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My question is what is the Fourier transform of the function

$f(t)=\exp(t/2)\text{sech}(t)$

Is there any closed form available? One possibility is that

$f(t)^2=\frac{1}{2} \text{sech}(t) + \frac{1}{4} (\dfrac{d}{dt}\operatorname{sech}t)$

and then the Fourier transform at frequency f is

$F(f(t)^2)=\frac{1}{2} F(\operatorname{sech}t)(1 + j\pi f )$

and $F(\operatorname{sech}t)=\pi \operatorname{sech}(\pi^2 f)$. But I am interested in $F(f(t))$ itself.

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  • $\begingroup$ CAS says: $\mathcal{F}_t\left[\exp \left(\frac{t}{2}\right) \text{sech}(t)\right](\omega )=\frac{1}{4} \pi \left(\cot \left(\frac{1}{8} \pi (1+2 i \omega )\right)+\cot \left(\frac{1}{8} \pi (3+2 i \omega )\right)-4 i \sqrt{2} \text{sech}(\pi \omega ) \sinh \left(\frac{\pi \omega }{2}\right)+\tan \left(\frac{1}{8} \pi (1+2 i \omega )\right)+\tan \left(\frac{1}{8} \pi (3+2 i \omega )\right)\right)$ $\endgroup$ Mar 15, 2018 at 10:53

4 Answers 4

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A very simple solution may be had using the residue theorem.

Consider the contour integral

$$\oint_C dz \, \frac{e^{z/2}}{\cosh{z}} e^{i k z} $$

where $C$ is the rectangle having vertices at the points $-R$, $R$, $R+i \pi$, and $-R+i \pi$. Then as $R \to \infty$ (where the integrals over the vertical sides vanish), the contour integral is equal to

$$\left (1+ i \, e^{-\pi k} \right ) \int_{-\infty}^{\infty} dx \, \frac{e^{x/2}}{\cosh{x}} e^{i k x} $$

(NB the relation $\cosh{(x+i \pi)} = -\cosh{x}$ was used.)

The contour integral is also equal to $i 2 \pi$ times the residue at the pole of the integrand of the contour integral, or at $z=i \pi/2$. This residue is equal to $-i e^{i \pi/4} e^{-\pi k/2}$. Accordingly, after a bit of algebra, we have the following result:

$$\int_{-\infty}^{\infty} dx \, \frac{e^{x/2}}{\cosh{x}} e^{i k x} = \sqrt{2} \pi \operatorname{sech}{(\pi k)} \left [\cosh{\left ( \frac{\pi k}{2} \right )} + i \sinh{\left ( \frac{\pi k}{2} \right )} \right ]$$

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    $\begingroup$ Every time I read one of your answer I recognise how much I still have to learn from you, and in general. +1 my dear. $\endgroup$
    – Laplacian
    Mar 16, 2018 at 8:19
  • $\begingroup$ Dear Ron, I admire your expertise in complex analysis and integration, so maybe you can help me with my question "Elusive Laplace transform" about the inverse LT of sinh{ks½}/s^½. If you don't want to go public, my email is jchoelgaard@yahoo.com. Thanks in advance. $\endgroup$
    – Jens
    Mar 21, 2018 at 10:33
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The best way to attack the problem is to use the definition of the hyperbolic secant, together with the Geometric Series.

$$\text{sech}(t) = \frac{2}{e^t + e^{-t}}$$

We then have:

$$\mathcal{F}\left(e^{t/2}\text{sech}(t)\right)(k) = \frac{2}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{t/2} e^{-i kt }\frac{\text{d}t}{e^t + e^{-t}}$$

Now we have to split the integral into the two ranges $(-\infty, 0]\cup[0, +\infty)$ in order to use geometric series.

