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

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(Created page with "We have :<math>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>H'_t(z) = \frac{i}{2} \int_...")
 
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For any <math>0 \leq \theta < \pi/8</math>, we may shift the contour to <math>i\theta + {\bf R}</math> and then use the even nature of <math>\Phi</math> to reflect the left half of that contour around the origin to obtain the identities
 
For any <math>0 \leq \theta < \pi/8</math>, we may shift the contour to <math>i\theta + {\bf R}</math> and then use the even nature of <math>\Phi</math> to reflect the left half of that contour around the origin to obtain the identities
  
:<math>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>
+
:<math>H_t(z) = \frac{1}{2} \int_{i\theta}^{i\theta+\infty} e^{tu^2} \Phi(u) e^{izu}\ du + \frac{1}{2} \int_{i\theta-\infty}^{i\theta} e^{tu^2} \Phi(u) e^{izu}\ du</math>
 
+
:<math>= \frac{1}{2} \int_{i\theta}^{i\theta+\infty} e^{tu^2} \Phi(u) e^{izu}\ du + \frac{1}{2} \int_{-i\theta}^{-i\theta+\infty} e^{tu^2} \Phi(u) e^{-izu}\ du</math>
 +
:<math>= \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
 
and similarly
  
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:<math>\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>
 
:<math>\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
+
we thus have for <math>z=x+iy</math>
  
 
:<math>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) +
 
:<math>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>
+
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)} </math>
 +
:<math> = \frac{1}{2} \sum_{n=1}^\infty 2\pi^2 n^4 I_{t,\theta}(x-(9-y)i, \pi n^2) - 3\pi n^2 I_{t,\theta}(x-(5-y)i, \pi n^2) +
 +
2\pi^2 n^4 \overline{I_{t,\theta}(x-(9+y)i, \pi n^2)} - 3\pi n^2 \overline{I_{t,\theta}(x-(5+y)i, \pi n^2)} \quad (0)</math>
  
 
and
 
and
  
:<math>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) -
+
:<math>H'_t(z) = \frac{i}{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>
+
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)} </math>
 +
:<math>= \frac{i}{2} \sum_{n=1}^\infty 2\pi^2 n^4 J_{t,\theta}(x-(9-y)i, \pi n^2) - 3\pi n^2 J_{t,\theta}(x-(5-y)i, \pi n^2) -
 +
2\pi^2 n^4 \overline{J_{t,\theta}(x-(9+y)i, \pi n^2)} + 3\pi n^2 \overline{J_{t,\theta}(x - (5+y)i, \pi n^2)} \quad (1)</math>
  
 
where
 
where
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:<math>J_{t,\theta}(b,\beta) := \int_{i\theta}^{i\theta+\infty} \exp( tu^2 - \beta e^{4u} + ibu) u\ du.</math>
 
:<math>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>I_{t,\theta}(b,\beta), J_{t,\theta}(b,\beta)</math>.
+
It is thus of interest to find good estimates for <math>I_{t,\theta}(b,\beta), J_{t,\theta}(b,\beta)</math> when <math>\beta = \pi n^2</math> for some <math>n</math>, and <math>b = x-ia</math> for <math>a = 9+y, 9-y, 5+y, 5-y</math>.
  
If <math>z = x+yi</math> with <math>y>0</math>, we expect the dominant term here to be <math>2\pi^2 n^4 \overline{J_{t,\theta}(\overline{z}-9i, \pi n^2)}</math>.  For <math>b = \overline{z}-9i = x - (y+9)i</math> and <math>\beta = \pi n^2</math>, the phase <math>-\beta e^{4u} + ibu</math> has a stationary point at  
+
If <math>x,y>0</math>, we expect the dominant term here to be <math>2\pi^2 n^4 \overline{J_{t,\theta}(x-(9+y)i, \pi n^2)}</math> (these eventually correspond to the "<math>B</math>" terms in the <math>H_t</math> expansion, with the <math>n=1</math> case corresponding to the "<math>B_0</math>" term).  For <math>b = \overline{z}-9i = x - (y+9)i</math> and <math>\beta = \pi n^2</math>, the phase <math>-\beta e^{4u} + ibu</math> has a stationary point at  
 
