Watson's Lemma

Derivation

Watson's Lemma

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Statement

Let \(I(\lambda)=\int_0^{T} f(t)\,e^{-\lambda t}\,dt\) with \(0<T\le\infty\), where \(f\) is locally integrable on \((0,T)\) and admits, as \(t\to 0^{+}\), the asymptotic expansion \(f(t)\sim\sum_{n=0}^{\infty}a_n\,t^{\alpha_n}\) with real exponents \(-1<\alpha_0<\alpha_1<\cdots\to\infty\). If \(T=\infty\) assume \(f(t)=O(e^{ct})\) for some real \(c\). Then Watson's lemma asserts the term-by-term Laplace transform of the expansion is itself the asymptotic expansion of \(I\) as \(\lambda\to+\infty\): \[ I(\lambda)\;\sim\;\sum_{n=0}^{\infty} a_n\,\frac{\Gamma(\alpha_n+1)}{\lambda^{\alpha_n+1}},\qquad \lambda\to+\infty . \]

Why it matters

Watson's lemma is the workhorse behind virtually every large-parameter asymptotic in mathematical physics: the Stirling expansion of \(\Gamma\), the large-argument behaviour of Bessel functions \(K_\nu\), the WKB connection integrals, and the tails of special-function representations. It converts a purely local datum — how the integrand behaves in an arbitrarily small neighbourhood of the endpoint \(t=0\) — into the global large-\(\lambda\) behaviour of the whole integral.

Its power is that it is mechanical: once you have the small-\(t\) expansion of \(f\), every coefficient of the answer is a Gamma-function value. The exponentially decaying weight \(e^{-\lambda t}\) concentrates the integral so tightly at the endpoint that everything away from it is negligible to all algebraic orders, which is exactly why an asymptotic (not convergent) series suffices.

Assumptions

Endpoint expansion exists.If \(f\) has no asymptotic power expansion at \(t=0\) (e.g. \(f(t)=\sin(1/t)\)), no coefficients \(a_n\) are defined and the lemma has nothing to assert. Leading exponent \(\alpha_0>-1\).If \(\alpha_0\le-1\) the integral \(\int_0 t^{\alpha_0}\,dt\) diverges at the origin and \(I(\lambda)\) is not even finite; the Gamma factor \(\Gamma(\alpha_0+1)\) also hits a pole. Exponents strictly increasing to \(+\infty\).If they accumulate or fail to grow, remainder terms are not successively smaller and the expression ceases to be an asymptotic series (no ordering of scales). Sub-exponential control of the tail.If \(T=\infty\) and \(f\) grows faster than any \(e^{ct}\) (say \(e^{t^2}\)), the Laplace integral diverges for every \(\lambda\) and the endpoint no longer dominates. The expansion is asymptotic, not assumed convergent.We only use \(f(t)-\sum_{n\le N}a_nt^{\alpha_n}=O(t^{\alpha_{N+1}})\) on a neighbourhood \([0,\delta]\); demanding convergence would exclude most physically relevant \(f\) (e.g. \(1/(1+t)\) has radius \(1\) yet the method still works on \([0,\infty)\)).

Derivation

\[ M_\alpha(\lambda)\;:=\;\int_0^{\infty} t^{\alpha}\,e^{-\lambda t}\,dt \;=\;\frac{\Gamma(\alpha+1)}{\lambda^{\alpha+1}},\qquad \alpha>-1,\ \lambda>0. \]

The single monomial "master integral." Substitute \(u=\lambda t\), \(du=\lambda\,dt\): \(\int_0^\infty (u/\lambda)^\alpha e^{-u}\,du/\lambda=\lambda^{-(\alpha+1)}\int_0^\infty u^\alpha e^{-u}\,du=\lambda^{-(\alpha+1)}\Gamma(\alpha+1)\), using the Euler integral definition of \(\Gamma\). Convergence at \(0\) needs \(\alpha>-1\). A

\[ I(\lambda)=\underbrace{\int_0^{\delta} f(t)\,e^{-\lambda t}\,dt}_{I_{\mathrm{near}}}\;+\;\underbrace{\int_{\delta}^{T} f(t)\,e^{-\lambda t}\,dt}_{I_{\mathrm{far}}},\qquad 0<\delta<T . \]

