# Difference between revisions of "Bounding the derivative of H t - second approach"

We have

$H_t(z) = \frac{1}{2} \int_{\bf R} e^{tu^2} \Phi(u) e^{izu}\ du$

and hence on differentiation under the integral sign

$H'_t(z) = \frac{i}{2} \int_{\bf R} e^{tu^2} u \Phi(u) e^{izu}\ du.$

For any $0 \leq \theta \lt \pi/8$, we may shift the contour to $i\theta + {\bf R}$ and then use the even nature of $\Phi$ to reflect the left half of that contour around the origin to obtain the identities

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

and similarly

$H'_t(z) = \frac{i}{2} \int_{i\theta}^{i\theta+\infty} e^{tu^2} \Phi(u) u e^{izu}\ du - \frac{i}{2} \overline{\int_{i\theta}^{i\theta+\infty} e^{tu^2} \Phi(u) u e^{i\overline{z}u}\ du}.$

Since

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

we thus have

$H_t(z) = \frac{1}{2} \sum_{n=1}^\infty 2\pi^2 n^4 I_{t,\theta}(z-9i, \pi n^2) - 3\pi n^2 I_{t,\theta}(z-5i, \pi n^2) + 2\pi^2 n^4 \overline{I_{t,\theta}(\overline{z}-9i, \pi n^2)} - 3\pi n^2 \overline{I_{t,\theta}(\overline{z}-5i, \pi n^2)} \quad (0)$

and

$H'_t(z) = \frac{1}{2} \sum_{n=1}^\infty 2\pi^2 n^4 J_{t,\theta}(z-9i, \pi n^2) - 3\pi n^2 J_{t,\theta}(z-5i, \pi n^2) - 2\pi^2 n^4 \overline{J_{t,\theta}(\overline{z}-9i, \pi n^2)} + 3\pi n^2 \overline{J_{t,\theta}(\overline{z}-5i, \pi n^2)} \quad (1)$

where

$I_{t,\theta}(b,\beta) := \int_{i\theta}^{i\theta+\infty} \exp( tu^2 - \beta e^{4u} + ibu)\ du$

and

$J_{t,\theta}(b,\beta) := \int_{i\theta}^{i\theta+\infty} \exp( tu^2 - \beta e^{4u} + ibu) u\ du.$

It is thus of interest to find good estimates for $I_{t,\theta}(b,\beta), J_{t,\theta}(b,\beta)$.

If $z = x+yi$ with $y\gt0$, we expect the dominant term here to be $2\pi^2 n^4 \overline{J_{t,\theta}(\overline{z}-9i, \pi n^2)}$. For $b = \overline{z}-9i = x - (y+9)i$ and $\beta = \pi n^2$, the phase $-\beta e^{4u} + ibu$ has a stationary point at

$u = \frac{1}{4} \log \frac{ix+y+9}{4\pi n^2} = \frac{1}{4} \log \frac{|ix+y+9|}{4\pi n^2} + i (\frac{\pi}{8} - \arctan \frac{y+9}{x}).$

It is thus natural to select $\theta$ to equal

$\theta := \frac{\pi}{8} - \arctan \frac{y+9}{x} \quad (1)$

as this should remove most of the oscillation. But one can select other values too. (For the purposes of numerically bounding $H'_t(z)$ for $x$ in some interval, one should probably pick a single $\theta$ for this interval that is close to, but not necessarily equal to, the value given in (1). Note though that one has to use the same value of $\theta$ to evaluate all the terms in the above sum (e.g. one needs the same $\theta$ to handle $b = z-9i, \overline{z}-9i, z-5i, \overline{z}-5i$).

