<s>[INST] 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.
 [/INST] 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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