Watson's lemma

Remarks

Idea of the proof: for $x$ large, the factor $e^{-xt}$ makes the integral over $[\delta, T)$ exponentially small in $x$, for any fixed $\delta > 0$. On $(0, \delta)$, replace $f$ by its first $N$ terms plus an error bounded by $C t^{N + \lambda}$. Each term gives $\int_0^\infty e^{-xt} t^{n + \lambda} \, dt = \Gamma(n + \lambda + 1) x^{-n-\lambda-1}$, up to an exponentially small change, and the error contributes $O(x^{-N-\lambda-1})$. Example: $\int_0^\infty \frac{e^{-xt}}{1 + t} \, dt \sim \sum_{n \ge 0} (-1)^n \frac{n!}{x^{n+1}}$, a divergent series obtained from $\frac{1}{1+t} = \sum (-1)^n t^n$ near $0$. The condition $\lambda > -1$ is needed: if $\lambda \le -1$ and $a_0 \ne 0$, then $\abs{f(t)}$ is of order $t^{\lambda}$ near $0$ and the integral diverges at $t = 0$.

Going deeper

For a student

The lemma gives the large-$x$ behaviour of the Laplace transform $F(x)$ of $f$ from the small-$t$ behaviour of $f$. Together with a change of variables it yields Laplace's method and the method of steepest descent for integrals $\int e^{x \phi(t)} g(t) \, dt$. The resulting series often diverge, and Borel summation can be seen as the inverse process that recovers $F$ from its series.