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I am looking for a method to calculate the Fourier transform of the function $\frac{1}{\cosh x}$. There're already two wonderful answers on Math.SE here dealing with the same problem, but both methods require the residue theorem.

The final answer is (up to some irrelevant constants) $$F\left[\frac {1}{\cosh x}\right](\xi)=\frac{1}{\cosh \xi}.$$

I wonder if there exists a method which does not involve the residues or any other topic beyond first two years of university mathematics.

Any help will be greatly appreciated.

Community
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TZakrevskiy
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  • Have you tried writing $$f(x) = \frac{1}{\cosh(x)} = sech(x) \implies \hat f(\xi) = \int_{-\infty}^{\infty} sech(x)\exp(-ix\xi)dx$$ and then converting $sech$ into exponential form? – Matthew Cassell Jan 06 '15 at 17:36
  • @Mattos of course, but I still haven't find anything useful yet from this approach (apart obvious passage to residues). – TZakrevskiy Jan 06 '15 at 17:38
  • It is too late here for me to help you, but you can do it by Fourier Series instead if you would like to try that.. See here for help on it https://www.physicsforums.com/threads/fourier-transform-of-the-hyperbolic-secant-function.662698/ – Matthew Cassell Jan 06 '15 at 17:49
  • Also, note the FT of $sech(\alpha x)$ is $\frac{1}{\alpha}\sqrt {\frac{\pi}{2}} sech\bigg(\frac{k\pi}{2\alpha}\bigg)$ – Matthew Cassell Jan 06 '15 at 17:53
  • @Mattos thank, I'll take a look at that link.\ – TZakrevskiy Jan 06 '15 at 18:52

1 Answers1

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It is kind of a hack, but if we have already guessed what the Fourier transform should be, it is enough to prove that $f(x)=\frac{1}{\cosh x}$ is an eigenfunction for the differential operator:

$$ \mathcal{F}: g(x) \to \widehat{g}(t)=\frac{1}{\pi}\int_{-\infty}^{+\infty}g(x)\cos\frac{2tx}{\pi}\,dx $$ defined over the Schwarz space $\mathfrak{S}$, with respect to the eigenvalue $\lambda=1$.

Since: $$\frac{1}{\cosh z} = \sum_{n=0}^{+\infty}\frac{(-1)^n (2n+1)\pi}{(2n+1)^2\pi^2+4z^2}\tag{1}$$ it follows that: $$\mathcal{F}\left(\frac{1}{\cosh x}\right)=\sum_{n=0}^{+\infty}2(-1)^n e^{-(2n+1)|t|}=\frac{1}{\cosh t},\tag{2}$$ as wanted. However, usual proofs of $(1)$ make use of the Weierstrass product for the $\cosh$ function, or some form of the residue theorem. So the problem boils down to proving $(1)$ without complex techniques: the Laplace transform is a chance, but I bet it is possible to find also other real-analytic proofs of $(1)$ on this site.

Jack D'Aurizio
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  • In fact, one can simply rewrite the integral $\int_{\Bbb R}\frac{\exp{-i\xi x}}{\cosh x}dx$ using the geometric series in $e^{-2x}$, then use Tonelli to change the order of integration. After some manipulations I arrived to the series (up to some constants) (1). And here I faced the same problem: how to prove that this series indeed represents $ \operatorname{sech} x$=) – TZakrevskiy Jan 06 '15 at 21:11