# 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

$I_0(s,b) = \sqrt{\frac{2\pi}{s}} + O_{\leq}(\frac{1}{C |s|^{3/2}}$

assuming for instance that $\mathrm{Re}(s) \geq 1/2$ (need a value for C, try Olver's "asymptotics and special functions"). As the phase $1+u-e^u$ is stationary at $u=0$, one has the heuristic approximation

$I_t(s,b) \approx I_0(s,b) = \exp( s - s \log s ) \Gamma(s) \approx \sqrt{\frac{2\pi}{s}}$

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 $\epsilon \gt 0$ be a parameter, and assume the inequalities

$T \geq 0.0234 t$
$\epsilon \leq 0.771$

1. We have

$|I_t(s,0)| \leq \sqrt{2} \frac{ \exp( -T (\epsilon - e^{-\epsilon} \sin \epsilon ) + \sigma (1 - \epsilon - e^{-\epsilon} \cos \epsilon ) )}{T ( 1 + e^{-\epsilon} \sin \epsilon - e^{-\epsilon} \cos \epsilon )}$
$+ \sqrt{2 \pi} \frac{\sqrt{T+\sigma}}{T-\sigma} \frac{2\sqrt{\epsilon - e^{-\epsilon} \sin \epsilon}}{1 + e^{-\epsilon} \sin \epsilon - e^{-\epsilon} \cos \epsilon}$
$+ 0.286 \frac{\exp(-2.21 T + 1.30 \sigma )}{T - 0.0234 t}.$

2. We have

$|I_t(s,0) - \sqrt{\frac{2\pi}{s}}| \leq \frac{1}{|s|} \frac{3\sqrt{2}t}{16} \frac{\exp( -T (\epsilon - e^{-\epsilon} \sin \epsilon ) + \sigma (1 - \epsilon - e^{-\epsilon} \cos \epsilon ) + \epsilon )}{T ( 1 + e^{-\epsilon} \sin \epsilon - e^{-\epsilon} \cos \epsilon ) - 1}$
$+ \frac{1}{|s|} 1.24 t \frac{\sqrt{T+\sigma}}{T-\sigma} \frac{\sqrt{\epsilon - e^{-\epsilon} \sin \epsilon}}{1 + e^{-\epsilon} \sin \epsilon - e^{-\epsilon} \cos \epsilon}$
$+ \frac{1}{|s|} (0.0188 t + 0.171) \frac{\exp(-2.21 T + 1.30 \sigma )}{T - 0.0234 t}.$
$+ \frac{1}{C|s|^{3/2}}.$

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

$|I_t(s,0)| = O( \frac{\exp( - T \epsilon^2 / 2 )}{\epsilon T} ) + (1 + O(\epsilon_-)) \sqrt{\frac{2\pi}{T}} + O( 1/T^{3/2} ).$

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

$|I_t(s,0)| \leq (1+o(1)) \sqrt{\frac{2\pi}{T}}.$

Similarly the estimates in part 2 give

$|I_t(s,0) - I_0(s,0)| \ll \frac{1}{T^{3/2}}.$

Proof Let $\epsilon_+ := 1.292$ (the significance of this constant being that $\exp(\epsilon_+) \cos(\epsilon_+)$ is close to 1, but one could easily run the argument below with another choice without significant change to the main terms in the estimates). 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 $1 - e^x \cos \epsilon_+ \leq 1 - e^{\epsilon_+} \cos \epsilon_+ \leq 0$ and using $x \leq e^{1-e} e^x$ we have

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

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

$|I_{t,3}(s,0)| \leq \frac{\exp( \Phi_3(\epsilon_+) - \epsilon_+ )}{T \sin \epsilon_+ - e^{1-e} \frac{t}{8})}$
$\leq 0.286 \frac{\exp(-2.21 T + 1.30 \sigma )}{T - 0.0234 t}.$

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

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

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 $\epsilon_+ = 1.292$, 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 $0 \lt x \lt \epsilon_+ = 1.292$, 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.