Geometric series will be used for the denominator as follows:

$$\frac{1}{e^t + e^{-t}} = \frac{1}{e^t(1 + e^{-2t})} = \frac{e^{-t}}{1 + e^{-2t}} = e^{-t}\sum_{j = 0}^{+\infty}(-1)^j e^{-2jt} ~~~~~~~ \text{for} ~~ [0, +\infty)$$

$$\frac{1}{e^t + e^{-t}} = \frac{1}{e^{-t}(1 + e^{2t})} = \frac{e^{t}}{1 + e^{2t}} = e^{t}\sum_{j = 0}^{+\infty}(-1)^j e^{2jt} ~~~~~~~ \text{for} ~~ (-\infty, 0]$$

Hence the two integrals we need to calculate are:

I

$$\frac{2}{\sqrt{2\pi}}\sum_{j = 0}^{+\infty}(-1)^j\int_0^{+\infty} e^{t/2} e^{-ikt} e^{-t}e^{-2jt}\ \text{d}t = \frac{2}{\sqrt{2\pi}}\sum_{j = 0}^{+\infty}(-1)^j \frac{2}{1 + 4j + 2ik}$$

Provided that $4 \Re(j)+1>2 \Im(k)$ which should be the case in question.

The sum can be obtained by some knowledge of Special Functions, and it sums:

$$\frac{4}{\sqrt{2\pi}}\sum_{j = 0}^{+\infty}(-1)^j \frac{1}{1 + 4j + 2ik} = \frac{\Phi \left(-1,1,\frac{1}{4} (2 i k+1)\right)}{\sqrt{2 \pi }}$$

The Hurwitz function that obtained can be expanded into "more elementary" functions, like the Polygamma Function:

$$\frac{\Phi \left(-1,1,\frac{1}{4} (2 i k+1)\right)}{\sqrt{2 \pi }} \equiv \frac{\psi ^{(0)}\left(\frac{i k}{4}+\frac{5}{8}\right)-\psi ^{(0)}\left(\frac{i k}{4}+\frac{1}{8}\right)}{2 \sqrt{2 \pi }}$$

Here $\Phi$ denotes the Hurwitz function and $\Psi$ denotes the Polygamma.

II

Part two is similar to the first one, except for the calculations. In any case, similarly, we get:

$$\frac{2}{\sqrt{2\pi}}\sum_{j = 0}^{+\infty}(-1)^j\int_0^{+\infty} e^{t/2} e^{-ikt} e^{t}e^{2jt}\ \text{d}t = \frac{2}{\sqrt{2\pi}}\sum_{j = 0}^{+\infty}(-1)^j \frac{2}{4 j-2 i k+3}$$

Provided that $\Im(k)+2 \Re(j)>-\frac{3}{2}$ which again holds.

The sum can be evaluated in the same manner:

$$\frac{4}{\sqrt{2\pi}}\sum_{j = 0}^{+\infty}(-1)^j \frac{1}{4 j-2 i k+3} = \frac{\Phi \left(-1,1,\frac{1}{4} (3-2 i k)\right)}{\sqrt{2 \pi }}$$

And again:

$$\frac{\Phi \left(-1,1,\frac{1}{4} (3-2 i k)\right)}{\sqrt{2 \pi }} \equiv \frac{\psi ^{(0)}\left(\frac{7}{8}-\frac{i k}{4}\right)-\psi ^{(0)}\left(\frac{3}{8}-\frac{i k}{4}\right)}{2 \sqrt{2 \pi }}$$

The final result is thence the sum of the two previous results, which means:

Final Result

$$\mathcal{F}\left(e^{t/2}\text{sech}(t)\right)(k) = \frac{\psi ^{(0)}\left(\frac{i k}{4}+\frac{5}{8}\right)-\psi ^{(0)}\left(\frac{i k}{4}+\frac{1}{8}\right)}{2 \sqrt{2 \pi }} + \frac{\psi ^{(0)}\left(\frac{7}{8}-\frac{i k}{4}\right)-\psi ^{(0)}\left(\frac{3}{8}-\frac{i k}{4}\right)}{2 \sqrt{2 \pi }} = $$ $$ = \frac{-\psi ^{(0)}\left(\frac{i k}{4}+\frac{1}{8}\right)+\psi ^{(0)}\left(\frac{i k}{4}+\frac{5}{8}\right)-\psi ^{(0)}\left(\frac{3}{8}-\frac{i k}{4}\right)+\psi ^{(0)}\left(\frac{7}{8}-\frac{i k}{4}\right)}{2 \sqrt{2 \pi }}$$

