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

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• Jul 22nd 2010, 09:23 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!
• Jul 22nd 2010, 12:29 PM
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]]
• Jul 22nd 2010, 12:34 PM
Ackbeet
Which version of MMA do you have? Mine doesn't compute that (it's version 4.0). Neither does WolframAlpha.
• Jul 22nd 2010, 12:53 PM
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}
• Jul 22nd 2010, 12:56 PM
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.
• Jul 22nd 2010, 11:24 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 $(x+1)^{2t}x^{-2t/2} = ((x+1)x^{-1/2})^{2t} = (x^{1/2}+x^{-1/2})^{2t}$
• Jul 22nd 2010, 11:29 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 $\text{sech}(\pi x)$
• Jul 23rd 2010, 02:36 AM
Ackbeet
Reply to AlexisM at post # 21:

Ah, got it. Looks good!
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