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For real $x\to +\infty$, deformation into the upper half-plane separates a logarithmically singular endpoint at zero from the smooth endpoint at one. The first contributes the exact logarithmic term displayed. The second yields its factorial asymptotic expansion by Watson's lemma. The two endpoints must both be retained; repeated integration by parts without treating the logarithmic singularity loses its leading contribution.

The Gamma reflection formula gives $(\sin \pi z/\pi )\Gamma (z)=1/\Gamma (1-z)$, analytic at zero with value one. A finite-parameter Watson's lemma representation of Ramanujan's function therefore gives the displayed coefficients. In particular $c_0=1$, $c_1=-\gamma$ and $c_2={\gamma }^2-{\pi }^2/6$, with $\gamma$ Euler's constant. Obtaining this representation requires taking an imaginary part before sending its Mellin exponent to zero; the original complex integral is divergent at that boundary.

For $g(s)=-(1/a)e^{ias}{\int }_0^{\infty }{x}^{s+\sigma -1}{e}^{-tx}/(ia+\log x)dx$ with positive $a,t,\sigma$ and $s\ge 0$, differentiation cancels the logarithmic denominator and gives $g_s=-(1/a)e^{ias}\Gamma (s+\sigma )t^{-s-\sigma }$. Integrating this derivative expresses $g(0)$ as a finite parameter-Laplace integral plus $g(s)$. The elementary lower bound $|ia+\log x|\ge a$ gives the displayed remainder, smaller than every logarithmic term at fixed $s>0$. Watson's lemma then determines all coefficients from the finite integral's endpoint Taylor series.

Fix a small $s>0$, for example $s<\sigma /2$, independently of $t$. Taking imaginary parts of the identity in part (a) givesThe parameter-shift remainder bound for a logarithmic Laplace integral makes the last term $O(t^{-\sigma -s})$, smaller than $t^{-\sigma }{L}^{-k}$ for every fixed $k$.

The endpoint function $h$ is analytic near zero. For any fixed $N$, its Taylor expansion isIntegrating this expansion against $e^{-Lr}$ useswhile the error is $O(L^{-N-2})$ and the extension from $s$ to infinity contributes an exponentially small tail. This is Watson's lemma, and givesEquivalently the coefficients have the stated exponential generating function $h(r)={\sum }_{n\ge 0}{\beta }_n{r}^n/n!$. In particular,Thus the apparent $t^{-\sigma }(\log t{)}^{-1}$ term vanishes: the first nonzero terms are $t^{-\sigma }[\Gamma (\sigma )(\log t{)}^{-2}+2{\Gamma }^{\prime }(\sigma )(\log t{)}^{-3}+\cdots ]$. The derivatives of the gamma function supply the successive logarithmic coefficients.

The relevant boundary values are $a=\pi$ and $\sigma =0$, but the original complex integral defining $g(0)$ requires $\sigma >0$. Substituting zero into it is invalid: its real part has a nonintegrable $dx/(x\log x)$ singularity near zero. Its imaginary part is different, since for positive $\sigma$This has a finite limit as $\sigma \downarrow 0$. On $(0,1)$, $x^{\sigma -1}\le x^{-1}$, whose displayed logarithmic denominator makes the integral finite; at infinity, a fixed upper bound on $\sigma$ and exponential damping provide an integrable dominator. Thus dominated convergence gives $N(t)={\lim }_{\sigma \downarrow 0}F(t,\pi ,\sigma )$.

For the asymptotics, take the imaginary-part parameter identity before passing to that limit, with a fixed $0<s<1/2$. Its endpoint factor converges toby the Gamma reflection formula. Near zero, $\sin (\pi r)\Gamma (\sigma +r)$ is bounded uniformly for small $\sigma \ge 0$, because the sine zero compensates the gamma pole. At $s>0$, the remainder integral itself remains absolutely convergent at $\sigma =0$ and is $O(t^{-s})$. ConsequentlyThe endpoint function is analytic with value one, so Watson's lemma proves the logarithmic expansion of Ramanujan's functionIn particular, with $\gamma$ Euler's constant,This use of imaginary-part regularization at a zero Mellin exponent matters: the positive-$\sigma$ coefficient ${\beta }_0$ in part (b) is always zero, whereas the boundary endpoint function has $h_0(0)=1$. The fixed-positive-parameter expansion is not uniform as $\sigma \downarrow 0$, so its coefficients cannot simply be sent to that boundary term by term.