Now we turn to part 2. We can write

$I_t(s,0) - I_0(s,0) = -\frac{1}{s} \int_{C_1+C_2+C_3} (\frac{d}{du} \exp( s(1 + u - e^u) )) \frac{\exp( t u^2 / 16 ) - 1}{e^u - 1}\ du$

and hence on integration by parts (noting that the quotient here has a removable singularity at $u=0$

$I_t(s,0) - I_0(s,0) = \frac{1}{s} (\tilde I_{t,1}(s,0) + \tilde I_{t,2}(s,0) + \tilde I_{t,3}(s,0))$

where

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

We now estimate $\tilde I_{t,j}$ in a similar fashion to the $I_{t,j}$. We begin with $\tilde I_{t,3}(s,0)$. We can write this as

$\int_{\epsilon_+}^\infty \exp( \phi_3(x) ) \psi_3(x)\ dx$

where

$\psi_3(x) := \exp( - tu^2/16) \frac{d}{du} \frac{\exp( t u^2 / 16 ) - 1}{e^u - 1}|_{u = x - i\epsilon_+}.$

We can explicitly write

$\psi_3(x) = \frac{t}{8} \frac{u}{e^u-1} - e^u \frac{1 - \exp(-t u^2/16)}{(e^u-1)^2}.$

One can check that $|\frac{u}{e^u-1}|$ and $|\frac{e^u}{(e^u-1)^2}|$ for $u = x-i\epsilon_+$, $x \geq \epsilon_+$ both attain their maximum at the endpoint $u = \epsilon_+ - i \epsilon_+$, so that

$|\frac{u}{e^u-1}| \leq 0.523$

and similarly

$|\frac{e^u}{(e^u-1)^2}| \leq 0.298.$

Also

$|\exp(-tu^2/16)| = \exp( t (\epsilon_+^2 - x^2)/16 ) \leq 1$

and hence $|1 - \exp(-t u^2/16)| \leq 2$. Thus from the triangle inequality

$\psi_3(x) \leq 0.0654 t + 0.596$

and hence by the analysis used to estimate $I_{t,3}$ we have

$|\tilde I_{t,3}(s,0)| \leq (0.0188 t + 0.171) \frac{\exp(-2.21 T + 1.30 \sigma )}{T - 0.0234 t}.$

Now we turn to $\tilde I_{t,1}(s,0)$. We write this as

$(1-i) \int_{-\infty}^{-\epsilon} \exp( \phi_1(x) - x) \psi_1(x)\ dx$

where

$\psi_1(x) := \exp( x - tu^2/16) \frac{d}{du} \frac{\exp( t u^2 / 16 ) - 1}{e^u - 1}|_{u = x - ix}.$

Calculating as before we have

$\psi_1(x) = \frac{t}{8} \frac{(x-ix) e^{x}}{e^{x-ix}-1} - e^{2x-ix} \frac{1 - \exp(it x^2/8)}{(e^{x-ix}-1)^2}.$

We bound $1 - \exp(itx^2/8)| \leq tx^2/8$. The quantities $|\frac{(x-ix) e^x}{e^{x-ix}-1}|$ and $\frac{e^{2x} x^2}{|e^{x-ix}-1|^2}$ can be verified to be increasing for $x\lt0$, so by the triangle inequality and comparing with $x=0$ we have

$|\psi_1(x)| \leq \frac{t}{8} + \frac{t}{16} = \frac{3t}{16}.$

and hence by the analysis used to estimate $I_{t,1}$ we have

$|I_{t,1}(s,0)| \leq \frac{3\sqrt{2}t}{16} \frac{\exp(\Phi_1(-\epsilon) + \epsilon)}{T ( 1 + e^{-\epsilon} \sin \epsilon - e^{-\epsilon} \cos \epsilon ) - 1}.$