:<math>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>
 
:<math>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>\theta</math> to equal
 
It is thus natural to select <math>\theta</math> to equal
 
:<math> \theta := \frac{\pi}{8} - \arctan \frac{y+9}{x} \quad (1)</math>
 
:<math> \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>H'_t(z)</math> for <math>x</math> in some interval, one should probably pick a single <math>\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>\theta</math> to evaluate all the terms in the above sum (e.g. one needs the same <math>\theta</math> to handle <math>b = z-9i, \overline{z}-9i, z-5i, \overline{z}-5i</math>).
+
as this should remove most of the oscillation.  But one can select other values too.  (For the purposes of numerically bounding <math>H'_t(z)</math> for <math>x</math> in some interval, one should probably pick a single <math>\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>\theta</math> to evaluate all the terms in the above sum (e.g. one needs the same <math>\theta</math> to handle <math>b = z-9i, \overline{z}-9i, z-5i, \overline{z}-5i</math>).
  
 
The integral <math>I_{t,\theta}(b,\beta)</math> is improper and so cannot be directly evaluated numerically.  However, we can always cut it off as
 
The integral <math>I_{t,\theta}(b,\beta)</math> is improper and so cannot be directly evaluated numerically.  However, we can always cut it off as
 
:<math>I_{t,\theta}(b,\beta) = I_{t,\theta,\leq X}(b,\beta) + I_{t,\theta, \geq X}(b,\beta)</math>
 
:<math>I_{t,\theta}(b,\beta) = I_{t,\theta,\leq X}(b,\beta) + I_{t,\theta, \geq X}(b,\beta)</math>
 
for any cutoff <math>X \geq 0</math>, where
 
for any cutoff <math>X \geq 0</math>, where
:<math>I_{t,\theta,\geq X}(b,\beta) := \int_{i\theta}^{i\theta+X} \exp( tu^2 - \beta e^{4u} + ibu}\ du</math>
+
:<math>I_{t,\theta,\leq X}(b,\beta) := \int_{i\theta}^{i\theta+X} \exp( tu^2 - \beta e^{4u} + ibu)\ du</math>
 
and  
 
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>
+
:<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>u = i\theta + X + v</math> and setting <math>b = x - i a</math> (where in practice <math>a = 9+y, 9-y, 7+y, 7-y</math>), the tail can be written as
+
Writing <math>u = i\theta + X + v</math> and setting <math>b = x - i a</math> (where in practice <math>a = 9+y, 9-y, 5+y, 5-y</math>), the tail can be written as
:<math>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>
+
:<math>I_{t,\theta,\geq X}(b,\beta) = \int_0^\infty \exp( t (-\theta^2 + X^2 + 2Xv + v^2 + 2i\theta X + 2i\theta v) - \beta e^{4X} e^{4v} \cos(4\theta) - i \beta e^{4X} e^{4v} \sin (4\theta) - \theta x + iX x + iv x+ aX + av + ia\theta)\ dv</math>
 
and hence by the triangle inequality
 
and hence by the triangle inequality
:<math>|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>
+
:<math>|I_{t,\theta,\geq X}(b,\beta)| \leq \exp( -t \theta^2 + tX^2 - \beta e^{4X} \cos(4\theta) - \theta x + aX ) \int_0^\infty \exp( tv^2 + 2tXv - \beta e^{4X} (e^{4v}-1) \cos (4\theta) + av )\ dv.</math>
The exponent <math>tv^2 - \beta e^{4X} (e^{4v}-1) \cos \theta + av</math> equals <math>0</math> when <math>v=0</math> and has derivative
+
(Note: the <math>\exp(-\theta x)</math> factor here gives the main decay factor <math>e^{-\pi x/8}</math> for <math>H_t</math> once one chooses <math>\theta</math> close to <math>\pi/8</math>.  But one cannot choose it ''too'' close, otherwise the <math>\cos(4\theta)</math> term becomes too small.)
:<math>2tv - 4 \beta e^{4X} \cos \theta e^{4v} + a.</math>
+
The exponent <math>tv^2 + 2tXv - \beta e^{4X} (e^{4v}-1) \cos(4\theta) + av</math> equals <math>0</math> when <math>v=0</math> and has derivative
 +
:<math>2tv + 2tX - 4 \beta e^{4X} \cos(4\theta) e^{4v} + a.</math>
 
