Bounding the derivative of H t - second approach

We have

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

and hence on differentiation under the integral sign

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

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

[math]\displaystyle{ 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} }[/math]

and similarly

[math]\displaystyle{ 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}. }[/math]

Since

[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 thus have

[math]\displaystyle{ 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) }[/math]

and

[math]\displaystyle{ 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) }[/math]

where

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

and

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

It is thus of interest to find good estimates for [math]\displaystyle{ I_{t,\theta}(b,\beta), J_{t,\theta}(b,\beta) }[/math].

If [math]\displaystyle{ z = x+yi }[/math] with [math]\displaystyle{ y\gt 0 }[/math], we expect the dominant term here to be [math]\displaystyle{ 2\pi^2 n^4 \overline{J_{t,\theta}(\overline{z}-9i, \pi n^2)} }[/math]. For [math]\displaystyle{ b = \overline{z}-9i = x - (y+9)i }[/math] and [math]\displaystyle{ \beta = \pi n^2 }[/math], the phase [math]\displaystyle{ -\beta e^{4u} + ibu }[/math] has a stationary point at

[math]\displaystyle{ 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}). }[/math]

It is thus natural to select [math]\displaystyle{ \theta }[/math] to equal

[math]\displaystyle{ \theta := \frac{\pi}{8} - \arctan \frac{y+9}{x} \quad (1) }[/math]

as this should remove most of the oscillation. But one can select other values too. (For the purposes of numerically bounding [math]\displaystyle{ H'_t(z) }[/math] for [math]\displaystyle{ x }[/math] in some interval, one should probably pick a single [math]\displaystyle{ \theta }[/math] 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 [math]\displaystyle{ \theta }[/math] to evaluate all the terms in the above sum (e.g. one needs the same [math]\displaystyle{ \theta }[/math] to handle [math]\displaystyle{ b = z-9i, \overline{z}-9i, z-5i, \overline{z}-5i }[/math]).

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

[math]\displaystyle{ I_{t,\theta}(b,\beta) = I_{t,\theta,\leq X}(b,\beta) + I_{t,\theta, \geq X}(b,\beta) }[/math]

for any cutoff [math]\displaystyle{ X \geq 0 }[/math], where

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

and

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

Writing [math]\displaystyle{ u = i\theta + X + v }[/math] and setting [math]\displaystyle{ b = x - i a }[/math] (where in practice [math]\displaystyle{ a = 9+y, 9-y, 7+y, 7-y }[/math]), the tail can be written as

[math]\displaystyle{ 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 }[/math]

and hence by the triangle inequality

[math]\displaystyle{ |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. }[/math]

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

[math]\displaystyle{ 2tv - 4 \beta e^{4X} \cos \theta e^{4v} + a. }[/math]

Note that [math]\displaystyle{ e^{4v} \geq 1+4v }[/math]. Thus if [math]\displaystyle{ X }[/math] is so large that

[math]\displaystyle{ \beta e^{4X} \cos \theta \gt \max( \frac{t}{2}, \frac{a}{4}) }[/math]

or equivalently

[math]\displaystyle{ X \gt \frac{1}{4} \log \frac{\max(\frac{t}{2}, \frac{a}{4})}{\beta \cos \theta} \quad (2) }[/math]

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

[math]\displaystyle{ \exp( tv^2 - \beta e^{4X} (e^{4v}-1) \cos \theta + av ) \leq \exp( - v (4 \beta e^{4X} \cos \theta - a) ) }[/math]

thus giving the tail bound

[math]\displaystyle{ |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) }[/math]

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

For large values of [math]\displaystyle{ n }[/math] (and hence of [math]\displaystyle{ \beta = \pi n^2 }[/math]), one can set [math]\displaystyle{ X=0 }[/math] and estimate the entire integral. Indeed, we conclude that if

[math]\displaystyle{ \pi n^2 \cos \theta \gt \max(\frac{t}{2}, \frac{a}{4} ) }[/math]

then

[math]\displaystyle{ |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}. }[/math]

For instance, if

[math]\displaystyle{ \pi n_0^2 \cos \theta \gt \max(\frac{t}{2}, \frac{a}{4} ) }[/math]

then

[math]\displaystyle{ |\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} }[/math]

[math]\displaystyle{ \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) }[/math]

[math]\displaystyle{ = \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 }[/math]

[math]\displaystyle{ = \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) }[/math]

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

[math]\displaystyle{ |\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} ). }[/math]

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

[math]\displaystyle{ |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. }[/math]

If (2) holds then again we have

[math]\displaystyle{ \exp( tv^2 - \beta e^{4X} (e^{4v}-1) \cos \theta + av ) \leq \exp( - v (4 \beta e^{4X} \cos \theta - a) ) }[/math]

and hence

[math]\displaystyle{ |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) }[/math]

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

[math]\displaystyle{ \pi n_0^2 \cos \theta \gt \max(\frac{t}{2}, \frac{a}{4} ) }[/math]

then

[math]\displaystyle{ |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 ) }[/math]

and by arguing as before

[math]\displaystyle{ |\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 ) }[/math]

[math]\displaystyle{ \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 ) }[/math]

and similarly

[math]\displaystyle{ |\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 ) }[/math]

[math]\displaystyle{ \times ( \frac{n_0^2}{1-\alpha} + \frac{2n_0 \alpha}{(1-\alpha)^2} + \frac{\alpha^2+\alpha}{(1-\alpha)^3} ). }[/math]