Bounding the derivative of H t - third approach

To compute $H_t(x+iy)$, we begin with the heat fundamental solution formula

$H_t(x+iy) = \frac{1}{8} \int_{-\infty}^\infty \xi( \frac{1-y+ix}{2} + \sqrt{t} v) \frac{1}{\sqrt{\pi}} e^{-v^2}\ dv$

which on making the change of variables $s = \frac{1-y+ix}{2} + \sqrt{t} v$ is

$H_t(x+iy) = \frac{1}{8\sqrt{\pi t}} \int_L \xi(s) \exp( - \frac{(s - \frac{1-y+ix}{2})^2}{t} )\ ds$

where $L$ is any horizontal line contour; differentiating under the integral sign, we also have

$H'_t(x+iy) = \frac{i}{8\sqrt{\pi} t^{3/2}} \int_L \xi(s) \exp( - \frac{(s - \frac{1-y+ix}{2})^2}{t} ) (s - \frac{1-y+ix}{2})\ ds.$

In particular, for any real number $T$, we have

$H_t(x+iy) = \frac{1}{8\sqrt{\pi t}} \int_{-\infty}^\infty \xi(\sigma+iT) \exp( - \frac{(\sigma+iT - \frac{1-y+ix}{2})^2}{t} )\ d\sigma$

and

$H'_t(x+iy) = \frac{i}{8\sqrt{\pi} t^{3/2}} \int_{-\infty}^\infty \xi(\sigma+iT) \exp( - \frac{(\sigma+iT - \frac{1-y+ix}{2})^2}{t} ) (\sigma+iT - \frac{1-y+ix}{2}) \ d\sigma.$

Making the change of variables $\sigma \mapsto 1-\sigma$ for $\sigma \lt 1/2$, and using the functional equation $\xi(\sigma+iT) =xi(1-\sigma-iT) = \overline{\xi(1-\sigma+iT)}$, one can write this as

$H_t(x+iy) = \frac{1}{8\sqrt{\pi t}} \int_{1/2}^\infty \xi(\sigma+iT) \exp( - \frac{(\sigma+iT - \frac{1-y+ix}{2})^2}{t} ) + \overline{\xi(\sigma+iT)} \exp( - \frac{(1-\sigma+iT - \frac{1-y+ix}{2})^2}{t} \ d\sigma$

and

$H'_t(x+iy) = \frac{i}{8\sqrt{\pi}t^{3/2}} \int_{1/2}^\infty \xi(\sigma+iT) \exp( - \frac{(\sigma+iT - \frac{1-y+ix}{2})^2}{t} ) (\sigma+iT - \frac{1-y+ix}{2}) + \overline{\xi(\sigma+iT)} \exp( - \frac{(1-\sigma+iT - \frac{1-y+ix}{2})^2}{t} ) (1-\sigma+iT - \frac{1-y+ix}{2}) \ d\sigma$

Cutting off the integral at some cutoff parameter

$X \gt 1 \quad (1),$

one has

$H_t(x+iy) = \frac{1}{8\sqrt{\pi t}} \int_{1/2}^X \xi(\sigma+iT) \exp( - \frac{(\sigma+iT - \frac{1-y+ix}{2})^2}{t} ) + \overline{\xi(\sigma+iT)} \exp( - \frac{(1-\sigma+iT - \frac{1-y+ix}{2})^2}{t} ) \ d\sigma + E$

and

$H'_t(x+iy) = \frac{i}{8\sqrt{\pi}t^{3/2}} \int_{1/2}^X \xi(\sigma+iT) \exp( - \frac{(\sigma+iT - \frac{1-y+ix}{2})^2}{t} ) (\sigma+iT - \frac{1-y+ix}{2}) + \overline{\xi(\sigma+iT)} \exp( - \frac{(1-\sigma+iT - \frac{1-y+ix}{2})^2}{t} ) (1-\sigma+iT - \frac{1-y+ix}{2}) \ d\sigma + E'$

where the error terms $E, E'$ can be bounded using the triangle inequality by

$|E| \leq \frac{1}{8\sqrt{\pi t}} \int_X^\infty |\xi(\sigma+iT)| (\exp( \frac{(T-\frac{x}{2})^2 - (\sigma - (1-y)/2)^2}{t}) + \exp( \frac{(T-\frac{x}{2})^2 - (1-\sigma - (1-y)/2)^2}{t})) \ d\sigma.$

and

$|E'| \leq \frac{1}{8\sqrt{\pi}t^{3/2}} \int_X^\infty |\xi(\sigma+iT)| (\exp( \frac{(T-\frac{x}{2})^2 - (\sigma - (1-y)/2)^2}{t}) |\sigma+iT - \frac{1-y+ix}{2}| + \exp( \frac{(T-\frac{x}{2})^2 - (1-\sigma - (1-y)/2)^2}{t}) |1-\sigma+iT - \frac{1-y+ix}{2}| ) \ d\sigma.$

If we assume

$y \geq 0, \quad (2)$

then

$(\sigma - (1+y)/2)^2 = (1-\sigma - (1-y)/2)^2 \leq (\sigma - (1-y)/2)^2$

and $|1-\sigma+iT - \frac{1-y+ix}{2}| \leq |\sigma+iT - \frac{1-y+ix}{2}| \leq |T-x/2| + \sigma - \frac{1-y}{2}$ hence