Finally we turn to $\tilde I_{t,2}(s,0)$. We write this as

$(1-i) \int_{-\infty}^{-\epsilon} \exp( \phi_2(x) ) \psi_2(x)\ dx$

where

$\psi_2(x) := \exp(- tu^2/16) \frac{d}{du} \frac{\exp( t u^2 / 16 ) - 1}{e^u - 1}|_{u = x - ix}.$

Calculating as before we have

$\psi_2(x) = \frac{t}{8} \frac{(x-ix)}{e^{x-ix}-1} - e^{x-ix} \frac{1 - \exp(it x^2/8)}{(e^{x-ix}-1)^2}.$

For $-0.771 \leq x \leq 0$, one can bound $|\frac{(x-ix)}{e^{x-ix}-1}|$ by 1.47 and $e^x \frac{x^2}{|e^{x-ix}-1|^2}$ by 0.5, hence by the triangle inequality

$\psi_2(x) \leq 0.184 t + \frac{1}{16} t \leq 0.247 t$

and hence by the analysis used to estimate $I_{t,2}$ we have

$|I_{t,2}(s,0)| \leq 1.24 t \frac{\sqrt{T+\sigma}}{T-\sigma} \frac{\sqrt{\epsilon - e^{-\epsilon} \sin \epsilon}}{1 + e^{-\epsilon} \sin \epsilon - e^{-\epsilon} \cos \epsilon}.$

Combining these estimates gives the claim. $\Box$

Setting $\epsilon = 0.771$ we have the more explicit bounds

$|I_t(s,0)| \leq 1.428 \frac{ \exp( -0.448 T - 0.102 \sigma ) }{T}$
$+ 3.39 \frac{\sqrt{T+\sigma}}{T-\sigma}$
$+ 0.286 \frac{\exp(-2.21 T + 1.30 \sigma )}{T - 0.0234 t}.$

and

$|I_t(s,0) - \sqrt{\frac{2\pi}{s}}| \leq \frac{0.402}{|s|} \frac{ \exp( -0.448 T - 0.102 \sigma ) }{T}$
$+ \frac{0.839}{|s|} t \frac{\sqrt{T+\sigma}}{T-\sigma}$
$+ \frac{1}{|s|} (0.0188 t + 0.171) \frac{\exp(-2.21 T + 1.30 \sigma )}{T - 0.0234 t}$
$+ \frac{1}{C |s|^{3/2}}.$

## Estimating $I_t(s,b)$

We will estimate $I_t(s,b)$ by transforming this integral into something of the form $I_t(\tilde s, 0)$ by a change of variables. Namely, let $\alpha$ be a complex number to be chosen later. Translating $u$ by $\alpha$, we conclude that

$I_t(s,b) = \int_C \exp( s(1 + u + \alpha - e^\alpha e^u) + \frac{t}{16} ((u + \alpha + b)^2 - b^2) )\ du$
$= \exp( s + s \alpha + \frac{t}{8} b \alpha + \frac{t}{16} \alpha^2 ) \int_C \exp( (s + \frac{t(\alpha+b)}{8}) u - s e^\alpha e^u )\ du.$

Suppose that we can find $\alpha$ solving the equation

$e^\alpha = 1 + \frac{t}{8s} (\alpha+b),$

then we have

$I_t(s,b) = \exp( s + s \alpha + \frac{t}{8} b \alpha + \frac{t}{16} \alpha^2 ) \int_C \exp( (s + \frac{t(\alpha+b)}{8}) (u - s e^\alpha e^u )\ du$
$= \exp( s \alpha + \frac{t}{8} b \alpha + \frac{t}{16} \alpha^2 - \frac{t(\alpha+b)}{8} ) I_t( s + \frac{t(\alpha+b)}{8}, 0 )$
$= \exp( s (1 + \alpha - e^\alpha) + \frac{t}{8} b \alpha + \frac{t}{16} \alpha^2 ) I_t( s + \frac{t(\alpha+b)}{8}, 0 ).$

