)\ dx = \int_{\Phi(b)}^0 e^{-y}\ dy \leq 1$

as desired. Part 2 is proven similarly.

To prove Part 3, we may normalise $\Phi(x_0) = x_0 = 0$ and the supremum to be 1, then $\Phi$ is negative on the rest of $[a,b]$ and by Taylor expansion we may write $\Phi(x) = - f(x)^2$ for some smooth $f: [a,b] \to {\bf R}$ with $f(0)=0$ and $f'(x) \gt 0$ for all $x \in [a,b]$. For any $x \in [a,b] \backslash \{x_0\}$, we have

$|\psi(x)| \leq \frac{|\Phi'(x)|}{\sqrt{-\Phi(x)}} = \frac{2 |f(x)| f'(x)}{|f(x)|} = 2 f'(x)$

and hence by the triangle inequality and change of variables

$|I| \leq 2 \int_a^b e^{-f(x)^2} f'(x)\ dx = 2 \int_{f(a)}^{f(b)} e^{-y^2}\ dy \leq 2 \sqrt{\pi}$

as desired.

Part 4 is proven similarly to Part 3, except that the upper bound of $|\psi|$ is now $2 f(x)^2 f'(x)$, and one uses the identity $\int_{-\infty}^{\infty} e^{-y^2} y^2\ dy = \frac{1}{2} \sqrt{\pi}$. $\Box$

Note that one can use monotone convergence to send $b$ to infinity in part 1, and similarly send $a$ to negative infinity in part 2. In parts 3 and 4 one can send either $a$ or $b$ or both to infinity. The bounds can be tight, as can be seen by setting $\psi(x)=1$ (for parts 1,2,3) or $\psi(x) = x^2$ (for part 4) and $\phi(x)$ equal to $-x$ (for part 1), $x$ (for part 2), or $-x^2$ (for parts 3,4), and sending as many endpoints of integration to infinity as possible.

## Estimating $I_t(s,0)$

For any $t \geq 0$, $s, b \in {\bf C}$ with $\mathrm{Im} s \gt \mathrm{Re} s \gt 0$, we consider the integral

$I_t(s,b) := \int_C \exp( s(1 + u - e^u) + \frac{t}{16} ((u+b)^2 - b^2) )\ du$

where $C$ is any contour from $+i\infty$ to $+\infty$ that stays a bounded distance from the upper imaginary and right real axes. This integral appears in several places in the Riemann-Siegel formula for $H_t$, so it will be of importance to estimate it efficiently. Note that the integral is absolutely convergent on the contour $C$

From a change of variables $x = \exp(su)$ we see that

$I_0(s,b) = \exp( s - s \log s ) \Gamma(s)$

and hence by Stirling's formula

$I_0(s,b) = (1 + O(\frac{1}{|s|}) \sqrt{\frac{2\pi}{s}}$

assuming for instance that $\mathrm{Re}(s) \geq 1/2$. As the phase $1+u-e^u$ is stationary at $u=0$, one has the heuristic approximation

$I_t(s,b) \approx I_0(s,b) = \exp( s - s \log s ) \Gamma(s)$

as long as $b$ is not too large. In this section we give effective bounds on this quantity in the $b=0$ case, that is to say we estimate

$I_t(s,0) = \int_C \exp( s(1 + u - e^u) + \frac{t}{16} u^2 )\ du.$

Proposition 2 Let $t,s$ be above. Write $s = \sigma+iT$. Let $\epsilon_-, \epsilon_+ \gt 0$ be parameters, and assume the inequalities

$1 - e^{\epsilon_+} \cos \epsilon_+, \frac{t}{8} \leq T \sin \epsilon_+$
$\epsilon_- \leq 0.771; \quad \epsilon_+ \leq 1.292$

1. We have

$|I_t(s,b)| \leq \sqrt{2} \frac{ \exp( -T (e^{-\epsilon_-} \sin \epsilon_- - \epsilon_- ) + \sigma (1 - \epsilon_- - e^{\epsilon_-} \cos \epsilon_- ) }{T ( 1 + e^{\epsilon_-} \sin \epsilon_- - e^{-\epsilon_-} \cos \epsilon_- )}$
$+ \sqrt{2 \pi} \frac{\sqrt{T+\sigma}}{T-\sigma} \frac{2\sqrt{\epsilon_- - e^{-\epsilon_-} \sin \epsilon_-}}{1 + e^{-\epsilon_-} \sin \epsilon_- - e^{-\epsilon_-} \cos \epsilon_-}$
$+ \frac{\exp(-T ( e^{\epsilon_+} \sin \epsilon_+ - \epsilon_+) + \sigma(1+\epsilon_+ - e^\epsilon_+ \cos \epsilon_+) - \epsilon_+)}{T \sin \epsilon_+ - \max(1 - e^{\epsilon_+} \cos \epsilon_+, \frac{t}{8})}.$

2. We have

$|I_t(s,b) - I_0(s,b)| \leq ???$.

