Bounding the derivative of H t - third approach

To compute [math]\displaystyle{ H_t(x+iy) }[/math], we begin with the heat fundamental solution formula

[math]\displaystyle{ 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 }[/math]

which on making the change of variables [math]\displaystyle{ s = \frac{1-y+ix}{2} + \sqrt{t} v }[/math] is

[math]\displaystyle{ 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 }[/math]

where [math]\displaystyle{ L }[/math] is any horizontal line contour; differentiating under the integral sign, we also have

[math]\displaystyle{ 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. }[/math]

In particular, for any real number [math]\displaystyle{ T }[/math], we have

[math]\displaystyle{ 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 }[/math]

and

[math]\displaystyle{ 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. }[/math]

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

[math]\displaystyle{ 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 }[/math]

and

[math]\displaystyle{ 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 }[/math]

Cutting off the integral at some cutoff parameter

[math]\displaystyle{ X \gt 1 \quad (1), }[/math]

one has

[math]\displaystyle{ 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 }[/math]

and

[math]\displaystyle{ 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' }[/math]

where the error terms [math]\displaystyle{ E, E' }[/math] can be bounded using the triangle inequality by

[math]\displaystyle{ |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. }[/math]

and

[math]\displaystyle{ |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. }[/math]

If we assume

[math]\displaystyle{ y \geq 0, \quad (2) }[/math]

then

[math]\displaystyle{ (\sigma - (1+y)/2)^2 = (1-\sigma - (1-y)/2)^2 \leq (\sigma - (1-y)/2)^2 }[/math]

and [math]\displaystyle{ |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} }[/math] hence

[math]\displaystyle{ |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 }[/math]

and

[math]\displaystyle{ |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. }[/math]

Since

[math]\displaystyle{ \xi(s) = \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \zeta(s) }[/math]

and [math]\displaystyle{ \zeta(\sigma+iT) \leq \zeta(X) }[/math] when [math]\displaystyle{ \sigma \geq X }[/math], we have

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

Assuming

[math]\displaystyle{ T \geq 2, \quad (3) }[/math]

we can use the Stirling approximation

[math]\displaystyle{ |\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)}) }[/math]

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

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

where

[math]\displaystyle{ F(s) := \log s + \log(s-1) - \frac{s}{2} \log \pi + \frac{s-1}{2} \log \frac{s}{2} - \frac{s}{2} }[/math]

and so

[math]\displaystyle{ |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 }[/math]

and

[math]\displaystyle{ |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. }[/math]

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

[math]\displaystyle{ \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}) }[/math]

[math]\displaystyle{ = \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} }[/math]

[math]\displaystyle{ \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}. }[/math]

The derivative of this latter expression is

[math]\displaystyle{ \frac{\sigma}{2(\sigma^2+T^2)} - \frac{1}{t} \leq \frac{1}{4T} - \frac{1}{t} \lt 0 }[/math]

if we assume say

[math]\displaystyle{ t \leq 1/2 \quad (4) }[/math]

and so the above expression is at most [math]\displaystyle{ -Q }[/math], where

[math]\displaystyle{ 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) }[/math]

We will assume

[math]\displaystyle{ Q \gt 0 \quad (6) }[/math].

We can then use the fundamental theorem of calculus to bound

[math]\displaystyle{ \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}) }[/math]

and hence

[math]\displaystyle{ |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}) }[/math]

and

[math]\displaystyle{ |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}). }[/math]

These bounds are valid under the hypotheses (1)-(6). In practice it seems good to choose [math]\displaystyle{ T }[/math] to be slightly larger than [math]\displaystyle{ x/2 }[/math] (not so large that the [math]\displaystyle{ \exp(\frac{(T-\frac{x}{2})^2}{4}) }[/math] factor starts hurting): ,[math]\displaystyle{ T = \frac{x}{2} + \frac{\pi}{4} }[/math] 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 [math]\displaystyle{ \log x }[/math]).