The integral $I_{t,\theta}(b,\beta)$ is improper and so cannot be directly evaluated numerically. However, we can always cut it off as

$I_{t,\theta}(b,\beta) = I_{t,\theta,\leq X}(b,\beta) + I_{t,\theta, \geq X}(b,\beta)$

for any cutoff $X \geq 0$, where

$I_{t,\theta,\geq X}(b,\beta) := \int_{i\theta}^{i\theta+X} \exp( tu^2 - \beta e^{4u} + ibu}\ du$

and

math>I_{t,\theta,\geq X}(b,\beta) := \int_{i\theta+X}^{i\theta+\infty} \exp( tu^2 - \beta e^{4u} + ibu}\ du.[/itex]

Writing $u = i\theta + X + v$ and setting $b = x - i a$ (where in practice $a = 9+y, 9-y, 7+y, 7-y$), the tail can be written as

$I_{t,\theta,\geq X}(b,\beta) = \int_0^\infty \exp( t (-\theta^2 + X^2 + v^2 + 2i\theta X + 2i\theta v) - \beta e^{4X} e^{4v} \cos \theta - i \beta e^{4X} e^{4v} \sin \theta - \theta x + iX x + iv x+ aX + av + ia\theta)\ dv$

and hence by the triangle inequality

$|I_{t,\theta,\geq X}(b,\beta)| \leq \exp( -t \theta^2 + tX^2 - \beta e^{4X} \cos \theta - \theta x + aX ) \int_0^\infty \exp( tv^2 - \beta e^{4X} (e^{4v}-1) \cos \theta + av )\ dv.$

The exponent $tv^2 - \beta e^{4X} (e^{4v}-1) \cos \theta + av$ equals $0$ when $v=0$ and has derivative

$2tv - 4 \beta e^{4X} \cos \theta e^{4v} + a.$

Note that $e^{4v} \geq 1+4v$. Thus if $X$ is so large that

$\beta e^{4X} \cos \theta \gt \max( \frac{t}{2}, \frac{a}{4})$

or equivalently

$X \gt \frac{1}{4} \log \frac{\max(\frac{t}{2}, \frac{a}{4})}{\beta \cos \theta} \quad (2)$

then this derivative will be bounded from below by the negative quantity $-(4 \beta e^{4X} \cos \theta - a)$. This implies that

$\exp( tv^2 - \beta e^{4X} (e^{4v}-1) \cos \theta + av ) \leq \exp( - v (4 \beta e^{4X} \cos \theta - a) )$

thus giving the tail bound

$|I_{t,\theta,\geq X}(b,\beta)| \leq \frac{\exp( -t \theta^2 + tX^2 - \beta e^{4X} \cos \theta - \theta x + aX )}{4 \beta e^{4X} \cos \theta - a}\quad (3)$

whenever (2) holds. This will start decaying rapidly when $X$ is large enough (basically one needs $e^{4X} \cos \theta$ to become large).

For large values of $n$ (and hence of $\beta = \pi n^2$), one can set $X=0$ and estimate the entire integral. Indeed, we conclude that if

$\pi n^2 \cos \theta \gt \max(\frac{t}{2}, \frac{a}{4} )$

then

$|I_{t,\theta}(b,\pi n^2)| \leq \frac{\exp( -t \theta^2 - \pi n^2 \cos \theta - \theta x )}{4 \pi n^2 \cos \theta - a}.$

For instance, if

$\pi n_0^2 \cos \theta \gt \max(\frac{t}{2}, \frac{a}{4} )$

then

$|\sum_{n \geq n_0} 2\pi^2 n^4 I_{t,\theta}(x-ia, \pi n^2)| \leq \sum_{n \geq n_0} 2\pi^2 n^4 \frac{\exp( -t \theta^2 - \pi n^2 \cos \theta - \theta x )}{4 \pi n^2 \cos \theta - a}$
$\leq \frac{2\pi^2 \exp(-t \theta^2 - \theta x)}{4\pi n_0^2 \cos \theta - a} \sum_{n \geq n_0} n^4 \exp( - \pi n n_0 \cos \theta)$
$= \frac{2\pi^2 \exp(-t \theta^2 - \theta x - \pi n_0^2 \cos \theta)}{4\pi n_0^2 \cos \theta - a} \sum_{m=0}^\infty (n_0^4 + 4 n_0^3 m + 6 n_0^2 m^2 + 4 n_0 m^3 + m^4) \alpha^m$
$= \frac{2\pi^2 \exp(-t \theta^2 - \theta x - \pi n_0^2 \cos \theta)}{4\pi n_0^2 \cos \theta - a} ( \frac{n_0^4}{1 - \alpha} + \frac{4 n_0^3 \alpha}{(1 - \alpha)^2} + \frac{6 n_0^2 (\alpha^2+\alpha)}{(1-\alpha)^3} + \frac{4n_0 (\alpha^3+4\alpha^2+\alpha)}{(1-\alpha)^4} + \frac{\alpha^4 + 11 \alpha^3 + 11 \alpha^2 + \alpha}{(1-\alpha)^5 ) \quad (4)$

where $\alpha := e^{-\pi n_0 \cos \theta}$ (one can do better here by expressing $\frac{n^4}{4\pi n^2 \cos \theta - a}$ as an improper fraction). A similar argument gives

