Bounding the derivative of H t - third approach

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

[math] 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]s = \frac{1-y+ix}{2} + \sqrt{t} v[/math] is

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

[math] 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]T[/math], we have

[math] 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]


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

[math] 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]


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

one has

[math] 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]


[math] 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]E, E'[/math] can be bounded using the triangle inequality by

[math] |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]


[math] |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]y \geq 0, \quad (2)[/math]


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

and [math]|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] |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]


[math] |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]


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

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

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


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

we can use the Stirling approximation

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

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


[math]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] |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]


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

[math]\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] = \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] \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] \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] t \leq 1/2 \quad (4)[/math]

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

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

We can then use the fundamental theorem of calculus to bound

[math] \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] |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]


[math] |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]T[/math] to be slightly larger than [math]x/2[/math] (not so large that the [math]\exp(\frac{(T-\frac{x}{2})^2}{4})[/math] factor starts hurting): ,[math]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]\log x[/math]).