Consider the sigmoid function
$$
h_k(t) = \frac{1}{1+e^{-kt}}
$$
as an approximation to the Heaviside step function $u(t)$ as $k \rightarrow \infty$.
Given
$$
f(t) = h_k(t) \times s(t) \equiv \frac{1}{1+e^{-kt}} \times (A \cos(\omega_0 t) + B \sin(\omega_0 t)) \exp(-\alpha t),
$$
where $\alpha > 0$ and I used $\times$ for clarity, I want to evaluate its Fourier transform (FT)
$$
F(\omega) = \int_{-\infty}^{\infty} f(t) \exp(-i\omega t) dt.
$$
This integral exists if $\alpha < k$, so we choose $k$ accordingly.
After failing to evaluate the integral directly, my approach was to calculate $F(\omega)$ as
$$
F(\omega) = (H_k * S)(\omega),
$$
where $*$ denotes convolution and $H_k(\omega)$ and $S(\omega)$ are the FTs of $h_k(t)$ and $s(t)$.
As this answer shows,
$$
H_k(\omega) = \pi \delta(\omega) – \frac{i\pi}{k \sinh(\pi\omega/k)}
$$
in the "distributional sense" (which I understand to be in the sense of Lighthill's generalized functions), which for $k \rightarrow \infty$ indeed reduces nicely to the FT of $u(t)$, i.e. $U(\omega)$:
$$
H_\infty(\omega) = \pi\delta(\omega) + \frac{1}{i\omega} = U(\omega).
$$
The problem with my approach is that I cannot calculate $S(\omega)$, as its FT does not converge ($s(t)$ does not vanish for $t \rightarrow
-\infty$), but perhaps it might still be possible "in the distributional sense".
My questions: Can $F(\omega)$ be evaluated in a nice closed form? If not, how does $|F(\omega)|$ fall off for $\omega \rightarrow \infty$? This should be slower faster [edit] than $1/\omega$ for $k < \infty$, am I correct?
Best Answer
If I don't miss something, for $0 < \alpha < k$ the function $f$ belongs to Schwartz space. This space is closed under Fourier transforms. Therefore $F=\mathcal{F}f$ is also in Schwartz space, meaning that it is $C^\infty$ and that all of its derivatives fall off faster than $|x|^{-n}$ for every $n \geq 0$ as $|x| \to \infty.$