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Math Help - Fourier transform of a hyperbolic secant

  1. #1
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    Fourier transform of a hyperbolic secant

    Hello.

    I'm trying to manually calculate the spectrum of a hyperbolic secant function. It's well known that
    \int_{-\infty }^{\infty } \text{sech}  (t \alpha) \exp (- i \omega t)d t = \frac{\pi}{\alpha} \text{sech} \left(\frac{\pi^2}{\alpha} \omega\right).
    How exactly is this expression obtained? All of my atempts to integrate this equation were unsuccessful. What measures should be taken in this particular case?
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  2. #2
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    I've never done that, but may result if you integrate by parts two times putting du=e^{-iwt} and then use hyperbolic identities.
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  3. #3
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    Quote Originally Posted by Nouvelle Vague View Post
    Hello.

    I'm trying to manually calculate the spectrum of a hyperbolic secant function. It's well known that
    \int_{-\infty }^{\infty } \text{sech}  (t \alpha) \exp (- i \omega t)d t = \frac{\pi}{\alpha} \text{sech} \left(\frac{\pi^2}{\alpha} \omega\right).
    How exactly is this expression obtained? All of my atempts to integrate this equation were unsuccessful. What measures should be taken in this particular case?
    Hi. I think you mean:

    \int_{-\infty}^{\infty} \text{sech}(\alpha x) e^{-2\pi i x \omega} dx=\frac{\pi}{\alpha}\text{sech}\left(\frac{\pi^2 \omega}{\alpha}\right)

    which we can evaluate via the Residue Theorem. Consider the contour integral:

    \mathop\oint\limits_{D_u} \text{sech}(\alpha z)e^{-2\pi i z\omega}dz

    where D_u is the path along the real-axis, then circling back in a large upper half-circle. As the radius of this contour approaches infinity, it encloses all the poles of the integrand in the upper half-plane at \frac{(2n+1)\pi i}{2\alpha}.

    Now, the integral tends to zero along the circular half-circle as the radius tends to infinity and we're left with just the line segment along the real axis or:

    \int_{-\infty}^{\infty} \text{sech}(\alpha z)e^{-2\pi i z\omega}dz=-\frac{2 \pi }{\alpha }\sum _{n=0}^{\infty } (-1)^{n+1}\text{Exp}\left[\frac{\pi ^2 \omega }{\alpha }(2n+1)\right]=\frac{\pi}{\alpha}\text{sech}\left(\frac{\pi^2 \omega}{\alpha}\right)
    Last edited by shawsend; January 5th 2010 at 03:04 PM. Reason: corrected expression for poles
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  4. #4
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    last step...?

    I can get to the alternating exponential series in your last post (albeit in a different way by calculating the residues by differentiating the denominator of a rational function and evaluating at the poles). However, I can't seem to follow your final step. The Taylor series expansion of a sech function, while alternating as this one does, has the Euler numbers as coefficients and I don't see how they are equivalent. Thanks in advance.
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  5. #5
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    Nevermind, I figured it out--its just a simple infinite geometric series with the common ratio r = -exp[b*Pi/a] so that it goes to (coeff)*series = (coeff)/(1-r) = (coeff)/(1+exp[b*Pi/a]) and then rearrange to get in canonical form for cosh as sum of exponentials in the denominator. Thanks anyway.
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