Fourier transform of $\operatorname{sech}^2(x)$

Asked Apr 01 '21 at 08:35

Active Apr 01 '21 at 08:38

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I'm trying to derive the Fourier transform of various hyperbolic functions and I don't see how to derive the transform of $\operatorname{sech}^2(x)$. I got as far as decomposing the function into exponentials but that's it.

$$\operatorname{sech}^2(x) = \dfrac{4}{\left(\mathrm{e}^x+\mathrm{e}^{-x}\right)^2} $$

edited Apr 01 '21 at 08:38

user10354138

33,887

asked Apr 01 '21 at 08:35

Dimitri_896

Since sech is (up to silly constants) an eigenfunction of Fourier transform, just take convolution square. – user10354138 Apr 01 '21 at 08:45
How do I do that? – Dimitri_896 Apr 01 '21 at 08:55
See this answer. – user10354138 Apr 01 '21 at 09:37

Fourier transform of $\operatorname{sech}^2(x)$

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