Asymptotics of H t

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Asymptotics for [math]t=0[/math]

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

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

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


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


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

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

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

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

and thus

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

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

Asymptotics for [math]t \gt 0[/math]

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

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


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

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


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

[math] \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]\exp( (ix \mp y + 9) (v - \frac{1}{4} e^{4v}) )[/math] decays rapidly away from [math]v=0[/math]. This suggests firstly that this integral is going to be very small when [math]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]n[/math] is now restricted to the range [math]n \leq N[/math]. Next, we approximate [math]\exp( \frac{t}{16} (\log \frac{ix \pm y+9}{4\pi n^2} + v)^2)[/math] by [math]\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 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]w := \frac{ix \mp y + 9}{4} e^{4v}[/math] we see that

[math]\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 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 \frac{ix \pm y+9}{4\pi n^2}^2 ) [/math]

which simplifies a bit to

[math] \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 \frac{ix \pm y+9}{4\pi n^2}^2 )}{n^{\frac{1 \mp y + ix}{2}}} [/math]

and thus we heuristically have

[math] 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]


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