There are two contributing regions: an endpoint of width $x^{-1}$ near $u=0$, and a very narrow peak near $u=1$. Both must be retained in the asymptotic expansion.

For the peak, set $a=e^{-x}$ and $u=1+av$. At fixed $v$, $\log u\sim av$ and $e^{-xu}\sim a$, so the integrand times $du$ approaches $dv/(1+v^2)$. This is the Lorentzian contribution from shrinking regularization. Its total contribution isFor example, cut off this region at $|u-1|=e^{-x/2}$: its scaled endpoints tend to infinity, while the changes in the exponential and logarithm inside the peak are exponentially small. The matched tails outside it are also exponentially small compared with the endpoint terms below.

At the endpoint, put $v=xu$ and $L=\log x$. To algebraic orders in $L^{-1}$, the regularizing denominator term is negligible there, giving the Laplace endpoint expansion with a logarithmic denominatorThis integral represents the endpoint expansion only; it is cut off before the original peak when making the argument precise. Since ${\Gamma }^{\prime }(1)=-{\gamma }_E$, with ${\gamma }_E$ the Euler--Mascheroni constant, the first two contributions to the original integral areThe endpoint correction refines to $[x{L}^2{]}^{-1}[1-2{\gamma }_E/L+3({\gamma }_E^2+{\pi }^2/6)/L^2+\cdots ]$, using derivatives of the gamma function. An application of Watson's lemma confined to $u=0$ would omit the leading peak contribution $\pi$.

First take $z>0$. Set $s=t^3-1$ in the tail integral. This puts it in the endpoint form needed for Watson's lemma:The binomial coefficients are $(-1{)}^r(2/3{)}_r/r!$, where $(a{)}_r=\Gamma (a+r)/\Gamma (a)$ is the rising factorial. Integrating each endpoint term givesAlso, scaling the complete positive-ray integral gives ${\int }_0^{\infty }{e}^{-z^3{t}^3}dt=\Gamma (1/3)/(3z)$. HenceThe endpoint term is exponentially small on this positive ray but must be retained for continuation to other sectors.

For complex $z$, use $s=zt$ so $G(z)=z^{-1}{\int }_0^z{e}^{-s^3}ds$. Put $C=\Gamma (1/3)/3$ and $\omega =e^{2\pi i/3}$. The complete steepest descent contours from the cubic saddle are the rays $\arg s=2\pi j/3$, whose integrals are $C{\omega }^j$. Deform the finite contour onto one of these rays followed by the endpoint contour. The cubic exponential integral Stokes sectors giveFor $0\le \arg z\le 2\pi$, take $j=0$ below $\pi /3$, $j=1$ between $\pi /3$ and $\pi$, $j=2$ between $\pi$ and $5\pi /3$, and $j=3$ above $5\pi /3$. The corresponding algebraic coefficients are $C,C\omega ,C{\omega }^2,C$. A negative-real $z$ provides a useful sign check: $G(z)=G(|z|e^{i\pi })$ is real and positive, so its two lateral algebraic coefficients must be complex conjugates.

The subdominant algebraic term switches across the Stokes linesThere $z^3$ is negative real and the endpoint exponential is dominant. The intervening anti-Stokes lines are $\arg z=\pi /6+j\pi /3$, where $\mathrm{Re}{z}^3=0$ and the exponential changes from growth to decay. The naming convention here distinguishes switching from equal exponential magnitude.

