We wish to calculate the Fourier Transform of $\text{Sech}(x)$. I have tried to do this using the same technique as similar questions. We attempt to directly calculate the integral using integration by parts twice:
$\mathcal{F}[\text{Sech}(x)](\xi) = \int^{\infty}_{-\infty} \text{Sech}(x) e^{-2\pi i x \xi} \text{d}x $
$= \int^{\infty}_{-\infty} \text{Sech}(x) \text{Cos}(2\pi x \xi) \text{d}x - i\int^{\infty}_{-\infty} \text{Sech}(x) \text{Sin}(2\pi x \xi) \text{d}x $
We consider first the integral $ \int^{\infty}_{-\infty} \text{Sech}(x) \text{Cos}(2\pi x \xi) \text{d}x $
$ = [-\text{Sech}(x) \frac{\text{Sin}(2 \pi x \xi)}{2 \pi \xi} ]^{x=\infty}_{x=-\infty} - \int^{\infty}_{-\infty} \text{Sech}(x) \text{Tanh(x)} \frac{\text{Sin}(2 \pi x \xi)}{2 \pi \xi} \text{d}x $
The term on the left is $0$, so we are left with just $ - \frac{1}{2\pi \xi} \int^{\infty}_{-\infty} \text{Sech}(x) \text{Tanh(x)} \text{Sin}(2 \pi x \xi) \text{d}x $
$ = -\frac{1}{2 \pi \xi} ( [\text{Sech}(x) \text{Tanh(x)} \frac{\text{Cos}(2 \pi x \xi)}{2 \pi \xi}]^{x=\infty}_{x=-\infty} - \int^{\infty}_{-\infty} \{\text{Sech}^3 (x) - \text{Sech}(x)\text{Tanh}^2(x) \} \frac{\text{Cos}(2 \pi x \xi)}{2 \pi \xi} \text{d}x ) $
Again, the term on the left is $0$, so we are left with:
$ +(\frac{1}{2 \pi \xi})^2 \int^{\infty}_{-\infty} \{\text{Sech}^3 (x) - \text{Sech}(x)\text{Tanh}^2(x) \}\text{Cos}(2 \pi x \xi) \text{d}x $
$ = (\frac{1}{2 \pi \xi})^2 \int^{\infty}_{-\infty} \text{Sech}(x)\text{Cos}(2 \pi x \xi) \{\text{Sech}^2 (x) - \text{Tanh}^2(x) \} \text{d}x $
So we have performed integration by parts twice and are left with the equation:
$ \int^{\infty}_{-\infty} \text{Sech}(x) \text{Cos}(2\pi x \xi) \text{d}x = (\frac{1}{2 \pi \xi})^2 \int^{\infty}_{-\infty} \text{Sech}(x)\text{Cos}(2 \pi x \xi) \{\text{Sech}^2 (x) - \text{Tanh}^2(x) \} \text{d}x $
Unfortunately, this does not seem usable for this situation, and it does not seem like a useful result will appear with any further uses of integration by parts. I am left to assume we need to use a totally different method. If anyone has any suggestions, I would be very grateful. Thank you.
$$\mathcal{F}[sech(x)](\xi)=2\int_{-\infty}^{\infty}\frac{e^{2\pi ix\xi}}{e^{2x}+1}e^xdx\overbrace{=}^{e^{2x}=z}\int_{0}^{\infty}\frac{z^{\pi i\xi+\frac{1}{2}-1}}{z+1}dz$$ $$\mathcal{F}[sech(x)](\xi)=\mathfrak{B}\left(\pi i\xi+\frac{1}{2},1-\pi i\xi-\frac{1}{2}\right)=\pi csc\left(\pi\left(\pi i\xi+\frac{1}{2}\right)\right)=\pi sech(\pi^2 \xi)$$