Note that <math>e^{4v} \geq 1+4v</math>.  Thus if <math>X</math> is so large that
 
Note that <math>e^{4v} \geq 1+4v</math>.  Thus if <math>X</math> is so large that
:<math> \beta e^{4X} \cos \theta > \max( \frac{t}{2}, \frac{a}{4}) </math>
+
:<math> \beta e^{4X} \cos(4\theta) > \max( \frac{t}{2}, \frac{a+2tX}{4}) \quad (2) </math>
or equivalently
+
then this derivative will be bounded from below by the negative quantity <math>-(4 \beta e^{4X} \cos(4\theta) - a - 2tX)</math>.  This implies that
:<math>X > \frac{1}{4} \log \frac{\max(\frac{t}{2}, \frac{a}{4})}{\beta \cos \theta} \quad (2)</math>
+
:<math>\exp( tv^2 - \beta e^{4X} (e^{4v}-1) \cos(4\theta) + av ) \leq \exp( - v (4 \beta e^{4X} \cos(4\theta) - a- 2tX) )</math>
then this derivative will be bounded from below by the negative quantity <math>-(4 \beta e^{4X} \cos \theta - a)</math>.  This implies that
+
:<math>\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
 
thus giving the tail bound
:<math>|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>
+
:<math>|I_{t,\theta,\geq X}(b,\beta)| \leq \frac{\exp( -t \theta^2 + tX^2 - \beta e^{4X} \cos(4\theta) - \theta x + aX )}{4 \beta e^{4X} \cos(4\theta) - a - 2tX}\quad (3) </math>
whenever (2) holds.  This will start decaying rapidly when <math>X</math> is large enough (basically one needs <math>e^{4X} \cos \theta</math> to become large).
+
whenever (2) holds.  This will start decaying rapidly when <math>X</math> is large enough (basically one needs <math>\beta e^{4X} \cos(4\theta)</math> to become large).
  
 
For large values of <math>n</math> (and hence of <math>\beta = \pi n^2</math>), one can set <math>X=0</math> and estimate the entire integral.  Indeed, we conclude that if
 
For large values of <math>n</math> (and hence of <math>\beta = \pi n^2</math>), one can set <math>X=0</math> and estimate the entire integral.  Indeed, we conclude that if
 
:<math>\pi n^2 \cos \theta > \max(\frac{t}{2}, \frac{a}{4} )</math>
 
:<math>\pi n^2 \cos \theta > \max(\frac{t}{2}, \frac{a}{4} )</math>
 
then
 
then
:<math>|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>
+
:<math>|I_{t,\theta}(b,\pi n^2)| \leq \frac{\exp( -t \theta^2 - \pi n^2 \cos(4\theta) - \theta x )}{4 \pi n^2 \cos(4\theta) - a}.</math>
  
 
For instance, if
 
For instance, if
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then
 
then
 
:<math> |\sum_{n \geq n_0} 2\pi^2 n^4 I_{t,\theta}(x-ia, \pi n^2)|
 
:<math> |\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>
+
\leq \sum_{n \geq n_0} 2\pi^2 n^4 \frac{\exp( -t \theta^2 - \pi n^2 \cos(4\theta) - \theta x )}{4 \pi n^2 \cos(4\theta) - a} </math>
:<math> \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> \leq \frac{2\pi^2 \exp(-t \theta^2 - \theta x)}{4\pi n_0^2 \cos(4\theta) - a- 2tX} \sum_{n \geq n_0} n^4 \exp( - \pi n n_0 \cos(4\theta)) </math>
:<math> = \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> = \frac{2\pi^2 \exp(-t \theta^2 - \theta x - \pi n_0^2 \cos(4\theta))}{4\pi n_0^2 \cos(4\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>= \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} +  
+
:<math>= \frac{2\pi^2 \exp(-t \theta^2 - \theta x - \pi n_0^2 \cos(4\theta))}{4\pi n_0^2 \cos(4\theta) - a} \times</math>
\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>
+
:<math> \times ( \frac{n_0^4}{1 - \alpha} +  
where <math>\alpha := e^{-\pi n_0 \cos \theta}</math> (one can do better here by expressing <math>\frac{n^4}{4\pi n^2 \cos \theta - a}</math> as an improper fraction).  A similar argument gives
+
\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>\alpha := e^{-\pi n_0 \cos \theta}</math> (one can do better here by expressing <math>\frac{n^4}{4\pi n^2 \cos \theta - a- 2tX}</math> as an improper fraction, but the formulae are messier).  A similar argument gives
 
