Inverse fourier transform of sech(mt/2)^(1/m)

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July 22nd 2010, 08:23 AM
Ackbeet

Are you sure this step:

$B(t+r,t-r)=\diplaystyle\frac{1}{2} \int_0^\infty \frac{x^r + x^{-r}}{(x+1)^{2t}x^{-t+1}}dx$ to
$B(t+r,t-r)=\diplaystyle \int_0^\infty \frac{x^r + x^{-r}}{(x^{\frac{1}{2}}+x^{-\frac{1}{2}})^{2t}}\frac{1}{2x}dx$ is correct? You skipped a lot of steps there. I'm not sure I followed them all.

Just double-check your work, is all.

The idea is very elegant. Nice work!
July 22nd 2010, 11:29 AM
shawsend

Quote:

Originally Posted by AlexisM

Hi,

could someone help me find a closed form for the inverse fourier transform of sech(mw/2)^(1/m)

Namely 1/(2*pi)\int_{-\infty}^{\infty} exp(itw) sech(mw/2)^(1/m) dw

sech is the hyperbolic secant :

sech(x) = 2/(exp(x) + exp(-x))

and m>0

Thanks for you help

AlexisM

If m is an integer, Mathematica gives a closed-form solution:

InverseFourierTransform[Power[Sech[n z/2], 1/n], z, t,
Assumptions -> Element[n, Integers]]
July 22nd 2010, 11:34 AM
Ackbeet

Which version of MMA do you have? Mine doesn't compute that (it's version 4.0). Neither does WolframAlpha.
July 22nd 2010, 11:53 AM
shawsend

I have ver. 7.0. Seems I don't even need to specify the assumption. The command:

InverseFourierTransform[Sech[m*(z/2)]^
(1/m), z, t]

gives:

$ \begin{aligned} &\frac{1}{\sqrt{\pi } \left(1+4 t^2\right)}2^{\frac{1}{2}+\frac{1}{m}}\\ &\left((1+2 i t) _2F_1\left[\frac{1}{m},\frac{1}{2 m}-\frac{i t}{m},1+\frac{1}{2 m}-\frac{i t}{m},-1\right]\\ &+\left.(1-2 i t)_2F_1\left[\frac{1}{m},\frac{1}{2 m}+\frac{i t}{m},1+\frac{1}{2 m}+\frac{i t}{m},-1\right]\right) \end{aligned} $
July 22nd 2010, 11:56 AM
Ackbeet

Ah, well. Those special functions, including the beta functions, are all related to the confluent hypergeometric functions. It's probably equivalent, through an amazing amount of difficult analysis. It even has similar arguments to the functions as what AlexisM has already computed.
July 22nd 2010, 10:24 PM
AlexisM

$x^{-t+1} = x^{-t}x = x^{-2t/2}x$
so I separate the lonely x which give me the 1/x and 1/2x including the 1/2 which is in factor.
and then I have $(x+1)^{2t}x^{-2t/2} = ((x+1)x^{-1/2})^{2t} = (x^{1/2}+x^{-1/2})^{2t}$
July 22nd 2010, 10:29 PM
AlexisM

Quote:

Originally Posted by shawsend

I have ver. 7.0. Seems I don't even need to specify the assumption. The command:

InverseFourierTransform[Sech[m*(z/2)]^
(1/m), z, t]

gives:

$ \begin{aligned} &\frac{1}{\sqrt{\pi } \left(1+4 t^2\right)}2^{\frac{1}{2}+\frac{1}{m}}\\ &\left((1+2 i t) _2F_1\left[\frac{1}{m},\frac{1}{2 m}-\frac{i t}{m},1+\frac{1}{2 m}-\frac{i t}{m},-1\right]\\ &+\left.(1-2 i t)_2F_1\left[\frac{1}{m},\frac{1}{2 m}+\frac{i t}{m},1+\frac{1}{2 m}+\frac{i t}{m},-1\right]\right) \end{aligned} $

When you set m=1, mathematica still gives a complicated formula whereas the FT of sech(x) has a known closed form $\text{sech}(\pi x)$
July 23rd 2010, 01:36 AM
Ackbeet

Reply to AlexisM at post # 21:

Ah, got it. Looks good!

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