Split at a fixed \(\delta\) small enough that the endpoint expansion of \(f\) is valid on \([0,\delta]\). This isolates the region that will dominate. A

\[ |I_{\mathrm{far}}|\;\le\; e^{-\lambda\delta/2}\int_{\delta}^{T}|f(t)|\,e^{-\lambda\delta/2}\,e^{-(\lambda-\lambda)t}\,dt \;\le\; C_\delta\,e^{-\lambda\delta/2}. \]

On \([\delta,T]\) write \(e^{-\lambda t}=e^{-\lambda t/1}\) and peel a factor \(e^{-\lambda\delta}\le e^{-\lambda t}\)-type bound: with the growth hypothesis \(|f(t)|e^{-t}\in L^1\), for \(\lambda\ge \lambda_0\) one has \(|I_{\mathrm{far}}|\le e^{-(\lambda-\lambda_0)\delta}\!\int_\delta^T |f|e^{-\lambda_0 t}dt\). The tail is thus \(O(e^{-a\lambda})\) with \(a=\delta/2>0\): smaller than every power \(\lambda^{-p}\). It contributes to no algebraic order and is dropped. C

\[ f(t)=\sum_{n=0}^{N} a_n\,t^{\alpha_n}+R_N(t),\qquad |R_N(t)|\le K_N\,t^{\alpha_{N+1}}\ \text{ on }[0,\delta]. \]

Definition of the asymptotic expansion at the endpoint: the \((N{+}1)\)-term truncation error is controlled by the first omitted power. This is the only regularity we use. B

\[ I_{\mathrm{near}}=\sum_{n=0}^{N} a_n\int_0^{\delta} t^{\alpha_n}e^{-\lambda t}dt\;+\;\int_0^{\delta} R_N(t)\,e^{-\lambda t}\,dt . \]

Linearity of the integral over the finite sum; the remainder integral is carried explicitly rather than discarded. A

\[ \int_0^{\delta} t^{\alpha_n}e^{-\lambda t}dt=\int_0^{\infty} t^{\alpha_n}e^{-\lambda t}dt-\int_{\delta}^{\infty} t^{\alpha_n}e^{-\lambda t}dt=\frac{\Gamma(\alpha_n+1)}{\lambda^{\alpha_n+1}}+O\!\big(e^{-\lambda\delta/2}\big). \]

Extend each monomial integral to \(\infty\) using Step 1; the correction \(\int_\delta^\infty t^{\alpha_n}e^{-\lambda t}dt\) is again exponentially small (same argument as Step 3, since \(t^{\alpha_n}e^{-\lambda t/2}\) is bounded on \([\delta,\infty)\)). Extending the range is what turns the finite integral into a clean Gamma value. C

\[ \left|\int_0^{\delta} R_N(t)\,e^{-\lambda t}dt\right|\le K_N\int_0^{\infty} t^{\alpha_{N+1}}e^{-\lambda t}dt=K_N\,\frac{\Gamma(\alpha_{N+1}+1)}{\lambda^{\alpha_{N+1}+1}}=o\!\left(\lambda^{-(\alpha_N+1)}\right). \]

Bound \(|R_N|\le K_N t^{\alpha_{N+1}}\), pull the constant out, extend to \(\infty\) (over-estimate), and apply Step 1. Because \(\alpha_{N+1}>\alpha_N\), the remainder is strictly smaller than the last kept term \(\lambda^{-(\alpha_N+1)}\): the defining property of an asymptotic series. C

\[ I(\lambda)=\sum_{n=0}^{N} a_n\,\frac{\Gamma(\alpha_n+1)}{\lambda^{\alpha_n+1}}\;+\;o\!\left(\lambda^{-(\alpha_N+1)}\right)\quad\text{for every fixed }N. \]

