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A Fourier transform of hyperbolic tangent

  1. Oct 5, 2016 #1

    jjr

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    Hello

    I am trying to determine the fourier transform of the hyperbolic tangent function. I don't have a lot of experience with Fourier transforms and after searching for a bit I've come up empty handed on this specific issue.

    So what I want to calculate is:

    ##\int\limits_{-\infty}^\infty e^{-it\omega}\text{tanh}(bt) dt##

    where ##b## is some constant.

    Using ##\text{tanh}(bt)=\frac{e^{bt}-e^{-bt}}{e^{bt}+e^{-bt}}## leads to a mess of exponential functions, and does not bring me closer to a solution. Perhaps there is some other way, using tricks specific to calculating Fourier transforms that could be helpful here?

    Any suggestions are most appreciated

    J
     
  2. jcsd
  3. Oct 5, 2016 #2

    blue_leaf77

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    I don't think the integral converges.
     
  4. Oct 5, 2016 #3

    jjr

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    Sorry, I should of course have added that running it through WolframAlpha returns

    ##\mathcal{F}_t[\text{tanh}(t)](\omega) = i\sqrt{\frac{\pi}{2}}\text{csch}\left(\frac{\pi\omega}{2}\right)##

    which I'm hoping to obtain analytically
     
  5. Oct 5, 2016 #4

    jasonRF

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    This is an interesting problem. The original integral does not exist as a classical Riemann integral, but the Fourier transform exists, at least as a generalized function (ie, in the same way that ##\mathcal{F}\left[ 1 \right](\omega) = 2\pi\delta(\omega)##, or ##\mathcal{F}\left[ sgn(t)\right](\omega)=2 \, \mathcal{P} \frac{1}{\omega}##, where sgn(t) = 1 for t>0 and -1 for t<0, and ##\mathcal{P}## refers to the principle part).

    One way to proceed may be to use the fact that ##\frac{d}{dt}\tanh(t) = \text{sech}^2(t)##, so that
    ##i\omega\mathcal{F}\left[ \tanh(t)\right](\omega) = \mathcal{F}\left[ \text{sech}^2(t)\right](\omega)##. Now you have an integral, ##\mathcal{F}\left[ \text{sech}^2(t)\right](\omega)## that is nicely behaved as in improper Riemann integral, and you may be able apply contour integration or some other technique.

    Note that in the above I have been assuming the definition ##\mathcal{F}[f(t)](\omega) = \int_{-\infty}^\infty dt\, e^{-i\omega t} \, f(t)##. I think Wolfram has the ##2\pi## in the exponent, so if you just ask for the fourier transform the constants won't be the same. So while the result from Wolframalpha may have constants that are not quite right for your definition of the Fourier transform, qualitatively the result seems plausible. This is because ## \text{csch}(\omega)## qualitatively looks like ##\frac{1}{\omega}## (especially near zero, which is what matters when b is large), and as ##b\rightarrow \infty##, ##\tanh(b t)## seems to "converge" (in the sense of distributions) to ##sgn(t)##.

    Jason
     
    Last edited: Oct 5, 2016
  6. Oct 6, 2016 #5

    jasonRF

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    Update: using contour integration I find
    ##\mathcal{F}\left[ \text{sech}^2(t)\right](\omega) = \int_{-\infty}^\infty dt\, e^{-i\omega t}\, \text{sech}^2(t) = \pi \omega \, \text{csch}\left(\frac{\pi \omega}{2} \right)##

    What do you get?

    Jason

    edit: by the way, I think (although I am not sure) that this means that the answer to your original problem is,
    ##\mathcal{F}\left[ \text{tanh}(t)\right](\omega) = -i \, \pi \, \mathcal{P}\, \text{csch}\left(\frac{\pi \omega}{2} \right)##.

    If you aren't worried about specifying how to treat the singularity at ##\omega=0## then you can get rid of the principle part and use
    ##\mathcal{F}\left[ \text{tanh}(t)\right](\omega) = -i \, \pi \, \text{csch}\left(\frac{\pi \omega}{2} \right)##
     
    Last edited: Oct 6, 2016
  7. Oct 7, 2016 #6

    jjr

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    I ended up going a different route. I was working with the ##\text{tanh}(t)## function because I needed something that closely resembles the error function ##\text{erf}(t)=\frac{1}{\sqrt{\pi}}\int_{-t}^t e^{-t'^2}dt'##, for which I initially couldn't find the F.T. I managed to calculate it after all, and so didn't need the F.T. of ##\text{tanh}(t)##.

    Thanks for your help. Since this seems to be a rather unique question hopefully the thread can help someone else at some point.

    J
     
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