To resolve the jump near ${\theta }_s=\pi /3$, optimally truncate the endpoint series at $N$ with $N+2/3=R+O(1)$, where $R=|z{|}^3$. Let $P_N$ denote that truncated endpoint contribution. The Borel remainder for a cubic exponential integral derived in solution, together with the Gaussian pole transition for a large gamma parameter in solution, givesThis estimate is on the transition scale $\theta -\pi /3=O(R^{-1/2})$ after optimal truncation. The error-function smoothing of a Stokes multiplier has width $O(|z{|}^{-3/2})$. Before the line $S\to 0$ and the coefficient is $C$; after it $S\to 1$ and the coefficient is $C\omega$, agreeing with the sector expansions. On the line the leading coefficient is their average. The original $G$ is an entire function; only its asymptotic decomposition has these sector-dependent coefficients.

Take the principal complex logarithm and deform the interval upwards, where $e^{ixt}$ decays for $x>0$. The small indentation at zero contributes nothing because $r\log r\to 0$; the distant horizontal segment vanishes exponentially. The two vertical edges therefore give the exact representationThis is an endpoint steepest descent contour, with distinct contributions from the singular endpoint and the smooth endpoint. Setting $u=xs$ in the first integral and using the given logarithmic moment givesFor the second edge, the Taylor series at $s=0$ is $\log (1+is)={\sum }_{n\ge 1}(-1{)}^{n+1}(is{)}^n/n$. Watson's lemma integrates it term by term as an asymptotic expansion, using ${\int }_0^{\infty }{s}^n{e}^{-xs}ds=n!/x^{n+1}$. ConsequentlyHere $\gamma$ is the Euler--Mascheroni constant. The first smooth-endpoint terms areFor every fixed $N\ge 1$, retaining $k=0,\dots ,N-1$ leaves $O(x^{-N-2})$. The factorial series is an asymptotic series, not a convergent sum. Treating the logarithmic endpoint by an ordinary smooth-endpoint integration formula would lose the leading logarithm; this is the logarithmic endpoint Fourier asymptotics mechanism.

Under local integrability and suitable bounds away from zero, Watson's lemma givesEach term follows by $u=\lambda t$ and the Gamma function integral; a finite local remainder is bounded in the same way, and the region away from zero is exponentially small.

For Laplace's method, assume an isolated nondegenerate global minimum $t_{\ast }$ of $p$, with $p^{\prime \prime }(t_{\ast })>0$ and suitable tail bounds. Expand $p,q$ near $t_{\ast }$ and set $t-t_{\ast }=v/\sqrt{z}$. The leading term is $e^{-zp(t_{\ast })}q(t_{\ast })\sqrt{2\pi /(z{p}^{\prime \prime }(t_{\ast }))}$; successive terms come from Gaussian moments of the Taylor expansions. Several equally low minima contribute additively, while degenerate minima need a different scale.

Here the integrand is entire in $t$, so deform the contour to the line at imaginary part $-\pi$, the vertical segment from $-i\pi$ to $i\pi$, and the line at imaginary part $\pi$. The two horizontal tails sum to $2{\int }_0^{\infty }{e}^{-z\cosh x}dx=O(e^{-z}{z}^{-1/2})$. The vertical segment contributes $i{\int }_{-\pi }^{\pi }{e}^{z\cos \theta }d\theta$. Its maximum is at zero; putting $\theta =v/\sqrt{z}$ and using $\cos \theta =1-{\theta }^2/2+{\theta }^4/24+\cdots$ givesThe fourth Gaussian moment is three times the zeroth, soSince $\Gamma (1/2)=\sqrt{\pi }$, this has the required normalization.

Watson's lemma gives, under integrability away from zero and the stated endpoint asymptotic expansion,Indeed each power integrates over the positive half-line to $\Gamma (n\beta +1){\lambda }^{-(n\beta +1)}$; replacing a fixed upper endpoint by infinity has exponentially small error. A remainder $o(t^{N\beta })$ near zero gives $o({\lambda }^{-N\beta -1})$ after splitting off a small endpoint neighbourhood and rescaling $u=\lambda t$.

