Asymptotics of H t

Asymptotics for [math]\displaystyle{ t=0 }[/math]

The approximate functional equation (see e.g. Titchmarsh equation (4.12.4)) asserts that

[math]\displaystyle{ \displaystyle \zeta(s) = \sum_{n \leq N} \frac{1}{n^s} + \pi^{s-1/2} \frac{\Gamma((1-s)/2)}{\Gamma(s/2)} \sum_{n \leq N} \frac{1}{n^{1-s}} + O( t^{-\sigma/2} ) }[/math]

for [math]\displaystyle{ s = \sigma +it }[/math] with [math]\displaystyle{ t }[/math] large, [math]\displaystyle{ 0 \lt \sigma \lt 1 }[/math], and [math]\displaystyle{ N := \sqrt{t/2\pi} }[/math]. This implies that

[math]\displaystyle{ \displaystyle \xi(s) = F(s) + F(1-s) + O( \Gamma(\frac{s+4}{2}) t^{-\sigma/2} ) }[/math]

where

[math]\displaystyle{ \displaystyle F(s) := \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \sum_{n=1}^N \frac{1}{n^s}. }[/math]

Writing

[math]\displaystyle{ \displaystyle \frac{s(s-1)}{2} \Gamma(s/2) = 2 \Gamma(\frac{s+4}{2}) - 3 \Gamma(\frac{s+2}{2}) }[/math]

we have [math]\displaystyle{ F(s) = 2 F_0(s) - 3 F_{-1}(s) }[/math], where

[math]\displaystyle{ \displaystyle F_j(s) := \pi^{-s/2} \Gamma(\frac{s+4}{2} + j) \sum_{n=1}^N \frac{1}{n^s}. }[/math]

The [math]\displaystyle{ F_{-1} }[/math] term sums to [math]\displaystyle{ O( \Gamma(\frac{s+4}{2}) t^{-\sigma/2} ) }[/math], hence

[math]\displaystyle{ \displaystyle \xi(s) = 2F_0(s) + 2F_0(1-s) + O( \Gamma(\frac{s+4}{2}) t^{-\sigma/2} ) }[/math]

and thus

[math]\displaystyle{ \displaystyle H(x+iy) = \frac{1}{4} F_0( \frac{1+ix-y}{2} ) + \frac{1}{4} \overline{F_0( \frac{1+ix+y}{2} )} + O( \Gamma(\frac{9+ix+y}{2}) x^{-(1+y)/2} ). }[/math]

One would expect the [math]\displaystyle{ \sum_{n=1}^N \frac{1}{n^s} }[/math] term to remain more or less bounded (this is basically the Lindelof hypothesis), leading to the heuristics

[math]\displaystyle{ \displaystyle |F_0(\frac{1+ix \pm y}{2})| \asymp \Gamma(\frac{9+ix \pm y}{2}). }[/math]

Since [math]\displaystyle{ \Gamma(\frac{9+ix - y}{2}) \approx \Gamma(\frac{9+ix+y}{2}) (ix)^{-y} }[/math], we expect the [math]\displaystyle{ F_0( \frac{1+ix+y}{2} ) }[/math] term to dominate once [math]\displaystyle{ y \gg \frac{1}{\log x} }[/math].

Asymptotics for [math]\displaystyle{ t \gt 0 }[/math]

Let [math]\displaystyle{ z=x+iy }[/math] for large [math]\displaystyle{ x }[/math] and positive bounded [math]\displaystyle{ y }[/math]. We have

[math]\displaystyle{ \displaystyle H_t(z) = \frac{1}{2} \int_{-\infty}^\infty e^{tu^2} \Phi(u) \exp(izu)\ du }[/math]

where

[math]\displaystyle{ \displaystyle \Phi(u) = \sum_{n=1}^\infty (2\pi^2 n^4 e^{9u} - 3\pi n^2 e^{5u}) \exp(-\pi n^2 e^{4u}). }[/math]

We can shift contours to

[math]\displaystyle{ \displaystyle H_t(z) = \frac{1}{2} \int_{i\theta-\infty}^{i\theta+\infty} e^{tu^2} \Phi(u) \exp(izu)\ du }[/math]

to any [math]\displaystyle{ -\pi/8 \lt \theta \lt \pi/8 }[/math] that we please; it seems that a good choice will be [math]\displaystyle{ \theta = \mathrm{arg} (ix+y+9) \approx \frac{\pi}{8} - \frac{y+9}{x} }[/math]. By symmetry, we thus have

[math]\displaystyle{ \displaystyle H_t(z) = G_t(x+iy) + \overline{G_t(x-iy)} }[/math]

where

[math]\displaystyle{ \displaystyle G_t(z) := \int_{i\theta}^{i\theta+\infty} e^{tu^2} \Phi(u) \exp(izu)\ du. }[/math]

