Converse of the Watson's lemma

Question

Watson's lemma basically says

$$ f(t) \sim t^{\alpha} \,\,\,(\text{for small } t) \implies \int_0^{\infty} f(t) e^{-st} dt \sim \frac{\Gamma(\alpha + 1)}{s^{\alpha + 1}} \,\,\,(\text{for large } s). $$

Under what condition is its converse true? Or more generally, if we know a function's asymptotic behavior and we know its inverse Laplace transformation, can we know the inverse's asymptotic behavior?

Answer 1

A very general converse of Watson's lemma is due to Feller [1].

Let $\mu$ be a measure concentrated on $[0,\infty)$ such that $$ \int_0^\infty e^{-s t}\,d\mu(t) $$ exists for all $s > 0$. Let $L$ be a real-valued function defined for large $x$ which satisfies $$ \lim_{x \to \infty} \frac{L(ax)}{L(x)} \to 1 $$ for all fixed $a > 0$ and let $-1 < \alpha < \infty$. Then $$ \int_0^\infty e^{-s t}\,d\mu(t) \sim \frac{\Gamma(\alpha+1)}{s^{\alpha+1}} L(s) \qquad \text{as } s \to \infty $$ if and only if $$ \mu([0,t]) \sim \frac{t^{\alpha+1}}{\alpha+1} L(1/t) \qquad \text{as } t \to 0^+. $$

To get a result similar to the usual statement of Watson's lemma, where we're integrating $e^{-st}$ against something like $f(x)\,dx$ rather than $d\mu(x)$, we can add on an additional monotonicity assumption:

If $\mu$ is as above and $f\colon(0,\infty)\to\mathbb R$ is a function which is monotone on some nonempty interval $(0,\delta)$ and which satisfies $$ d\mu(t) = f(t)\,dt $$ then $$ \int_0^\infty f(t) e^{-s t}\,dt \sim \frac{\Gamma(\alpha+1)}{s^{\alpha+1}} L(s) \qquad \text{as } s \to \infty $$ if and only if $$ f(t) \sim t^{\alpha} L(1/t) \qquad \text{as } t \to 0^+. $$

Common examples of the function $L$ are $L(x) = 1$ (which gives the usual form of Watson's lemma) and $L(x) = \log x$.

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[1] W. Feller, An Introduction to Probability Theory, Vol. II, Second edition, New York: John Wiley & Sons, 1971, pp. 442-446.