Collect Steps 3, 5, 6, 7: the exponentially small pieces and the remainder are absorbed into the error term. Since \(N\) is arbitrary, letting \(N\) range gives the full asymptotic series (written with "\(\sim\)"). B

Result

\[ \int_0^{T} f(t)\,e^{-\lambda t}\,dt\;\sim\;\sum_{n=0}^{\infty} a_n\,\frac{\Gamma(\alpha_n+1)}{\lambda^{\alpha_n+1}},\qquad \lambda\to+\infty,\quad f(t)\sim\sum_n a_n t^{\alpha_n}\ (t\to0^{+}). \]

Reading. Each power \(t^{\alpha_n}\) in the integrand's endpoint expansion is replaced by its own Laplace transform \(\Gamma(\alpha_n+1)\,\lambda^{-(\alpha_n+1)}\), and you simply add them up. The leading term \(a_0\,\Gamma(\alpha_0+1)\,\lambda^{-(\alpha_0+1)}\) fixes both the rate of decay of \(I(\lambda)\) and its amplitude; higher terms are successive corrections in inverse powers of \(\lambda\). Only the behaviour of \(f\) at \(t=0\) matters — the rest of the profile is exponentially invisible.

Units check. Let \([t]=T_u\) (a "time") so \([\lambda]=T_u^{-1}\) and \([dt]=T_u\). If \([f]=F\), then \([a_n]=F\,T_u^{-\alpha_n}\) since \(f\sim a_n t^{\alpha_n}\). The \(n\)-th term carries \([a_n]\cdot[\lambda]^{-(\alpha_n+1)}=F\,T_u^{-\alpha_n}\cdot T_u^{\alpha_n+1}=F\,T_u\), which matches \(\big[\int f\,dt\big]=F\,T_u\). Every term has the same dimension \(F\,T_u\), as it must (\(\Gamma\) is dimensionless).

Limiting cases

Analytic \(f\) with \(f(0)\ne0\): \(\alpha_n=n\), \(a_n=f^{(n)}(0)/n!\), giving \(I\sim\sum_n f^{(n)}(0)\,\lambda^{-(n+1)}\) since \(\Gamma(n+1)/n!=1\). The leading term is just \(f(0)/\lambda\).
Pure monomial \(f(t)=t^{\alpha}\): the series truncates to one exact term \(\Gamma(\alpha+1)\lambda^{-(\alpha+1)}\) — Watson reduces to the master integral of Step 1.
Half-integer exponents (\(\alpha_0=-\tfrac12\)): \(\Gamma(\tfrac12)=\sqrt{\pi}\) appears; leading decay \(\sim a_0\sqrt{\pi}\,\lambda^{-1/2}\), the universal signature of a square-root endpoint singularity.
Finite \(T\): the answer is identical to \(T=\infty\); the upper endpoint only adds \(O(e^{-\lambda(T-\delta)})\), beyond all orders.
Only the leading term wanted: Watson collapses to the elementary statement that \(I(\lambda)\sim a_0\,\Gamma(\alpha_0+1)\,\lambda^{-(\alpha_0+1)}\), the endpoint-dominance heuristic made rigorous.

Breaks when

No endpoint power expansion. If \(f\) oscillates or has an essential singularity as \(t\to0^+\) (e.g. \(e^{-1/t}\), \(\sin(1/t)\)), the coefficients \(a_n\) do not exist. For \(f=e^{-1/t}\) every \(a_n=0\) yet \(I(\lambda)\ne0\): the true behaviour is \(\sim e^{-2\sqrt{\lambda}}\), beyond all algebraic orders, and Watson (correctly) predicts only "smaller than every power."
Endpoint singularity too strong, \(\alpha_0\le-1\). The integral diverges at \(t=0\); \(\Gamma(\alpha_0+1)\) sits at or past a pole and the formula is meaningless.
Oscillatory (imaginary) exponent. For \(\int_0^T f(t)\,e^{i\lambda t}\,dt\) there is no exponential damping; the dominant balance is set by stationary phase and endpoint contributions on equal footing — one needs the method of stationary phase / steepest descent (a contour-rotated cousin of Watson), not the lemma as stated.
Maximum of the phase is interior, not at the endpoint. If the exponent is \(-\lambda\,\phi(t)\) with \(\phi\) minimised inside \((0,T)\) rather than at \(0\), the dominant contribution is a saddle; Laplace's method around the interior point supersedes Watson.

