Watson's lemma
/ WOT-sun /
Watson's lemma is the precise, reusable engine inside many asymptotic calculations. It answers a sharp question: if you have a Laplace-type integral I(s) = (integral from 0 to infinity of f(t) e^(-s t) dt) and you let s grow large, what is the full asymptotic expansion of I(s), term by term, not just its leading size? The answer is delightfully mechanical: it depends only on the behaviour of f(t) near t = 0, because the factor e^(-s t) kills off everything away from the origin when s is large.
Here is the statement. Suppose near t = 0 the function f has an expansion in powers (possibly fractional) f(t) approximately sum a_n t^(b_n) with exponents b_n increasing to infinity and each b_n > -1, and suppose f does not grow too fast for large t (so the integral converges). Then as s -> infinity, I(s) has the asymptotic expansion I(s) approximately sum a_n times Gamma(b_n + 1) / s^(b_n + 1). In words: integrate each term of the small-t series against e^(-s t) using the gamma-function formula (integral from 0 to infinity of t^b e^(-s t) dt = Gamma(b+1)/s^(b+1)), and the resulting series is the asymptotics of the whole integral. You may differentiate the local series and the lemma still applies term by term.
Watson's lemma is the workhorse that makes Laplace's method and the method of steepest descent rigorous and gives them all their higher-order corrections, not just the leading term. After you have reduced an integral to a standard Laplace form by locating the dominant point and changing variables, Watson's lemma turns out the full power series in 1/s automatically. The honest caveat: the result is an asymptotic series, which typically does not converge — adding more terms eventually makes the approximation worse — so you stop at the optimal number of terms for a given s.
Take I(s) = (integral from 0 to infinity of e^(-s t) / (1 + t) dt). Near t = 0 we have 1/(1 + t) = 1 - t + t^2 - t^3 + ..., so Watson's lemma gives I(s) approximately Gamma(1)/s - Gamma(2)/s^2 + Gamma(3)/s^3 - ... = 1/s - 1/s^2 + 2/s^3 - 6/s^4 + .... The factorials Gamma(n+1) = n! make this a famously divergent asymptotic series, yet truncating it gives an excellent estimate of the exact exponential-integral value for large s.
The small-t series is integrated term by term against e^(-s t) via the gamma function.
The hypothesis b_n > -1 is essential — it keeps each Gamma(b_n + 1) finite and each piece integrable at t = 0; a term like t^(-1) would make the original integral diverge at the origin.