To solve for $\alpha$, suppose that

$2|b| + 3 \leq \frac{8|s|}{t},$

then the map $\alpha \mapsto \log( 1 + \frac{t}{8s} (\alpha + b) )$ maps the unit disk to itself (since $1 + \frac{t}{8s} (\alpha + b)$ stays within 1/2 of the identity), and the derivative $\frac{t}{8s} \frac{1}{1 + \frac{t}{8s} (\alpha+b)}$ has magnitude at most 1/2, since one can check that

$\frac{t}{8|s|} \frac{1}{1 - \frac{t}{8|s|} (|b|+1)} \leq \frac{1}{2}$

(this is equivalent to $|b|+3 \leq \frac{8|s|}{t}$). Thus by the contraction mapping theorem, $\alpha$ exists and is unique in the unit disk. Since $0$ maps to $\log(1 + \frac{tb}{8s})$, we thus have the bound

$|\alpha| \leq 2 |\log(1 + \frac{tb}{8s})|.$

As $1 + \frac{tb}{8s}$ stays within 1/2 of the unit disk, and the logarithm function has derivative at most 2 there, we thus have

$|\alpha| \leq \frac{t|b|}{2|s|}.$

By Taylor's theorem with remainder we then have

$|1 + \alpha - e^\alpha| \leq \frac{1}{2} |\alpha|^2 \exp( \frac{t|b|}{2|s|} ) \leq \frac{t^2 |b|^2}{8|s|^2} \exp( \frac{t|b|}{2|s|} )$

and hence

$I_t(s,b) = \exp( O_{\leq}( X ) ) I_t( s + \frac{t(\alpha+b)}{8}, 0 )$

where

$X := \frac{t^2 |b|^2}{8|s|} \exp( \frac{t|b|}{2|s|} ) + \frac{t^2 |b|^2}{16 |s|} + \frac{t^3 |b|^2}{64 |s|^2}$

and $O_{\leq}(X)$ denotes a quantity bounded in magnitude by $X$. We can then bound

$|I_t(s,b)| \leq \exp( X ) |I_t( s + \frac{t(\alpha+b)}{8}, 0 )|$

and also

$|I_t(s,b) - I_0(s,0)| \leq (\exp( X )-1) |I_t( s + \frac{t(\alpha+b)}{8}, 0 )| + |I_t( s + \frac{t(\alpha+b)}{8}, 0 ) - \sqrt{\frac{2\pi}{s + \frac{t(\alpha+b)}{8}}}| + |\sqrt{\frac{2\pi}{s + \frac{t(\alpha+b)}{8}}} - \sqrt{\frac{2\pi}{s}}|.$

The first two terms can be controlled using Lemma 2. The third term can be bounded using the mean value theorem as $O_{\leq}( \frac{t|\alpha+b|}{8} \frac{\sqrt{2\pi}}{2 (|s| - t(|\alpha|+|b|)/8)^{3/2}})$.

## Estimating $F_{t,N}$

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

## Estimating $R_{t,N,M}$

$R_{t,N,M}(s) = 2 \sum_{n=0}^\infty \pi^{-s/2} \frac{e^{-i\pi s/2} e^{i \pi s n}}{2^{s}\pi^{1/2} \sin(\pi s/2)} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{5-s}{2} \log \frac{5-s}{2} - \frac{5-s}{2} + \frac{t}{16} \alpha^2) I_t(\frac{5-s}{2}, \alpha )\ dw$
$-3 \sum_{n=0}^\infty \pi^{-s/2} \frac{e^{-i\pi s/2} e^{i \pi s n}}{2^{s}\pi^{1/2} \sin(\pi s/2)} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{3-s}{2} \log \frac{3-s}{2} - \frac{3-s}{2} + \frac{t}{16} \beta^2 ) I_t(\frac{3-s}{2}, \beta )\ dw,$
$\alpha := \log \frac{5-s}{2} - 2 \log \frac{w}{2\sqrt{\pi}} - i\pi(2n-1)$
$\beta := \log \frac{3-s}{2} - 2 \log \frac{w}{2\sqrt{\pi}} - i\pi(2n-1).$