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Under the corresponding holomorphic and uniform remainder hypotheses, the expansion in Watson lemma remains valid for a complex large parameter $z$, uniformly in every closed subsectorwith $0<\delta <\pi /2$. A rotation of an endpoint contour reduces oscillatory factors to this decaying Laplace form.

A logarithmic singularity behaves locally like $a\log (z-z_0)$ with $a\ne 0$, on a chosen branch. Although unbounded at the endpoint, it is integrable along a finite ray. In a rapidly oscillating endpoint integral its contour rotation produces a logarithm of the large parameter and a branch-angle contribution; this differs from the ordinary smooth-amplitude Watson lemma.

Use the entire function $e^{ik{z}^3}$. The steepest descent contour starting at $1$ iswith the cube-root branch starting at $1$. Along it $ik{z}^3=ik-ku$, so the real part decreases and the imaginary part stays constant. It approaches the ray $\arg z=\pi /6$, on which $z=e^{i\pi /6}{u}^{1/3}$ and $ik{z}^3=-ku$.

Close a contour using the interval $[0,1]$, these two outgoing paths truncated at parameter $R$, and a short connecting arc. The two endpoints differ by $O(R^{-2/3})$, and the integrand on the connector decays exponentially in $R$. Cauchy integral theorem therefore gives ${\int }_0^1{e}^{ik{t}^3}dt=I_1-I_2$.

The specified substitutions giveandExpand $(1+iu{)}^{-2/3}=1-2iu/3+O(u^2)$ near zero, and apply Watson lemma, or split the integral at a fixed small $u$ and integrate the Taylor remainder. This gives the first two asymptotic termsKeeping the next term gives $-2e^{ik}/(9k^2)+O(k^{-3})$. The fractional power comes from the degenerate stationary endpoint at zero; the oscillatory integer-power terms come from the ordinary endpoint at one.

Let $S(t)=a\sinh t-t$. For $a>1$, $S^{\prime }(t)=a\cosh t-1>0$ on the integration interval, so its minimum is the endpoint $t=0$. The Taylor series isOn the contributing scale $t=O({\nu }^{-1})$, the higher terms are negligible. The endpoint form of Laplace's method, or the Watson lemma, therefore gives

Watson lemma states that ifand the Laplace integral has suitable growth control away from zero, then

The Taylor series of the amplitude isTermwise application of Watson lemma therefore gives the asymptotic expansionIn particular, the leading asymptotic approximation is $I(x)\sim 2x^{-3}$.

SetThenThe Taylor series at the endpoint isand henceThe Watson lemma, together withtherefore yields

One form of Watson lemma is this. If $f(t)$ has the asymptotic expansion $f(t)\sim {\sum }_{n\ge 0}{a}_n{t}^{{\lambda }_n-1}$ as $t\downarrow 0$, with $0<{\lambda }_0<{\lambda }_1<\cdots$, and the Laplace integral has suitable growth control away from zero, thenas $x\to \infty$, with truncation after any fixed number of terms justified by the corresponding remainder estimate.

Complete the square: $z^2-2iz=(z-i{)}^2+1$. The pole of $(1-iz{)}^{-1}$ is at $z=-i$, so the contour may be shifted upward through the saddle point $z=i$ without crossing a singularity. Setting $z=t+i$ gives the exact steepest-descent representationNear the saddle,Odd terms integrate to zero, whileThe Watson lemma therefore yields the full asymptotic expansion

Watson lemma states that if, near zero,and the tail is suitably controlled, thenas $x\to +\infty$.

Near the endpoint $t=0$,The Watson lemma integrates this local expansion term by term, andTherefore the reciprocal-amplitude Laplace integral has expansionso

Writing $t=x+u$ gives $E(x)=e^{-x}{x}^{-1}{\int }_0^{\infty }{e}^{-u}(1+u/x{)}^{-1}du$. Expanding the last factor and applying Watson lemma yields the same coefficients $(-1{)}^n\Gamma (n+1)$.