For Laplace's method, assume the phase and amplitude are smooth near the nondegenerate maximum, as required for a full expansion. Localize to a small neighbourhood of $c$: outside it the unique maximum and continuity make the exponential smaller by a factor $e^{-\lambda \epsilon }$. SetNear $c$ this is a smooth increasing coordinate, with ${\zeta }^{\prime }(c)=\sqrt{|{\varphi }^{\prime \prime }(c)|}$. It need not be monotonic on the entire integration interval. In the localized integral put $A(\zeta )=F(t(\zeta ))dt/d\zeta$. Extending its Gaussian integration to both infinite endpoints incurs exponentially small error. Taylor expansion of $A$ and integration of even Gaussian moments givesHere $A(0)=F(c)/\sqrt{|{\varphi }^{\prime \prime }(c)|}$, giving the leading term $\boxed{e^{\lambda \varphi (c)}F(c)\sqrt{2\pi /(\lambda |{\varphi }^{\prime \prime }(c)|)}}$. The asymptotic-equivalence notation for that term presumes $F(c)\ne 0$; if it vanishes the series determines the first nonzero term. The maximum condition alone, without amplitude regularity, would not imply a full expansion.

For the Gamma function, set $t=xs$:The phase has unique maximum $-1$ at $s=1$, with second derivative $-1$. To obtain the correction, put $s=1+u/\sqrt{x}$. Expansion givesand hence a Gaussian multiplierThe odd term integrates to zero. The normalized Gaussian fourth and sixth moments are 3 and 15, so the correction is $-3/4+15/18=1/12$. Localization near one, with the exponentially suppressed endpoint tails, justifies termwise expansion. Thus Stirling's formula is

One convenient version of Watson's lemma is the following. Suppose $G(t)$ is integrable away from zero and has, as $t\downarrow 0$, an asymptotic expansion $G(t)\sim {\sum }_{j\ge 0}{c}_j{t}^{\beta +j-1}$, where $\beta >0$. On a finite interval, or on an infinite interval with suitable exponential growth control,To prove a finite truncation, split the integral at a fixed small $\delta >0$. The part away from zero is exponentially small. On $(0,\delta )$ subtract the first $N$ local terms and bound the remainder by $C{t}^{\beta +N-1}$; its integral is $O({\lambda }^{-\beta -N})$. Extending each polynomial integral to infinity changes it only exponentially, and the substitution $u=\lambda t$ produces the Gamma integral. A little-o local remainder gives the corresponding little-o asymptotic remainder. This is a proof by localization and remainder bounds, rather than merely formal integration of a series.

For the monotone-phase integral, assume the derivatives needed below exist and are bounded near $a$, and the rest of the integral is controlled. Put $t=\varphi (a)-\varphi (x)$. Since ${\varphi }^{\prime }(a)<0$, this is a regular endpoint coordinate andDifferentiating with $dx/dt=-1/{\varphi }^{\prime }$ gives $G(0)=-g(a)/{\varphi }^{\prime }(a)$ and $G^{\prime }(0)=g^{\prime }(a)/{\varphi }^{\prime }(a{)}^2-g(a){\varphi }^{\prime \prime }(a)/{\varphi }^{\prime }(a{)}^3$. Watson's lemma therefore gives the two-term monotone-endpoint Laplace expansionAn $O(e^{\lambda \varphi (a)}{\lambda }^{-3})$ remainder follows, for example, from a sufficiently smooth transformed amplitude with bounded second derivative near zero. The displayed second coefficient can vanish for special amplitudes without changing the expansion.

The unheaded transition request concerns a cubic endpoint-to-saddle transition. Near $x=0$ its exponent has the expansionAt the cubic endpoint $x$ has size ${\lambda }^{-1/3}$. Requiring the linear term to contribute on the same scale gives $1-\alpha =O({\lambda }^{-2/3})$. Thus the distinguished limit and its uniform leading integral areFor fixed $\nu$, the substitution $x={\lambda }^{-1/3}t$ makes the omitted exponent terms $O({\lambda }^{-2/3})$ on bounded $t$ intervals. Cubic decay controls the remaining tail, while the original interval away from zero is exponentially negligible. The estimate is uniform for $\nu$ in compact sets. In the printed range $\alpha \le 1$, $\nu \ge 0$; the same transition integral also makes sense for negative fixed $\nu$.

