# Asymptotics of H t

### Asymptotics for $t=0$

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

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

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

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

where

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

Writing

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

we have $F(s) = 2 F_0(s) - 3 F_{-1}(s)$, where

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

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

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

and thus

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

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

$\displaystyle |F_0(\frac{1+ix \pm y}{2})| \asymp \Gamma(\frac{9+ix \pm y}{2}).$

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

### Asymptotics for $t \gt 0$

Let $z=x+iy$ for large $x$ and positive bounded $y$. We have

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

where

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

We can shift contours to

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

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

$\displaystyle H_t(z) = G_t(x+iy) + \overline{G_t(x-iy)}$

where

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

By Fubini's theorem we have

$\displaystyle G_{t}(x \pm i y) = \sum_{n=1}^\infty \pi^2 n^4 \int_{i\theta}^{i\theta+\infty} \exp( tu^2 - \pi n^2 e^{4u} + (ix \mp y + 9) u)\ du$
$\displaystyle - \sum_{n=1}^\infty \frac{3}{2} \pi n^2 \int_{i\theta}^{i\theta+\infty} \exp( tu^2 - \pi n^2 e^{4u} + (ix \mp y + 5) u)\ du.$

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

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

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

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

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

$\int_{-\infty}^\infty \exp( (ix \mp y + 9) (v - \frac{1}{4} e^{4v}) )\ dv = \frac{1}{4} \Gamma(\frac{ix \mp y + 9}{4}) (\frac{4}{ix \mp y + 9})^{\frac{ix \mp y+9}{4}}$

and thus

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

which simplifies a bit to

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

and thus we heuristically have

$H_t(x+iy) \approx \frac{1}{4} F_t( \frac{1+ix-y}{2} ) + \frac{1}{4} \overline{F_t( \frac{1+ix+y}{2} )}$

where

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