Inverse Laplace transform of $\frac{1}{s^a + 1}$ with $0 < a \leq 1$

Question

Problem

I am looking for the following inverse Laplace transform, $$ f(t) = \mathcal{L}^{-1}\left[\frac{1}{s^a + 1}\right] \;\;\;\;\; \text{with} \;\;\;\;\; 0 < a \leq 1. $$

What I understand

• with $a = 1$, $f(t) = e^{-t}$

• for small $t \ll 1$, $f(t)$ should behave $$ f(t)\approx \frac{1}{\Gamma(a)} t^{a-1} $$ as $$ \frac{1}{s^a + 1} \approx s^{-a} $$ with large $s\gg 1$

• for large $t \gg 1$, $f(t)$ should behave $$ f(t) \approx \frac{a}{\Gamma(1-a)} t^{-(a+1)} $$ from $$ t f(t) = \mathcal{L}^{-1}\left[\frac{a s^{a-1}}{(s^a + 1)^2}\right] \approx \mathcal{L}^{-1}\left[a s^{a-1}\right] $$ with small $s \ll 1$.

Numerical solution

I have a master equation for $f(t)$ and the above Laplace transform is obtained from this master equation. By numerically solving the master equation, I obtained the following numerical solution,

But I want an analytical representation (even an approximated one), not the numerical solution.

Candidates?

Since the solution may smoothly connect between these two limit forms, and should be similar to the exponential function if $a\approx 1$, I expected something similar to the kappa distribution, $t^{a-1}\left(1 + \frac{t}{a+1}\right)^{-(a+1)}$. However, the exponent does not match, and this does not reproduce both the Laplace transform and the numerical solution.

Wolfram alpha does not give a solution, either.

Any idea would be appreciated. Thank you in advance.

Answer 1

As @vitamin pointed out, the distribution having the Laplace transform of $\frac{1}{1 + s^\alpha}$ is called Mittag-Leffler distribution.

Although there are no known closed forms, many of its properties are investigated, such as its exponential-mixture representation 1, $$ f(x; \alpha) = \frac{\sin(\pi \alpha)}{\pi} \int_0^\infty \frac{1}{\beta}\; e^{-x/\beta}\; \frac{\beta^\alpha}{1 + \beta^{2\alpha} + 2\beta^\alpha \cos(\pi \alpha)} \; d\beta $$