By Fubini's theorem we have

[math]\displaystyle{ \displaystyle G_{t}(x \pm i y) = \sum_{n=1}^\infty \pi^2 n^4 \int_{i\theta}^{i\theta+\infty} \exp( tu^2 - \pi n^2 e^{4u} + (ix \mp y + 9) u)\ du }[/math]

[math]\displaystyle{ \displaystyle - \sum_{n=1}^\infty \frac{3}{2} \pi n^2 \int_{i\theta}^{i\theta+\infty} \exp( tu^2 - \pi n^2 e^{4u} + (ix \mp y + 5) u)\ du. }[/math]

The second terms end up being about [math]\displaystyle{ O(1/x) }[/math] the size of the first terms and we will ignore them for now. Making the change of variables [math]\displaystyle{ u = \frac{1}{4} \log \frac{ix \pm y + 9}{4\pi n^2} + v }[/math], we basically have

[math]\displaystyle{ \displaystyle G_t(x \pm iy) \approx \sum_{n=1}^\infty \pi^2 n^4 (\frac{ix \pm y+9}{4\pi n^2})^{\frac{ix \mp y+9}{4}} \int_{-\frac{1}{4} \log \frac{|ix\pm y+9|}{4\pi n^2}}^\infty \exp( \frac{t}{16} (\log \frac{ix \pm y+9}{4\pi n^2} + v)^2 + (ix \mp y + 9) (v - \frac{1}{4} e^{4v}) )\ dv. }[/math]

The function [math]\displaystyle{ \exp( (ix \mp y + 9) (v - \frac{1}{4} e^{4v}) ) }[/math] decays rapidly away from [math]\displaystyle{ v=0 }[/math]. This suggests firstly that this integral is going to be very small when [math]\displaystyle{ n \gg N := \sqrt{x/4\pi} }[/math] (since the left limit of integration will then be to the right of the origin), so we will assume heuristically that [math]\displaystyle{ n }[/math] is now restricted to the range [math]\displaystyle{ n \leq N }[/math]. Next, we approximate [math]\displaystyle{ \exp( \frac{t}{16} (\log \frac{ix \pm y+9}{4\pi n^2} + v)^2) }[/math] by [math]\displaystyle{ \exp( \frac{t}{16} \log^2 \frac{ix \pm y+9}{4\pi n^2} ) }[/math], and then send the left limit off to infinity to obtain (heuristically)

[math]\displaystyle{ \displaystyle G_t(x \pm iy) \approx \sum_{n \leq N} \pi^2 n^4 (\frac{ix \pm y+9}{4\pi n^2})^{\frac{ix \mp y+9}{4}} \exp( \frac{t}{16} \log^2 \frac{ix \pm y+9}{4\pi n^2} ) \int_{-\infty}^\infty \exp( (ix \mp y + 9) (v - \frac{1}{4} e^{4v}) )\ dv. }[/math]

Making the change of variables [math]\displaystyle{ w := \frac{ix \mp y + 9}{4} e^{4v} }[/math] we see that

[math]\displaystyle{ \int_{-\infty}^\infty \exp( (ix \mp y + 9) (v - \frac{1}{4} e^{4v}) )\ dv = \frac{1}{4} \Gamma(\frac{ix \mp y + 9}{4}) (\frac{4}{ix \mp y + 9})^{\frac{ix \mp y+9}{4}} }[/math]

and thus

[math]\displaystyle{ \displaystyle G_t(x \pm iy) \approx \Gamma(\frac{ix \mp y + 9}{4}) \sum_{n \leq N} \frac{\pi^2}{4} n^4 (\frac{1}{\pi n^2})^{\frac{ix \mp y+9}{4}} \exp( \frac{t}{16} \log^2 \frac{ix \pm y+9}{4\pi n^2} ) }[/math]

which simplifies a bit to

[math]\displaystyle{ \displaystyle G_t(x \pm iy) \approx \frac{1}{4} \pi^{-\frac{ix \mp y + 1}{4}} \Gamma(\frac{ix \mp y + 9}{4}) \sum_{n \leq N} \frac{\exp( \frac{t}{16} \log^2 \frac{ix \pm y+9}{4\pi n^2} )}{n^{\frac{1 \mp y + ix}{2}}} }[/math]

and thus we heuristically have

[math]\displaystyle{ H_t(x+iy) \approx \frac{1}{4} F_t( \frac{1+ix-y}{2} ) + \frac{1}{4} \overline{F_t( \frac{1+ix+y}{2} )} }[/math]

where

[math]\displaystyle{ F_t( s ) := \pi^{-s/2} \Gamma(\frac{s+4}{2}) \sum_{n \leq N} \frac{\exp( \frac{t}{16} \log^2 \frac{s}{2\pi n^2} )}{n^{s}}. }[/math]