What is the Laplace transform of the Logistic function?

Question

What is the Laplace transform

$$F(s) = \mathcal{L} \left\{f\right\}(s) = \int_{0}^{\infty} f(t) e^{-st} \,\mathrm{d}t, \qquad s\in\mathbb{C}$$

of the Logistic function

$$f(x) = \frac{L}{1 + e^{-k(x-x_0)}}$$

with constant $k,x_0,L$?

Answer 1

The Maclaurin series of the logistic (sigmoid) function is expressed in terms of $x$ as

$$ f(x) = \frac{L}{1+e^{-k(x-x_0)}} $$ $$ = L\sum_{n=0}^{\infty} \frac{(-k)^nE_n(0)}{2n!}(x-x_0)^n $$

Placing this into the Laplace transform as a function of $t$ gives

$$F(s) = \int_{0}^{\infty} f(t) e^{-st} \,\mathrm{d}t = \int_{0}^{\infty} \frac{L}{1+e^{-k(t-x_0)}} e^{-st} \,\mathrm{d}t $$

$$ = \int_{0}^{\infty} L\sum_{n=0}^{\infty} \frac{(-k)^nE_n(0)}{2n!}(t-x_0)^n e^{-st} \,\mathrm{d}t $$

Where $ E_n(0) $ is an Euler polynomial of degree $n$, evaluated at $0$.

Rearranging terms gives $$ F(s) = L\sum_{n=0}^{\infty} \frac{(-k)^nE_n(0)}{2n!}\int_{0}^{\infty} (t-x_0)^n e^{-st} \,\mathrm{d}t $$

The integral term closely resembles tabulated values of the Laplace transform, but it still requires some extra work. By using integration by parts (IBP) for n=0,1,2,3..., a pattern emerges. This comes out to be

$$ \int_{0}^{\infty} (t-x_0)^n e^{-st} \,\mathrm{d}t = \sum_{g=0}^{n} \frac{n!}{(n-g)!} \frac{x_0^{(n-g)}}{s^{(g+1)}}, $$

where the ratio of factorials comes from a reduced form of the binomial coefficient. $k$ would normally be used here, but I substituted in $g$ because $k$ is already used as a coefficient in the logistic equation. If $x_0=0$ and if we assume $0^0=1$, then the summation will reduce to the tabulated expression of $$ \mathcal{L} \left\{f\right\}(s) = \int_{0}^{\infty} t^n e^{-st} \,\mathrm{d}t = \frac{n!}{s^{(n+1)}}$$

The above summation can then be substituted into the earlier expression, giving the Laplace transform of the logistic function with coefficients $L$, $k$, and $x_0$.

$$ F(s) = L\sum_{n=0}^{\infty} \frac{(-k)^nE_n(0)}{2n!} \sum_{g=0}^{n} \frac{n!}{(n-g)!} \frac{x_0^{(n-g)}}{s^{(g+1)}} $$