When is Laplace transform of a function power-law and relation to the behavior of the function near zero?

Question

I want to see when the Laplace transform of a non-negative function $f$ defined on $[0, +\infty)$ is a power function in the loose sense, i.e.,

$$g(s) = \mathcal L\{f\}(x) = \int_0^\infty f(x) e^{-sx} dx \stackrel{?}{=} L(s)s^{-\alpha}$$

where $L(s)$ is a slowly varying function, i.e., for all $a > 0$,

$$\lim_{s\to +\infty} \frac{L(as)}{L(s)} = 1$$

Intuitively it should have something to do with the behavior of $f$ near 0.

Using integration by parts we get

$$g(s) = \int_0^\infty f(x) e^{-sx} dx = s^{-1}(f(0) + \int_0^\infty f'(x) e^{-sx} dx)$$

• If $f(0) > 0$, $f(0) + \int_0^\infty f'(x) e^{-sx} dx$ is a slowly varying function, thus $g$ is power law.

• If $f(0) = 0$, we repeat the integration by parts until we get some $f^{\{n\}}(0) > 0$.

• If $f^{\{n\}}(0) = 0$ for all $n$, for instance $f$ is a shifted unit step function $f(x)=u(x−1)$ or $f$ is not $n$-time differentiable, $g$ is not power law.

However with this method I can only deal with situations where $s$ is raised to integer power. I'd like to know if it makes sense to examine the behavior of $f$ near 0 and how to better describe it. Thanks.

Answer 1

Intuitively it should have something to do with the behavior of $f$ near $0$.

This is true. Indeed, let $U(x):=\int_0^x du\,f(u)$. Then, according to (say) Tauberian Theorems 3 and 2 in Section XIII.5 of Vol. II of "An Introduction to Probability Theory and Its Applications", Second Ed., 1971, by Feller, for any real $a\ge0$ and any slowly varying at $\infty$ function $L$, $$g(s)\underset{s\to\infty}\sim s^{-a}L(s)\iff U(x)\underset{x\downarrow0}\sim \frac{x^a L(1/x)}{\Gamma(a+1)}.$$