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Aug
13
comment Fourier transform of $\Gamma \left (\frac{1}{2}-i \frac{p}{2 \pi} \right) /\sqrt{ \cosh(p/2)}$
@RandomVariable This does not spoil tired's asymptotics though (he's essentially taking only one branching point contribution at $i \pi$).
Aug
13
accepted Intriguing polynomials coming from a combinatorial physics problem
Aug
13
comment Fourier transform of $\Gamma \left (\frac{1}{2}-i \frac{p}{2 \pi} \right) /\sqrt{ \cosh(p/2)}$
I see what you did, excellent!! Simultaneously with your post, I have update the plot which confirms your asymptotics beautifully ($I(x)$ defined by @Leucippus equals $\psi(x)/\sqrt{\pi}$).
Aug
13
revised Fourier transform of $\Gamma \left (\frac{1}{2}-i \frac{p}{2 \pi} \right) /\sqrt{ \cosh(p/2)}$
graphs confirming the asymptotics
Aug
13
comment Fourier transform of $\Gamma \left (\frac{1}{2}-i \frac{p}{2 \pi} \right) /\sqrt{ \cosh(p/2)}$
@tired No, I am not sure! I was checking a single exponential. For positive $x$ your $\exp[ - \pi x]/\sqrt{x}$ fits much better (less than 0.1% difference for x>5). For negative $x$, adding $\sqrt{|x|}$ makes the agreement worse. Would you like me to update the plot?
Aug
12
reviewed No Action Needed If $D_1$ and $D_2$ are derivations of $A$, show that $D_1 ◦ D_2$ is not necessarily a derivation of $A$
Aug
12
reviewed No Action Needed How exactly do you measure circumference or diameter?
Aug
12
comment Master equation to Fokker Planck equation
I do not think so - I think one can construct a transition matrix with sufficiently non-local jumps that would defy Fokker-Plank equation. What kind of vairable/transition matrix do you have in mind?
Aug
12
answered Master equation to Fokker Planck equation
Aug
12
comment Fourier transform of $\Gamma \left (\frac{1}{2}-i \frac{p}{2 \pi} \right) /\sqrt{ \cosh(p/2)}$
@tired I would very much appreciate your effort! I have edited my post to explain more the motivation and potential valuable results, even if they fall short of an explicit analytic formula for $\psi(x)$.
Aug
12
revised Fourier transform of $\Gamma \left (\frac{1}{2}-i \frac{p}{2 \pi} \right) /\sqrt{ \cosh(p/2)}$
extendend motivation for the at least partial progress on the problem
Aug
11
revised Fourier transform of $\Gamma \left (\frac{1}{2}-i \frac{p}{2 \pi} \right) /\sqrt{ \cosh(p/2)}$
added 309 characters in body
Aug
11
comment Fourier transform of $\Gamma \left (\frac{1}{2}-i \frac{p}{2 \pi} \right) /\sqrt{ \cosh(p/2)}$
Aha, and since the Gamma function is never zero, I can close the contour either in the upper or the lower half plane of $z$, depending on the sign of $x$, right? But then I think the poles are not in general limited to $m>0$
Aug
11
comment Fourier transform of $\Gamma \left (\frac{1}{2}-i \frac{p}{2 \pi} \right) /\sqrt{ \cosh(p/2)}$
Where does your second condition for the poles (with $z_m$) come from?
Aug
11
revised Fourier transform of $\Gamma \left (\frac{1}{2}-i \frac{p}{2 \pi} \right) /\sqrt{ \cosh(p/2)}$
small technical correction
Aug
11
revised Fourier transform of $\Gamma \left (\frac{1}{2}-i \frac{p}{2 \pi} \right) /\sqrt{ \cosh(p/2)}$
added 111 characters in body
Aug
11
comment Fourier transform of $\Gamma \left (\frac{1}{2}-i \frac{p}{2 \pi} \right) /\sqrt{ \cosh(p/2)}$
@Alizter I am reasonably well-versed in Mathematica, Integrate and FourierTransform return the integral unevaluated.
Aug
11
comment Fourier transform of $\Gamma \left (\frac{1}{2}-i \frac{p}{2 \pi} \right) /\sqrt{ \cosh(p/2)}$
Shaping of on-demand electron wave-packets, here is an abstract I have presented on a conference few weeks ago: fqmt.fzu.cz/15/func/viewpdf.php?reg=650&num=1
Aug
11
asked Fourier transform of $\Gamma \left (\frac{1}{2}-i \frac{p}{2 \pi} \right) /\sqrt{ \cosh(p/2)}$
Dec
20
awarded  Constituent