:<math> |\sum_{n \geq n_0} 3\pi n^2 I_{t,\theta}(x-ia, \pi n^2)|
 
:<math> |\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}  
+
\leq \frac{3\pi \exp(-t \theta^2 - \theta x - \pi n_0^2 \cos(4\theta))}{4\pi n_0^2 \cos(4\theta) - a}  
 
( \frac{n_0^2}{1-\alpha} + \frac{2n_0 \alpha}{(1-\alpha)^2} + \frac{\alpha^2+\alpha}{(1-\alpha)^3} ).</math>
 
( \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).
 
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
+
Similar arguments let us handle the derivative.  Indeed we can again split
:<math>|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>
+
:<math>J_{t,\theta}(b,\beta) = J_{t,\theta,\leq X}(b,\beta) + J_{t,\theta, \geq X}(b,\beta)</math>
 +
for any cutoff <math>X \geq 0</math>, where
 +
:<math>J_{t,\theta,\leq X}(b,\beta) := \int_{i\theta}^{i\theta+X} \exp( tu^2 - \beta e^{4u} + ibu) u\ du</math>
 +
and
 +
:<math>J_{t,\theta,\geq X}(b,\beta) := \int_{i\theta+X}^{i\theta+\infty} \exp( tu^2 - \beta e^{4u} + ibu)u \ du.</math>
 +
From the triangle inequality as before we have
 +
:<math>|J_{t,\theta,\leq X}(b,\beta)| \leq \exp( -t \theta^2 - \theta x ) \int_0^X \exp( tu^2 - \beta e^{4u} \cos \theta + au ) (\theta + u)\ du</math>
 +
and
 +
:<math>|J_{t,\theta,\geq X}(b,\beta)| \leq \exp( -t \theta^2 + tX^2 - \beta e^{4X} \cos(4\theta) - \theta x + aX ) \int_0^\infty \exp( tv^2 + 2tXv - \beta e^{4X} (e^{4v}-1) \cos \theta + av ) (|X+i\theta| + v)\ dv. \quad (5)</math>
 +
[Side remark: one nice feature of these bound is that the dependence on <math>x</math> becomes very simple if we do not let <math>\theta</math> or <math>X</math> vary with <math>x</math>.  This could be important for obtaining bounds on <math>|H'_t(x+iy)|</math> that are uniform for <math>x</math> in an interval.]
 
If (2) holds then again we have
 
If (2) holds then again we have
:<math>\exp( tv^2 - \beta e^{4X} (e^{4v}-1) \cos \theta + av ) \leq \exp( - v (4 \beta e^{4X} \cos \theta - a) )</math>
+
:<math>\exp( tv^2 - \beta e^{4X} (e^{4v}-1) \cos(4\theta) + av ) \leq \exp( - v (4 \beta e^{4X} \cos(4\theta) - a- 2tX) )</math>
 
and hence
 
and hence
:<math>|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>
+
:<math>|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- 2tX} + \frac{1}{4 \beta e^{4X} \cos(4\theta) - a- 2tX)^2}) \quad (6) </math>
 
which lets us handle the tail for any <math>J_{t,\theta}(b,\beta)</math>.  In particular, if <math>n \geq n_0</math> and  
 
which lets us handle the tail for any <math>J_{t,\theta}(b,\beta)</math>.  In particular, if <math>n \geq n_0</math> and  
 