$|\sum_{n \geq n_0} 3\pi n^2 I_{t,\theta}(x-ia, \pi n^2)| \leq \frac{3\pi \exp(-t \theta^2 - \theta x - \pi n_0^2 \cos \theta)}{4\pi n_0^2 \cos \theta - a} ( \frac{n_0^2}{1-\alpha} + \frac{2n_0 \alpha}{(1-\alpha)^2} + \frac{\alpha^2+\alpha}{(1-\alpha)^3} ).$

This lets us control the tail of the series in (0).

Similar arguments let us handle the derivative. Indeed by arguing as before we have

$|J_{t,\theta,\geq X}(b,\beta)| \leq \exp( -t \theta^2 + tX^2 - \beta e^{4X} \cos \theta - \theta x + aX ) \int_0^\infty \exp( tv^2 - \beta e^{4X} (e^{4v}-1) \cos \theta + av ) (|X+i\theta| + v)\ dv.$

If (2) holds then again we have

$\exp( tv^2 - \beta e^{4X} (e^{4v}-1) \cos \theta + av ) \leq \exp( - v (4 \beta e^{4X} \cos \theta - a) )$

and hence

$|J_{t,\theta,\geq X}(b,\beta)| \leq \exp( -t \theta^2 + tX^2 - \beta e^{4X} \cos \theta - \theta x + aX ) ({|X+i\theta|}{4 \beta e^{4X} \cos \theta - a} + \frac{1}{4 \beta e^{4X} \cos \theta - a)^2}) \quad (5)$

which lets us handle the tail for any $J_{t,\theta}(b,\beta)$. In particular, if $n \geq n_0$ and

$\pi n_0^2 \cos \theta \gt \max(\frac{t}{2}, \frac{a}{4} )$

then

$|J_{t,\theta}(b,\pi n^2)| \leq \exp( -t \theta^2 - \pi n^2 \cos \theta - \theta x ) ( \frac{\theta}{4 \pi n^2 \cos \theta - a} + \frac{1}{(4 \pi n^2 \cos \theta - a)^2 )$

and by arguing as before

$|\sum_{n \geq n_0} 2\pi^2 n^4 J_{t,\theta}(x-ia, \pi n^2)| \leq 2\pi^2 \exp( -t \theta^2 - \pi n^2 \cos \theta - \theta x ) ( \frac{\theta}{4 \pi n_0^2 \cos \theta - a} + \frac{1}{(4 \pi n_0^2 \cos \theta - a)^2 )$
$\times ( \frac{n_0^4}{1 - \alpha} + \frac{4 n_0^3 \alpha}{(1 - \alpha)^2} + \frac{6 n_0^2 (\alpha^2+\alpha)}{(1-\alpha)^3} + \frac{4n_0 (\alpha^3+4\alpha^2+\alpha)}{(1-\alpha)^4} + \frac{\alpha^4 + 11 \alpha^3 + 11 \alpha^2 + \alpha}{(1-\alpha)^5 )$

and similarly

$|\sum_{n \geq n_0} 3\pi n^2 J_{t,\theta}(x-ia, \pi n^2)| \leq 3\pi \exp( -t \theta^2 - \pi n^2 \cos \theta - \theta x ) ( \frac{\theta}{4 \pi n_0^2 \cos \theta - a} + \frac{1}{(4 \pi n_0^2 \cos \theta - a)^2 )$
$\times ( \frac{n_0^2}{1-\alpha} + \frac{2n_0 \alpha}{(1-\alpha)^2} + \frac{\alpha^2+\alpha}{(1-\alpha)^3} ).$