7.2 Watson’s Lemma
Integrals of the form \(\int ^{\infty }_{0}e^{-xt}f(t)\,dt\) are exactly Laplace transforms, and their behaviour for large \(x\) is governed entirely by the behaviour of \(f\) near \(t=0\) — because the factor \(e^{-xt}\) suppresses everything else.
Theorem 7.2.1 (Watson’s lemma). Suppose \(f\) is locally integrable on \([0,\infty )\), of at most exponential order at infinity, and has the asymptotic expansion \[f(t)\ \sim \ \sum ^{\infty }_{n=0}a_n\,t^{\lambda _n} \qquad (t\to 0^{+}),\] with \(-1<\lambda _0<\lambda _1<\cdots \to \infty \). Then \[\int ^{\infty }_{0}e^{-xt}f(t)\,dt \ \sim \ \sum ^{\infty }_{n=0}\frac {a_n\,\Gamma \left (\lambda _n+1\right )} {x^{\lambda _n+1}} \qquad (x\to \infty ).\]
Sketch. Term by term, \(\int ^{\infty }_{0}e^{-xt}t^{\lambda }dt = \Gamma (\lambda +1)/x^{\lambda +1}\) by the substitution of Example 6.1.5. The content of the lemma is that the error committed in truncating \(f\) transfers to an error of the same order after integration, which follows from splitting the range at any fixed \(\delta >0\): on \([\delta ,\infty )\) the factor \(e^{-xt}\) makes the contribution exponentially small, and on \([0,\delta ]\) the truncation error is bounded by a constant times \(t^{\lambda _{N+1}}\). □
Note 7.2.2. Every term of the answer is a gamma value, and the condition \(\lambda _0>-1\) is once more the convergence condition of Problem 1.6.9. The mechanism is worth stating plainly: because \(e^{-xt}\) concentrates the integral near \(t=0\) as \(x\) grows, only the behaviour of \(f\) at the origin survives, and the gamma function is what converts each power of \(t\) there into a power of \(1/x\) here.
Example 7.2.3. For the complementary error function, substitute \(u = t^{2}-x^{2}\) in \[\operatorname {erfc}(x) = \frac {2}{\sqrt {\pi }}\int ^{\infty }_{x}e^{-t^{2}}dt = \frac {e^{-x^{2}}}{\sqrt {\pi }}\int ^{\infty }_{0} \frac {e^{-u}}{\sqrt {u+x^{2}}}\,du .\] Expanding \(\left (u+x^{2}\right )^{-1/2} = x^{-1}\left (1 - \frac {u}{2x^{2}} + \frac {3u^{2}}{8x^{4}} - \cdots \right )\) and applying Theorem 7.2.1 with \(\Gamma (n+1)=n!\), \[\operatorname {erfc}(x)\ \sim \ \frac {e^{-x^{2}}}{x\sqrt {\pi }} \left (1 - \frac {1}{2x^{2}} + \frac {3}{4x^{4}} - \cdots \right ).\] The leading term is the estimate obtained by a single integration by parts in Problem 2.5.2; Watson’s lemma supplies the whole expansion at once.
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