Failure modes

Treating the series as convergent. Summing all terms of \(\sum a_n\Gamma(\alpha_n+1)\lambda^{-(\alpha_n+1)}\) as if it equalled \(I\). It is asymptotic: for fixed \(\lambda\) the terms eventually grow. Truncate near the least term (see D-... optimal truncation).
Off-by-one in the Gamma argument. Writing \(\Gamma(\alpha_n)\) instead of \(\Gamma(\alpha_n+1)\), or using \(\alpha_n!\) for non-integer \(\alpha_n\). The \(+1\) comes from \(\int_0^\infty u^{\alpha}e^{-u}du=\Gamma(\alpha+1)\).
Using \(f\)'s Taylor series over the whole range. The expansion need only hold as \(t\to0^+\); students integrate a finite-radius series over \([0,\infty)\) and worry it "leaves the disc." Watson never uses convergence — only the endpoint estimate.
Calling the finite-\(T\) correction "algebraically small." The tail is \(e^{-\lambda(T-\delta)}\), exponentially small, so it changes no coefficient; mistaking it for a \(\lambda^{-p}\) correction corrupts the series.
Demanding analyticity of \(f\). A merely asymptotic (even non-smooth) expansion at \(0\) is enough; requiring \(f\in C^\infty\) needlessly rejects valid cases like \(f(t)=t^{1/3}g(t)\).
Forgetting to check \(\alpha_0>-1\). Applying the formula to a non-integrable endpoint singularity produces a finite-looking but nonsensical answer.

Discussion

The physical content of Watson's lemma is endpoint dominance. The weight \(e^{-\lambda t}\) is a probe of width \(1/\lambda\): as \(\lambda\to\infty\) it samples \(f\) only within \(O(1/\lambda)\) of the origin. Consequently the integral cannot "know" anything about \(f\) except its Taylor/Puiseux data at \(t=0\), and the entire large-\(\lambda\) expansion is built from that germ. This is the integral-transform analogue of the statement that, in a strongly damped system, the long-time or large-rate response is governed by the local structure at the boundary of the accessible region.

The appearance of \(\Gamma\) is not incidental: \(\Gamma(\alpha+1)=\int_0^\infty u^{\alpha}e^{-u}du\) is exactly the Laplace transform of a power at unit rate, so Watson is "Laplace transform, term by term." Because the reflection/continuation of \(\Gamma\) (assumed prior result) extends \(\Gamma(\alpha+1)\) to all non-integer and negative-but-\({>}\,{-1}\) exponents, fractional endpoint singularities \(t^{-1/2}\), \(t^{1/3}\), \(t^{\pi-3}\) are handled uniformly, delivering \(\sqrt{\pi}\), \(\Gamma(4/3)\), etc., with no special-casing.

The series is generically divergent, and this is a feature, not a defect. For \(I(\lambda)=\int_0^\infty e^{-\lambda t}/(1+t)\,dt\) one gets \(\sum_n(-1)^n n!\,\lambda^{-(n+1)}\), whose terms grow like \(n!\). Optimal truncation (assumed prior result) stops just before the least term, near \(n\approx\lambda\), yielding an error \(\sim e^{-\lambda}\) — exponentially accurate despite formal divergence. Watson's lemma is therefore the natural setting in which to see that an asymptotic series can outperform any convergent one at large \(\lambda\), and the exponentially small remainder it leaves behind is precisely the "beyond-all-orders" physics (Stokes phenomena, tunnelling amplitudes) that resurgence theory later organises.

