Fourier Transform of $ e^ {t/2} \operatorname{sech}(t) $ 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.
 A: 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
$$\color{blue}{\frac{\pi  \tan \left(\frac{1}{8} (2 \pi  i k+\pi )\right)+\pi  \cot \left(\frac{1}{8} (2 \pi  i k+\pi )\right)}{2 \sqrt{2 \pi }}}$$
A: 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:

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

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

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

\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}
A: 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
$$\mathcal F[e ^{t/2} \operatorname{sech} t] =
\int_{-\infty}^{\infty} e^{i (p - i/2) t} \operatorname{sech} t \,dt = \\
F(p - i/2) = \pi \operatorname{sech} \frac {\pi (2p - i)} 4.$$
