Page 2 of 2 FirstFirst 12
Results 16 to 23 of 23

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

  1. #16
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    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!
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Quote Originally Posted by AlexisM View Post
    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]]
    Follow Math Help Forum on Facebook and Google+

  3. #18
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Which version of MMA do you have? Mine doesn't compute that (it's version 4.0). Neither does WolframAlpha.
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Super Member
    Joined
    Aug 2008
    Posts
    903
    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:

    <br />
\begin{aligned}<br />
&\frac{1}{\sqrt{\pi } \left(1+4 t^2\right)}2^{\frac{1}{2}+\frac{1}{m}}\\<br />
&\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]\\<br />
&+\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)<br />
\end{aligned}<br />
    Follow Math Help Forum on Facebook and Google+

  5. #20
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #21
    Newbie
    Joined
    Apr 2009
    Posts
    21
    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}
    Follow Math Help Forum on Facebook and Google+

  7. #22
    Newbie
    Joined
    Apr 2009
    Posts
    21
    Quote Originally Posted by shawsend View Post
    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:

    <br />
\begin{aligned}<br />
&\frac{1}{\sqrt{\pi } \left(1+4 t^2\right)}2^{\frac{1}{2}+\frac{1}{m}}\\<br />
&\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]\\<br />
&+\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)<br />
\end{aligned}<br />
    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)
    Follow Math Help Forum on Facebook and Google+

  8. #23
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Reply to AlexisM at post # 21:

    Ah, got it. Looks good!
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Inverse fourier Transform of 1/(1+.5jw)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 21st 2011, 10:30 AM
  2. inverse Fourier transform
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: May 27th 2011, 08:23 AM
  3. Inverse fourier transform
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: May 15th 2010, 11:19 PM
  4. Inverse Fourier Transform
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: November 12th 2009, 04:09 AM
  5. Inverse Fourier Transform
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 6th 2008, 09:51 PM

Search Tags


/mathhelpforum @mathhelpforum