:<math>\pi n_0^2 \cos \theta > \max(\frac{t}{2}, \frac{a}{4} )</math>
 
:<math>\pi n_0^2 \cos \theta > \max(\frac{t}{2}, \frac{a}{4} )</math>
 
then
 
then
:<math>|J_{t,\theta}(b,\pi n^2)| \leq \exp( -t \theta^2 - \pi n^2 \cos \theta - \theta x )
+
:<math>|J_{t,\theta}(b,\pi n^2)| \leq \exp( -t \theta^2 - \pi n^2 \cos(4\theta) - \theta x )
( \frac{\theta}{4 \pi n^2 \cos \theta - a} + \frac{1}{(4 \pi n^2 \cos \theta - a)^2 )</math>
+
( \frac{\theta}{4 \pi n^2 \cos(4\theta) - a} + \frac{1}{(4 \pi n^2 \cos(4\theta) - a)^2} )</math>
 
and by arguing as before
 
and by arguing as before
:<math> |\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 )
+
:<math> |\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(4\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>
+
( \frac{\theta}{4 \pi n_0^2 \cos(4\theta) - a} + \frac{1}{(4 \pi n_0^2 \cos \theta - a)^2} )</math>
 
:<math> \times ( \frac{n_0^4}{1 - \alpha} +  
 
:<math> \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>
+
\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
 
and similarly
:<math> |\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 )
+
:<math> |\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(4\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>
+
( \frac{\theta}{4 \pi n_0^2 \cos(4\theta) - a} + \frac{1}{(4 \pi n_0^2 \cos(4\theta) - a)^2} )</math>
 
:<math> \times ( \frac{n_0^2}{1-\alpha} + \frac{2n_0 \alpha}{(1-\alpha)^2} + \frac{\alpha^2+\alpha}{(1-\alpha)^3} ).</math>
 
:<math> \times ( \frac{n_0^2}{1-\alpha} + \frac{2n_0 \alpha}{(1-\alpha)^2} + \frac{\alpha^2+\alpha}{(1-\alpha)^3} ).</math>
 +
This controls the tail of the sum in (1).  The main terms can be controlled by (5) and (6) for a suitable choice of cutoff <math>X</math>.  After fixing <math>n_0, X, \theta</math>, the only <math>x</math> dependence in these terms is a factor of <math>e^{-\theta x}</math>, so one gets uniform estimates for any <math>x \geq x_0</math>.

Revision as of 07:28, 13 March 2018

We have

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

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

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

[math]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)} [/math]
[math] = \frac{1}{2} \sum_{n=1}^\infty 2\pi^2 n^4 I_{t,\theta}(x-(9-y)i, \pi n^2) - 3\pi n^2 I_{t,\theta}(x-(5-y)i, \pi n^2) + 2\pi^2 n^4 \overline{I_{t,\theta}(x-(9+y)i, \pi n^2)} - 3\pi n^2 \overline{I_{t,\theta}(x-(5+y)i, \pi n^2)} \quad (0)[/math]

and

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

where

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

and

[math]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]I_{t,\theta}(b,\beta), J_{t,\theta}(b,\beta)[/math] when [math]\beta = \pi n^2[/math] for some [math]n[/math], and [math]b = x-ia[/math] for [math]a = 9+y, 9-y, 5+y, 5-y[/math].

If [math]x,y\gt0[/math], we expect the dominant term here to be [math]2\pi^2 n^4 \overline{J_{t,\theta}(x-(9+y)i, \pi n^2)}[/math] (these eventually correspond to the "[math]B[/math]" terms in the [math]H_t[/math] expansion, with the [math]n=1[/math] case corresponding to the "[math]B_0[/math]" term). For [math]b = \overline{z}-9i = x - (y+9)i[/math] and [math]\beta = \pi n^2[/math], the phase [math]-\beta e^{4u} + ibu[/math] has a stationary point at

[math]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]\theta[/math] to equal

[math] \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]H'_t(z)[/math] for [math]x[/math] in some interval, one should probably pick a single [math]\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]\theta[/math] to evaluate all the terms in the above sum (e.g. one needs the same [math]\theta[/math] to handle [math]b = z-9i, \overline{z}-9i, z-5i, \overline{z}-5i[/math]).