A sharper viewpoint: the map \(f\mapsto I\) is the Laplace transform \(\mathcal{L}\), and Watson's lemma is the statement that \(\mathcal{L}\) intertwines the two natural gradations — the small-\(t\) filtration by powers \(t^{\alpha}\) on one side and the large-\(\lambda\) filtration by powers \(\lambda^{-(\alpha+1)}\) on the other — with \(\Gamma\) as the "structure constant." In the Borel–Laplace theory of resurgent functions this is the elementary link \(\widehat{t^{\alpha}}=\Gamma(\alpha+1)^{-1}\zeta^{\alpha}\) up to normalisation; the singularities of the Borel transform in the \(\zeta\)-plane are the endpoint powers themselves, and Watson's term-by-term rule is the leading, singularity-free part of the full alien calculus.

Common misconceptions. (i) "More terms always help" — false; past the least term the partial sums diverge. (ii) "The answer depends on the whole shape of \(f\)" — false to all algebraic orders; only the germ at \(t=0\) enters, the rest is exponentially small. (iii) "\(\Gamma(\alpha_n+1)=\alpha_n!\)" — only for non-negative integers; for fractional exponents use the analytically continued \(\Gamma\).

Worked examples

Example 1 — the Stieltjes / exponential-integral series (integer powers, divergent). Evaluate the large-\(\lambda\) expansion of \(\displaystyle I(\lambda)=\int_0^{\infty}\frac{e^{-\lambda t}}{1+t}\,dt=e^{\lambda}E_1(\lambda)\), then truncate optimally at \(\lambda=5\).

\[ f(t)=\frac{1}{1+t}=\sum_{n=0}^{\infty}(-1)^n t^{n},\qquad t\to0^{+}. \]

Geometric expansion of the integrand at the endpoint: \(a_n=(-1)^n\), \(\alpha_n=n\) (integer powers, \(\alpha_0=0>-1\)). A

\[ I(\lambda)\sim\sum_{n=0}^{\infty}(-1)^n\,\frac{\Gamma(n+1)}{\lambda^{\,n+1}}=\sum_{n=0}^{\infty}\frac{(-1)^n\,n!}{\lambda^{\,n+1}}. \]

Apply Watson term by term with \(\Gamma(n+1)=n!\). A

\[ \lambda=5:\quad \frac{n!}{5^{\,n+1}}:\ 0.2,\ 0.04,\ 0.016,\ 0.0096,\ 0.00768,\ 0.00768,\ 0.009216,\dots \]

Term magnitudes; the least term is at \(n\approx\lambda=5\) (here \(n=4\) and \(n=5\) tie), where optimal truncation stops. B

\[ \sum_{n=0}^{4}\frac{(-1)^n n!}{5^{\,n+1}}=0.2-0.04+0.016-0.0096+0.00768=0.17376. \]

Partial sum through the least term. A

\[ e^{5}E_1(5)=0.17048\ldots,\qquad I_{\text{trunc}}=0.1738,\qquad |\text{error}|\approx3.3\times10^{-3}. \]

Reading. A four-term optimally truncated divergent series already reproduces the true value to \(\sim2\%\); the residual error is of the order of the least term \(0.0077\), consistent with the least-term bound.

Units check. \(f\) is dimensionless, \([\lambda]=T_u^{-1}\); each term \(n!\,\lambda^{-(n+1)}\) has dimension \(T_u^{\,n+1}\)... all terms carry \([\int f\,dt]=T_u\) once the implicit \([f]=1\) and \([a_n]=T_u^{-n}\) are restored. Numerically \(\lambda\) is taken dimensionless here.

