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I am trying to find the Fourier transform $F(k)$ of the function $f(x) = \text{sech}^4(x)$. I put it into wolfram alpha (result), and we get that the transform is:

$$F(k) = \frac{1}{6} \sqrt{\frac{\pi}{2}}k(k^2+4) \text{csch}(\pi k/2)$$

Note that when $k = 0$ we get a problem. However in wolframs graph, there is not a problem, that is there is a value for $k=0$ (note that in wolfram they use $\omega$ instead of $k$):

enter image description here

How do we know what the value should be at $k = 0$?

Note the limit as $k\rightarrow 0$ in the expression for $F(k)$ is given as $\frac{2}{3}\sqrt{\frac{2}{\pi}}$

fosho
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    Note that you have a factor proportional to $k$ in the result, so you can look at the limiting behavior as $k\to 0$ using your favorite method of evaluating limits. The most straightforward is to probably plug in the first three terms for the Taylor series of $e^y$ and $e^{-y}$ in the definition of $\mathrm{csch}(y)$ and then evaluate. – Steven Stadnicki May 25 '17 at 19:46
  • So we have just plugged in the value of the limit at $k = 0$? – fosho May 25 '17 at 19:47
  • Something like that (or at least, Wolfram has). Since your function has a well-defined limit as $k\to 0$, Wolfram is presumably happy to define its value there as being that limit for purposes of plotting. – Steven Stadnicki May 25 '17 at 19:48
  • But if we write $F(k) = \frac{1}{2\pi} \int \text{sech}^4(x) e^{ikx} dx$ and evaluate at $k = 0$, we get just the integral across the reals line of $\text{sech}^4$ which should be $\frac{1}{2\pi} \frac{1}{4}$, no? This is much smaller than what wolfram is putting there. – fosho May 25 '17 at 19:51
  • I mean $\frac{4}{3} \cdot \frac{1}{2\pi}$ – fosho May 25 '17 at 20:03
  • There are different definitions of the Fourier transform; the one WA is using is this one. The key point is that the factor of $2\pi$ is in the exponent, not overall. Hence $4/3$ is correct for the definition. – Semiclassical May 25 '17 at 20:42
  • is it possible to recover the formula using my definition? – fosho May 25 '17 at 21:00
  • Alpha normalizes the Fourier transform with $1/\sqrt{2\pi},$ not $1/(2\pi),$ giving $4/3/\sqrt{2\pi}\approx 0.532$ for the height of the curve at $k=0.$ – whuber Apr 30 '21 at 17:47

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