FINAL BEAUTY

Through definitions of the Polygamma, and few algebra, we can transform that pretty ugly expression into something cute and simple. The "true" final result is then:

$$\color{red}{\frac{\pi \tan \left(\frac{\pi i k}{4}+\frac{\pi }{8}\right)+\pi \cot \left(\frac{\pi i k}{4}+\frac{\pi }{8}\right)}{2 \sqrt{2 \pi }}}$$

Which can be written also as

$$\color{blue}{\frac{\pi \tan \left(\frac{1}{8} (2 \pi i k+\pi )\right)+\pi \cot \left(\frac{1}{8} (2 \pi i k+\pi )\right)}{2 \sqrt{2 \pi }}}$$

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\mrm{f}\pars{t} \equiv \expo{t/2}\mrm{sech}\pars{t} = \int_{-\infty}^{\infty}\hat{\mrm{f}}\pars{\omega}\expo{-\ic\omega t} {\dd\omega \over 2\pi} \color{red}{\iff} \hat{\mrm{f}}\pars{\omega} = \int_{-\infty}^{\infty}\mrm{f}\pars{t}\expo{\ic\omega t}\dd t:\ {\Large ?}}$.

\begin{align} \hat{\mrm{f}}\pars{\omega} & = \int_{-\infty}^{\infty}\mrm{f}\pars{t}\expo{\ic\omega t}\dd t = \int_{-\infty}^{\infty}\expo{t/2}\mrm{sech}\pars{t}\expo{\ic\omega t}\dd t \\[5mm] &= 2\int_{-\infty}^{\infty}{\expo{t/2}\expo{\ic\omega t} \over \expo{t} + \expo{-t}}\,\dd t\qquad \pars{~\mbox{Set}\quad x \equiv \expo{t} \color{red}{\iff} t = \ln\pars{x}~} \\[5mm] & = 2\int_{0}^{\infty}{x^{1/2}\, x^{\omega\ic} \over x^{2} + 1}\,\dd x \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, \int_{0}^{\infty}{x^{-1/4 + \omega\ic/2} \over x + 1}\,\dd x \,\,\,\stackrel{x + 1\ \mapsto\ x}{=}\,\,\, \int_{1}^{\infty}{\pars{x - 1}^{-1/4 + \omega\ic/2} \over x}\,\dd x \\[5mm] \stackrel{x\ \mapsto\ 1/x}{=}\,\,\,& \int_{0}^{1}x^{-3/4 - \omega\ic/2}\,\pars{1 - x}^{-1/4 + \omega\ic/2}\,\,\dd x = {\Gamma\pars{1/4 -\omega\ic/2}\Gamma\pars{3/4 -\omega\ic/2} \over \Gamma\pars{1}} = {\pi \over \sin\pars{\pi\bracks{1/4 -\omega\ic/2}}} \\[5mm] & = {\pi \over \sin\pars{\pi/4}\cos\pars{\pi\omega\ic/2} - \cos\pars{\pi/4}\sin\pars{\pi\omega\ic/2}} = {\root{2}\pi \over \cosh\pars{\pi\omega/2} - \sinh\pars{\pi\omega/2}\ic} \\[5mm] & = \bbx{\root{2}\pi\,\mrm{sech}\pars{\pi\omega} \bracks{\cosh\pars{{\pi \over 2}\,\omega} + \sinh\pars{{\pi \over 2}\,\omega}\ic}} \end{align}

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If you already know that $$\mathcal F[\operatorname{sech} t] = F(p) = \int_{-\infty}^{\infty} e^{i p t} \operatorname{sech} t \,dt = \pi \operatorname{sech} \frac {\pi p} 2,$$ then $$\mathcal F[e ^{t/2} \operatorname{sech} t] = \int_{-\infty}^{\infty} e^{i (p - i/2) t} \operatorname{sech} t \,dt = \\ F(p - i/2) = \pi \operatorname{sech} \frac {\pi (2p - i)} 4.$$

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