Example 2 — a half-integer endpoint singularity. Find the first two terms of \(\displaystyle I(\lambda)=\int_0^{\infty} t^{-1/2}\,e^{-t^{2}}\,e^{-\lambda t}\,dt\) as \(\lambda\to\infty\).

\[ f(t)=t^{-1/2}e^{-t^{2}}=t^{-1/2}\Big(1-t^{2}+\tfrac12 t^{4}-\cdots\Big)=t^{-1/2}-t^{3/2}+\tfrac12 t^{7/2}-\cdots \]

Endpoint expansion: multiply the entire series \(e^{-t^2}=\sum(-1)^k t^{2k}/k!\) by \(t^{-1/2}\). Exponents \(\alpha_k=2k-\tfrac12\), coefficients \(a_k=(-1)^k/k!\). Note \(\alpha_0=-\tfrac12>-1\), so the integral converges. B

\[ I(\lambda)\sim \frac{\Gamma(\tfrac12)}{\lambda^{1/2}}-\frac{\Gamma(\tfrac52)}{\lambda^{5/2}}+\frac{1}{2}\frac{\Gamma(\tfrac92)}{\lambda^{9/2}}-\cdots \]

Watson term by term; \(\alpha_k+1=2k+\tfrac12\). A

\[ \Gamma(\tfrac12)=\sqrt{\pi},\quad \Gamma(\tfrac52)=\tfrac34\sqrt{\pi},\quad\Rightarrow\quad I(\lambda)\sim\sqrt{\pi}\,\lambda^{-1/2}-\tfrac34\sqrt{\pi}\,\lambda^{-5/2}+\cdots \]

Evaluate half-integer Gammas via \(\Gamma(z+1)=z\Gamma(z)\) from \(\Gamma(\tfrac12)=\sqrt\pi\) (analytic-continuation/reflection prior result). A

\[ \lambda=8:\quad \sqrt{\pi}=1.7725,\ \lambda^{-1/2}=0.35355,\ \lambda^{-5/2}=5.525\times10^{-3}. \]

Insert numbers. A

\[ I(8)\approx 1.7725(0.35355)-1.3293(0.005525)\approx 0.6267-0.0073=0.6194. \]

Reading. The square-root singularity at the origin dictates a \(\lambda^{-1/2}\) decay with amplitude \(\sqrt{\pi}\); the first correction is already \(\sim1\%\) at \(\lambda=8\). The \(e^{-t^2}\) factor is invisible except through its Taylor coefficients at \(t=0\).

Units check. With \([t]=T_u\): \([f]=T_u^{-1/2}\) (from \(t^{-1/2}\)), so \([\int f\,dt]=T_u^{1/2}\). Term \(k\): \([a_k]\,[\lambda]^{-(\alpha_k+1)}=T_u^{-\alpha_k}\cdot T_u^{\alpha_k+1}=T_u\)... restoring \([f]\): each term is \(T_u^{-1/2}\cdot T_u=T_u^{1/2}\), matching. Consistent.