Remark For $T$ large, $\sigma,t$ bounded, and $\epsilon_+, \epsilon_-$ small, the requirements of the proposition are obeyed, and part 1 essentially yields the bounds

$|I_t(s,b)| = O( \frac{\exp( - T \epsilon_-^2 / 2 )}{\epsilon_- T} ) + (1 + O(\epsilon_-)) \sqrt{\frac{2\pi}{T}} + \frac{\exp( -T \epsilon_+^2 / 2 )}{\epsilon_+ T}.$

Setting $\epsilon_-$ to be a large multiple of $T^{-1/2}$ and $\epsilon_+$ to be 1.292 we obtain a near-optimal bound of the shape

$|I_t(s,b)| \leq (1+o(1)) \sqrt{\frac{2\pi}}{T}}.$

...

Proof We begin with part 1. By shifting the contour we have

$I_t(s,0) = I_{t,1}(s,0) + I_{t,2}(s,0) + I_{t,3}(s,0)$

where

$I_{t,j}(s,0) := \int_{C_j} \exp( s(1 + u - e^u) + \frac{t}{16} u^2 )\ du$

for $j=1,2,3$, $C_1$ is the diagonal line parameterised by

$x \mapsto x - ix: -\infty \leq x \leq -\epsilon_-;$

$C_2$ is the diagonal line parameterised by

$x \mapsto x - ix: -\epsilon_- \leq x \leq \epsilon_+;$

and $C_3$ is the horizontal line parameterised by

$x \mapsto x - i\epsilon_+: \epsilon_+ \leq x \leq \infty;$

We will use Lemma 1.3 to estimate the first integral, Lemma 1.2 to estimate the second, and Lemma 1.1 to estimate the third.

We begin with the $C_3$ integral, which we can write as

$\int_{\epsilon_+}^\infty \exp( \phi_3(x) )\ dx$

where

$\phi_3(x) := (\sigma+iT) (1 + x - i\epsilon_+ - e^x e^{-i\epsilon_+}) + \frac{t}{16} (x - i\epsilon_+)^2.$

The real part $\Phi_3$ of $\phi_3$ is

$\Phi_3(x) = \sigma( 1 + x - e^x \cos \epsilon_+ ) + T (\epsilon_+ - e^x \sin \epsilon_+ ) + \frac{t}{16} (x^2 - \epsilon_+^2)$

and the first derivative is

$\Phi'_3(x) = \sigma( 1 - e^x \cos \epsilon_+ ) - T e^x \sin \epsilon_+ + \frac{t}{8} x.$

Bounding $1 - e^x \cos \epsilon_+ \leq 1 - e^{\epsilon_+} \cos \epsilon_+$ and using $1 + x \leq e^x$ we have

$\Phi'_3(x) \leq - (T \sin \epsilon_+ - \max(1 - e^{\epsilon_+} \cos \epsilon_+, \frac{t}{8})) e^x$.

Bounding $e^x \geq e^{\epsilon_+}$ and applying Lemma 1.1, we conclude that

$|I_{t,3}(s,0)| \leq \frac{\exp( \Phi_3(\epsilon_+) - \epsilon_+ )}{T \sin \epsilon_+ - \max(1 - e^{\epsilon_+} \cos \epsilon_+, \frac{t}{8})}.$

Note that

$\Phi_3(\epsilon_+) = -T ( e^{\epsilon_+} \sin \epsilon_+ - \epsilon_+) + \sigma(1+\epsilon_+ - e^\epsilon_+ \cos \epsilon_+)$.

Now we estimate $I_{t,1}(s,0)$. We write this term as

$(1-i) \int_{-\infty}^{-\epsilon_-} \exp( \phi_1(x) )\ dx$

where

$\phi_1(x) := (\sigma+iT) (1 + x - ix - e^x e^{-ix}) + \frac{t}{4} (x - ix)^2.$

The real part $\Phi_1$ of $\phi_1$ is

$\Phi_1(x) = \sigma( 1 + x - e^x \cos x ) + T (x - e^x \sin x )$

and the first derivative is

$\Phi'_1(x) = \sigma( 1 - e^x \cos x + e^x \sin x ) + T (1 - e^x \sin x - e^x \cos x ).$

The expression $1 - e^x \cos x + e^x \sin x$ is non-negative for $x \leq 0$. The expression $1 - e^x \sin x - e^x \cos x$ exceeds $1$ for $x \leq -0.771$ and is decreasing for $-0.771 \leq x \leq 0$, hence

$1 - e^x \sin x - e^x \cos x \geq 1 + e^{\epsilon_-} \sin \epsilon_- - e^{-\epsilon_-} \cos \epsilon_-$

and hence by Lemma 1.3

$|I_{t,1}(s,0)| \leq \sqrt{2} \frac{\exp(\Phi_1(-\epsilon_-))}{T ( 1 + e^{\epsilon_-} \sin \epsilon_- - e^{-\epsilon_-} \cos \epsilon_- )}.$