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

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

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

[math]I_{t,\theta,\leq 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]u = i\theta + X + v[/math] and setting [math]b = x - i a[/math] (where in practice [math]a = 9+y, 9-y, 5+y, 5-y[/math]), the tail can be written as

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

and hence by the triangle inequality

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

(Note: the [math]\exp(-\theta x)[/math] factor here gives the main decay factor [math]e^{-\pi x/8}[/math] for [math]H_t[/math] once one chooses [math]\theta[/math] close to [math]\pi/8[/math]. But one cannot choose it too close, otherwise the [math]\cos(4\theta)[/math] term becomes too small.) The exponent [math]tv^2 + 2tXv - \beta e^{4X} (e^{4v}-1) \cos(4\theta) + av[/math] equals [math]0[/math] when [math]v=0[/math] and has derivative

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

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

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

then this derivative will be bounded from below by the negative quantity [math]-(4 \beta e^{4X} \cos(4\theta) - a - 2tX)[/math]. This implies that

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

thus giving the tail bound

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

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

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

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

then

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

For instance, if

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

then

[math] |\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(4\theta) - \theta x )}{4 \pi n^2 \cos(4\theta) - a} [/math]
[math] \leq \frac{2\pi^2 \exp(-t \theta^2 - \theta x)}{4\pi n_0^2 \cos(4\theta) - a- 2tX} \sum_{n \geq n_0} n^4 \exp( - \pi n n_0 \cos(4\theta)) [/math]
[math] = \frac{2\pi^2 \exp(-t \theta^2 - \theta x - \pi n_0^2 \cos(4\theta))}{4\pi n_0^2 \cos(4\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]= \frac{2\pi^2 \exp(-t \theta^2 - \theta x - \pi n_0^2 \cos(4\theta))}{4\pi n_0^2 \cos(4\theta) - a} \times[/math]
[math] \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} ) \quad (4)[/math]

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

[math] |\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(4\theta))}{4\pi n_0^2 \cos(4\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 we can again split

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

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

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

and

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

From the triangle inequality as before we have

[math]|J_{t,\theta,\leq X}(b,\beta)| \leq \exp( -t \theta^2 - \theta x ) \int_0^X \exp( tu^2 - \beta e^{4u} \cos \theta + au ) (\theta + u)\ du[/math]

and

[math]|J_{t,\theta,\geq X}(b,\beta)| \leq \exp( -t \theta^2 + tX^2 - \beta e^{4X} \cos(4\theta) - \theta x + aX ) \int_0^\infty \exp( tv^2 + 2tXv - \beta e^{4X} (e^{4v}-1) \cos \theta + av ) (|X+i\theta| + v)\ dv. \quad (5)[/math]

[Side remark: one nice feature of these bound is that the dependence on [math]x[/math] becomes very simple if we do not let [math]\theta[/math] or [math]X[/math] vary with [math]x[/math]. This could be important for obtaining bounds on [math]|H'_t(x+iy)|[/math] that are uniform for [math]x[/math] in an interval.] If (2) holds then again we have

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

and hence

[math]|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- 2tX} + \frac{1}{4 \beta e^{4X} \cos(4\theta) - a- 2tX)^2}) \quad (6) [/math]

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

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

then

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

and by arguing as before

[math] |\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(4\theta) - \theta x ) ( \frac{\theta}{4 \pi n_0^2 \cos(4\theta) - a} + \frac{1}{(4 \pi n_0^2 \cos \theta - a)^2} )[/math]
[math] \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] |\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(4\theta) - \theta x ) ( \frac{\theta}{4 \pi n_0^2 \cos(4\theta) - a} + \frac{1}{(4 \pi n_0^2 \cos(4\theta) - a)^2} )[/math]
[math] \times ( \frac{n_0^2}{1-\alpha} + \frac{2n_0 \alpha}{(1-\alpha)^2} + \frac{\alpha^2+\alpha}{(1-\alpha)^3} ).[/math]

This controls the tail of the sum in (1). The main terms can be controlled by (5) and (6) for a suitable choice of cutoff [math]X[/math]. After fixing [math]n_0, X, \theta[/math], the only [math]x[/math] dependence in these terms is a factor of [math]e^{-\theta x}[/math], so one gets uniform estimates for any [math]x \geq x_0[/math].