At $\nu =0$, the Gamma integral gives $h(0)=6^{1/3}\Gamma (1/3)/3$, recovering the cubic endpoint regime. For $\nu \to +\infty$, scale $t=u/\nu$ and expand $e^{-u^3/(6{\nu }^3)}$ to obtain $h(\nu )={\nu }^{-1}-{\nu }^{-4}+O({\nu }^{-7})$. Its leading term gives ${\lambda }^{-1/3}/\nu =1/[\lambda (1-\alpha )]$, recovering the ordinary endpoint regime in their overlap. Both matching limits are therefore checked explicitly.

Watson's lemma applies to Laplace-type endpoint integrals such as ${\int }_0^b{e}^{-\lambda t}f(t)dt$ as $\lambda \to +\infty$, when $f$ has a local algebraic expansion $f(t)\sim {\sum }_j{c}_j{t}^{p+j-1}$ with $p>0$ and suitable integrability or exponential-growth control away from the endpoint. Each term contributes $c_j\Gamma (p+j){\lambda }^{-p-j}$. The exponential selects a neighborhood of $t=0$.

Laplace method instead treats ${\int }_a^{b}g(t)e^{\lambda \varphi (t)}dt$ by locating the dominant maxima of a real phase $\varphi$. A smooth nondegenerate interior maximum produces a neighborhood governed by a Gaussian function of width ${\lambda }^{-1/2}$; endpoint or degenerate maxima require their corresponding local scaling. Contributions away from a strict dominant maximum must be smaller. These methods overlap after suitable changes of variable, but their usual starting forms emphasize an endpoint amplitude expansion and an exponential maximum, respectively.

For the Gamma function, put $t=xu$. ThenThe phase has its unique maximum at $u=1$, with value $-1$ and second derivative $-1$. Set $u=1+s/\sqrt{x}$. Near this maximum,Combining the amplitude and exponential expansions in Laplace method givesLocalization near $u=1$ justifies extending the limits for the Gaussian integral; the distant endpoint is not being expanded uniformly by this local series. Odd terms integrate to zero. The moments of the Gaussian integral satisfy $I_1=I_0$, $I_2=3I_0$, $I_3=15I_0$, so the relative correction isThe two terms of the Stirling formula, obtained by the Stirling correction by Gaussian moments, are therefore

Use the principal complex logarithm in the upper half-plane. Deform the interval upward into a half-strip: the vertical side at zero is traversed upward, and the side at one downward. The top side vanishes exponentially; a small indentation at zero contributes $O(r|\log r|)$ and vanishes. Thus the contour deformation gives the exact representation, for $x>0$,Since $\log (is)=\log s+i\pi /2$, scaling $u=xs$ and using the integral defining Euler's constant givesFor the second endpoint, $\log (1+is)=is+s^2/2+O(s^3)$. Watson's lemma, or integration of these powers against $e^{-xs}$ with an exponentially small tail, yieldsIn particular the required first terms areThe two different endpoint scales come from the logarithmic singularity at zero and the simple zero of the amplitude at one.

Near zero, write the amplitude as $t^{-1/4}(1+t{)}^{-1/4}={\sum }_{m\ge 0}(\genfrac{}{}{0ex}{}{-1/4}{m})t^{m-1/4}$. The large parameter localizes the integral at $t=0$. Termwise integration of the local expansion givesConsequentlyHere $(1/4{)}_m$ is the rising factorial. This is an asymptotic series, not an assertion that the full integrated Taylor series converges.

The needed result is Watson's lemma: if a locally integrable amplitude has an expansion $g(s)\sim \sum c_m{s}^{{\rho }_m-1}$ as $s\downarrow 0$, with ${\rho }_m>0$ increasing and suitable growth at infinity so its Laplace integral converges, then ${\int }_0^{\infty }{e}^{-ks}g(s)ds\sim \sum c_m\Gamma ({\rho }_m)k^{-{\rho }_m}$, with the corresponding finite-order remainder bounds. Apply it after $s=t^2$, when $g(s)=\frac{1}{2}{s}^{-5/8}(1+\sqrt{s}{)}^{-1/4}$ and ${\rho }_m=3/8+m/2$. The tail is exponentially small after any fixed positive cutoff.