Problems

(A) State the master integral of Watson's lemma and use it to evaluate \(\int_0^{\infty} t^{3}e^{-2t}\,dt\).
Solution
The master integral is \(\int_0^\infty t^{\alpha}e^{-\lambda t}dt=\Gamma(\alpha+1)/\lambda^{\alpha+1}\) for \(\alpha>-1\). Here \(\alpha=3,\ \lambda=2\): \(\Gamma(4)/2^{4}=3!/16=6/16=\boxed{3/8=0.375}\).
(B) Give the first two terms of \(I(\lambda)=\int_0^{\infty}\dfrac{e^{-\lambda t}}{1+t^{2}}\,dt\) as \(\lambda\to\infty\).
Solution
Endpoint expansion \(1/(1+t^2)=1-t^2+t^4-\cdots\), so \(a_0=1\ (\alpha_0=0)\), \(a_1=-1\ (\alpha_1=2)\). Watson: \(I\sim \Gamma(1)/\lambda-\Gamma(3)/\lambda^{3}+\cdots=\dfrac{1}{\lambda}-\dfrac{2}{\lambda^{3}}+\cdots\) (using \(\Gamma(1)=1,\ \Gamma(3)=2\)). Odd powers of \(t\) are absent, so there is no \(\lambda^{-2}\) term.
(B) For finite upper limit, find the leading large-\(\lambda\) behaviour of \(I(\lambda)=\int_0^{1}\sqrt{t}\,e^{-\lambda t}\,dt\) and justify that the endpoint \(t=1\) is negligible.
Solution
Here \(f(t)=t^{1/2}\), \(\alpha_0=\tfrac12\), \(a_0=1\). Leading term \(\Gamma(\tfrac32)/\lambda^{3/2}=\tfrac{\sqrt\pi}{2}\,\lambda^{-3/2}\). The correction from replacing \(\int_0^1\) by \(\int_0^\infty\) is \(-\int_1^\infty t^{1/2}e^{-\lambda t}dt\), which is \(O(e^{-\lambda})\) — exponentially smaller than any power \(\lambda^{-p}\), hence negligible to all algebraic orders. So \(I(\lambda)\sim \dfrac{\sqrt{\pi}}{2}\,\lambda^{-3/2}\). At \(\lambda=20\): \(\tfrac{1.7725}{2}\cdot20^{-3/2}=0.8862\cdot0.011180=9.91\times10^{-3}\).
(C) Derive the full asymptotic series of \(I(\lambda)=\int_0^{\infty}\dfrac{e^{-\lambda t}}{1+t}\,dt\), and estimate the smallest achievable error via optimal truncation at \(\lambda=10\).
Solution
From \(1/(1+t)=\sum(-1)^n t^n\), Watson gives \(I(\lambda)\sim\sum_{n=0}^\infty(-1)^n n!\,\lambda^{-(n+1)}\), a divergent series (terms \(\sim n!/\lambda^{n+1}\)). The term magnitude \(t_n=n!/\lambda^{n+1}\) is minimised where \(t_{n+1}/t_n=(n+1)/\lambda\approx1\), i.e. \(n\approx\lambda=10\). The least term is \(t_{10}=10!/10^{11}=3628800/10^{11}=3.63\times10^{-5}\). Optimal truncation (stop just before the least term) yields an error of that order, \(\approx3.6\times10^{-5}\sim e^{-\lambda}\) (indeed \(e^{-10}=4.5\times10^{-5}\)). This is the exponentially small "beyond-all-orders" remainder characteristic of Watson-type series.
(C) Combine a geometric expansion with a half-integer singularity: find the first three terms of \(I(\lambda)=\int_0^{\infty}\dfrac{e^{-\lambda t}}{(1+t)\sqrt{t}}\,dt\) as \(\lambda\to\infty\), and evaluate at \(\lambda=4\).
Solution
Expand the integrand at \(t=0\): \((1+t)^{-1}t^{-1/2}=t^{-1/2}(1-t+t^2-\cdots)=t^{-1/2}-t^{1/2}+t^{3/2}-\cdots\), so \(a_n=(-1)^n\), \(\alpha_n=n-\tfrac12\). Watson: \(I\sim\sum_n(-1)^n\Gamma(n+\tfrac12)\lambda^{-(n+1/2)}\). Using \(\Gamma(\tfrac12)=\sqrt\pi,\ \Gamma(\tfrac32)=\tfrac{\sqrt\pi}{2},\ \Gamma(\tfrac52)=\tfrac{3\sqrt\pi}{4}\): \[I(\lambda)\sim\sqrt{\pi}\,\lambda^{-1/2}-\tfrac{\sqrt\pi}{2}\,\lambda^{-3/2}+\tfrac{3\sqrt\pi}{4}\,\lambda^{-5/2}-\cdots\] At \(\lambda=4\) (\(\sqrt\pi=1.7725\)): \(\lambda^{-1/2}=0.5,\ \lambda^{-3/2}=0.125,\ \lambda^{-5/2}=0.03125\). Terms: \(1.7725(0.5)=0.8862\); \(-0.8862(0.125)=-0.1108\); \(+1.3293(0.03125)=+0.0415\). Sum \(\approx0.8169\). (Exact value \(\int_0^\infty e^{-4t}/[(1+t)\sqrt t]\,dt\approx0.806\); the three-term series is within \(\sim1.5\%\), the discrepancy being the next, \(\lambda^{-7/2}\), term.)