$|E| \leq \frac{\exp(\frac{(T-\frac{x}{2})^2}{4t})}{4\sqrt{\pi t}} \int_X^\infty |\xi(\sigma+iT)| \exp( -\frac{(\sigma - (1-y)/2)^2}{t})\ d\sigma$

and

$|E'| \leq \frac{\exp(\frac{(T-\frac{x}{2})^2}{4t})}{4\sqrt{\pi} t^{3/2}} \int_X^\infty |\xi(\sigma+iT)| \exp( -\frac{(\sigma - (1-y)/2)^2}{t}) (|T-x/2| + \sigma - \frac{1-y}{2})\ d\sigma.$

Since

$\xi(s) = \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \zeta(s)$

and $\zeta(\sigma+iT) \leq \zeta(X)$ when $\sigma \geq X$, we have

$|\xi(s)| \leq \zeta(X) |\frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2)|.$

Assuming

$T \geq 2, \quad (3)$

we can use the Stirling approximation

$|\Gamma(s/2)| \leq \sqrt{2\pi} \exp( \frac{s-1}{2} \log \frac{s}{2} - \frac{s}{2} + \frac{1}{6(T-0.33)})$

whenever $\mathrm{Im}(s)=T$ (see Lemma 1.5 of [1]). Thus

$|\xi(s)| \leq \frac{\sqrt{2\pi} \exp( \frac{1}{6(T-0.33)}) \zeta(X)}{2} \exp( \mathrm{Re} F(s) )$

where

$F(s) := \log s + \log(s-1) - \frac{s}{2} \log \pi + \frac{s-1}{2} \log \frac{s}{2} - \frac{s}{2}$

and so

$|E| \leq \frac{\exp(\frac{(T-\frac{x}{2})^2}{4t} + \frac{1}{6(T-0.33)})}{\sqrt{32 t}} \int_X^\infty \exp(\mathrm{Re} F(\sigma+iT) - \frac{(\sigma - (1-y)/2)^2}{t})\ d\sigma$

and

$|E'| \leq \frac{\exp(\frac{(T-\frac{x}{2})^2}{4t} + \frac{1}{6(T-0.33)})}{\sqrt{32 t^3}} \int_X^\infty \exp(\mathrm{Re} F(\sigma+iT) - \frac{(\sigma - (1-y)/2)^2}{t}) (|T-x/2| + \sigma - \frac{1-y}{2})\ d\sigma.$

The $\sigma$ derivative of $\mathrm{Re} F(s) - \frac{(\sigma - (1-y)/2)^2}{t})$ is

$\mathrm{Re} \frac{1}{\sigma+iT} + \frac{1}{\sigma-1+iT} - \frac{1}{2} \log \pi + \frac{1}{2} \log \frac{\sigma+iT}{2} - \frac{1}{2(\sigma+iT)} - \frac{2(\sigma - (1-y)/2)}{t})$
$= \frac{\sigma}{2(\sigma^2+T^2)} + \frac{\sigma-1}{(\sigma-1)^2+T^2} + \frac{1}{2} \log \frac{\sqrt{\sigma^2+T^2}}{2\pi} - \frac{2(\sigma - (1-y)/2)}{t}$
$\leq \frac{1}{4T} + \frac{1}{2T} + \frac{1-y}{t} + \frac{1}{2} \log \frac{\sqrt{\sigma^2+T^2}}{2\pi} - \frac{2\sigma}{t}.$

The derivative of this latter expression is

$\frac{\sigma}{2(\sigma^2+T^2)} - \frac{1}{t} \leq \frac{1}{4T} - \frac{1}{t} \lt 0$

if we assume say

$t \leq 1/2 \quad (4)$

and so the above expression is at most $-Q$, where

$Q := \frac{2X}{t} - \frac{3}{4T} - \frac{1-y}{t} - \frac{1}{2} \log \frac{\sqrt{X^2+T^2}}{2\pi}. \quad (5)$

We will assume

$Q \gt 0 \quad (6)$.

We can then use the fundamental theorem of calculus to bound

$\exp(\mathrm{Re} F(\sigma+iT) - \frac{(\sigma - (1-y)/2)^2}{t}) \leq e^{-Q(\sigma-X)} \exp(\mathrm{Re} F(X+iT) - \frac{(X - (1-y)/2)^2}{t})$

and hence

$|E| \leq \frac{\exp(\frac{(T-\frac{x}{2})^2}{4t} + \frac{1}{6(T-0.33)})}{\sqrt{32 t} Q} \exp(\mathrm{Re} F(X+iT) - \frac{(X - (1-y)/2)^2}{t})$

and

$|E'| \leq \frac{\exp(\frac{(T-\frac{x}{2})^2}{4t} + \frac{1}{6(T-0.33)})}{\sqrt{32 t^3}} \exp(\mathrm{Re} F(X+iT) - \frac{(X - (1-y)/2)^2}{t}) (\frac{|T-x/2| + X - \frac{1-y}{2}}{Q} + \frac{1}{Q^2}).$

These bounds are valid under the hypotheses (1)-(6). In practice it seems good to choose $T$ to be slightly larger than $x/2$ (not so large that the $\exp(\frac{(T-\frac{x}{2})^2}{4})$ factor starts hurting): ,$T = \frac{x}{2} + \frac{\pi}{4}$ may be a good choice. X needs to be only moderately large (e.g. X = 8 for medium-sized x might be enough; for larger x, for (6) to hold one needs X to be somewhat larger than $\log x$).