What is the derivation of the inverse Laplace transform of sech(x)? Mathematica tells me the inverse Laplace transform of $sech(x)$ is as follows.
(1) $\quad\mathcal{L}_x^{-1}[sech(x)](y)=2\sum\limits_{k=0}^\infty (-1)^k\,\delta(y-2\,k-1)$
I thought perhaps the result illustrated in (1) above was based on the Taylor series for $sech(x)$ defined in (2) below, but this seems inconsistent with the inverse Laplace transform of $x^k$ illustrated in (3) below and the Laplace transform of $\delta(y-2\,k-1)$ illustrated in (4) below.
(2) $\quad sech(x)=\sum\limits_{k=0}^\infty\frac{E_k\,x^k}{k!}$
(3) $\quad\mathcal{L}_x^{-1}\left[x^k\right](y)=\frac{y^{-k-1}}{\Gamma(-k)}$
(4) $\quad\mathcal{L}_y[\delta(y-2\,k-1)](x)=\theta(2\,k+1)\,e^{-(2\,k+1)\,x}$
Questions: Is the inverse Laplace transform of $sech(x)$ illustrated in (1) above correct? If so, how is this result derived? If not, what is the correct inverse Laplace transform of $sech(x)$?
 A: Taking the forward transform of the right-hand side gives
$$ \mathcal{L}_y\left( 2\sum_{k=0}^{\infty} (-1)^k \delta(y-2k-1) \right)(s) = 2\sum_{k=0}^{\infty} (-1)^k e^{-(2k+1)s} = 2e^{-s} \sum_{k=0}^{\infty} (-e^{-2s})^k \\
= \frac{2}{e^s(1+e^{-2s})} = \operatorname{sech}{s}, $$
so the answer is correct. Working the other way is tricky: one would really have to invoke the convolution theorem using a smooth function to do it rigorously in the sense of distributions, so the default would be to carry out the series expansion above in terms of exponentials and essentially do this calculation in reverse. One may ask why this expansion is chosen: the heuristic can come from the Bromwich integral, which is
$$ \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{2e^{st-s}}{1+e^{2s}} \, ds. $$
Expanding the denominator gives integrals like $ \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} e^{s(t-2k-1)} \, ds, $ which look like those that represent the delta-function in Fourier analysis. Yet another approach is to push the contour to the left and add up the residues at $\pm i\pi/2,\pm 3i\pi/2, \dotsc$, which gives the Fourier series of a modified Dirac comb, as one might expect; only the positive part of the comb contributes to the unilateral Laplace transform space, of course.