Note that

$\Phi_1(-\epsilon_-) = -T (e^{-\epsilon_-} \sin \epsilon_- - \epsilon_- ) + \sigma (1 - \epsilon_- - e^{\epsilon_-} \cos \epsilon_- ).$

Finally we estimate $I_{t,2}(s,0)$. We write this as

$(1-i) \int_{-\epsilon_-}^{\epsilon_+} \exp( \phi_2(x) )\ dx$

where $\phi_2 = \phi_1$. The real part $\Phi_2 = \Phi_1$ and its derivative are as before.

For $x$ between -0.771 and 1.292, one can check that

$|1 - e^x \cos x + e^x \sin x| \leq |1 - e^x \sin x - e^x \cos x|$

so

$|\Phi'_2(x)| \geq (T-\sigma) |1 - e^x \sin x - e^x \cos x|.$

Similarly one can check that in this range that

$|1 + x - e^x \cos x| \leq |x - e^x \sin x|$

and so

$\Phi_2(0) - \Phi_2(x) \leq (T+\sigma) (e^x \sin x - x).$

The quantity

$\frac{\sqrt{(e^x \sin x-x)}}{|1 - e^x \sin x - e^x \cos x|}$

is decreasing for $-0.771 \leq x \leq 0$, equals $1/2$ at $x=0$, and lies below 1/2 for $0 \lt x \lt 1.292$, hence is bounded by

$\frac{\sqrt{\epsilon_- - e^{-\epsilon_-} \sin \epsilon_-}}{1 + e^{-\epsilon_-} \sin \epsilon_- - e^{-\epsilon_-} \cos \epsilon_-}.$

We thus have from Lemma 3.3 that the contribution of this integral is at most

$\sqrt{2 \pi} \frac{\sqrt{T+\sigma}}{T-\sigma} \frac{2\sqrt{\epsilon_- - e^{-\epsilon_-} \sin \epsilon_-}}{1 + e^{-\epsilon_-} \sin \epsilon_- - e^{-\epsilon_-} \cos \epsilon_-}.$

Combining these estimates we obtain part 1.

\frac{\sqrt{T+\sigma}}{T-\sigma} \exp( \frac{t}{16} \max( \epsilon (\alpha-\beta), \epsilon' (\beta-\alpha) ) )\quad (7.8).[/itex]


Thus

## Estimating $F_{t,N}$

From (5.2) we have

$F_{t,N}(s) = \sum_{n=1}^N \frac{2}{(\pi n^2)^{s/2}} \exp( \frac{s+4}{2} \log \frac{s+4}{2} - \frac{s+4}{2} + \frac{t}{16} \log^2 \frac{s+4}{2\pi n^2} ) I_t( \frac{s+4}{2}, \log \frac{s+4}{2\pi n^2} ) - \frac{3}{(\pi n^2)^{s/2}} \exp( \frac{s+2}{2} \log \frac{s+2}{2} - \frac{s+2}{2} \frac{t}{16} \log^2 \frac{s+2}{2\pi n^2} ) I_t( \frac{s+2}{2}, \log \frac{s+2}{2\pi n^2} )$

$\tilde \psi(u) = \frac{t}{8} (u+b) \frac{\exp( \frac{t}{16} ((u+b)^2 -b^2) )}{e^u - 1} - \frac{e^u (\exp( \frac{t}{16} ((u+b)^2 -b^2) )-1)}{(e^u-1)^2} \quad (7.9).$

This expression is tractable as long as $e^u-1$ stays away from zero. Otherwise we can do the following. First observe from the fundamental theorem of calculus that

$\exp( \frac{t}{16} ((u+b)^2 -b^2) ) - 1 = u \int_0^1 \frac{t}{8} (\theta u+b) \exp(\frac{t}{16} ((\theta u+b)^2 -b^2) )\ d\theta$

and

$\frac{u}{e^u-1} = \int_0^1 \frac{1}{1 + \sigma(e^u-1)}\ d\sigma$

and hence

$\tilde \psi(u) = \int_0^1 \int_0^1 \frac{d}{du} [ (\theta u+b) \frac{\exp(\frac{t}{16} ((\theta u+b)^2 -b^2) )}{1+\sigma(e^u-1)} ]\ d\sigma d\theta.$
$\tilde \psi(u) = \int_0^1 \int_0^1 \frac{\exp(\frac{t}{16} ((\theta u+b)^2 -b^2) )}{1+\sigma(e^u-1)} [ \theta + \frac{t}{8} (\theta u + b)^2 - (\theta u+b) \frac{\sigma e^u}{1+\sigma(e^u-1)} ]\ d\sigma d\theta.$