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For real $z>0$, defineThe integrand is positive and at most $e^{-t}$, so $0<F(z)<1$ and the improper integral converges. Substituting $s=z+t$ also gives $F(z)=z{e}^z{E}_1(z)$ in terms of the exponential integral; the original representation makes positivity and the large-$z$ behavior transparent.

Apply the exact finite geometric series identity inside the integral, without assuming that an infinite series can be integrated term by term:The Gamma integral gives ${\int }_0^{\infty }{e}^{-t}{t}^{k}dt=k!$. Therefore, with $N$ retained terms,This remainder formula proves the asymptotic expansion at every fixed order: $R_N=O(z^{-N})=o(z^{-N+1})$ as $z\to +\infty$. Its sign also gives alternating bounds: an odd $N$ overestimates $F(z)$, and an even $N$ underestimates it.

At $z=10$, an independent evaluation of the integral gives $F(10)\approx 0.9156333394$. Three retained terms giveFive retained terms give $S_5(10)=0.9164$ with error bounded by $5!/10^5=0.0012$. The actual errors are about $0.004367$ and $0.000767$, respectively. The error bound is conservative, but established without relying on the numerical integral evaluation.

Nevertheless, the formal infinite series diverges at every fixed $z>0$. The magnitudes $T_k=k!/z^k$ satisfy $T_{k+1}/T_k=(k+1)/z$, so they eventually increase without bound. In particular, the terms fail the necessary zero-term condition for a convergent series. The finite remainder identity remains valid after convergence has failed; this is precisely what makes the example useful.

For $p,q>0$, express the product of the two Gamma integrals as the nonnegative double integralUse $r=x+y$ and $u=x/(x+y)$, with inverse $x=ru$, $y=r(1-u)$ and Jacobian determinant of absolute value $r$. The new domain of integration is $r>0$, $0<u<1$, and the integrand with this factor becomes $r^{p+q-1}{e}^{-r}{u}^{p-1}(1-u{)}^{q-1}$. Tonelli's theorem and separation of the two integrals therefore give $\Gamma (p+q)B(p,q)$. The positive real parameter hypothesis ensures that all the original integrals are finite.

For complex parameters with positive real parts, the same proof first gives the undivided identity $\Gamma (p+q)B(p,q)=\Gamma (p)\Gamma (q)$. Absolute integrability follows by taking real parts in the exponents, so Fubini's theorem applies. The nonvanishing of the Gamma function then permits division by $\Gamma (p+q)$.

The integral $I(x)={\int }_0^{\infty }{e}^{xt-t^3/3}dt$ is a rescaled Scorer Hi function. It has the three useful limitsFor the negative-argument limit, scale $t=s/M$ and use the dominated convergence theorem. At zero, substitute $s=t^3/3$ and use the Gamma integral. For the positive-argument limit, the exponent has its maximum at $t=\sqrt{M}$, with second derivative $-2\sqrt{M}$; Laplace's method gives the displayed factor. These estimates describe a cubic endpoint-to-saddle transition.

Suppose a real phase has its unique minimum at the left endpoint and $\varphi (t)-\varphi (0)\sim c{t}^m$, with $c>0$ and integer $m\ge 1$. For a continuous amplitude $a$ with $a(0)\ne 0$, localization followed by $s=(\nu c{)}^{1/m}t$ givesThis uses the Gamma integral. It assumes sufficient decay away from the minimum to make the remaining integral negligible. A cubic endpoint has width ${\nu }^{-1/3}$; an ordinary quadratic endpoint has width ${\nu }^{-1/2}$.

The Gamma function extends factorials to a continuous argument. Its defining Gamma integral iswhere $t^{z-1}=\exp ((z-1)\log t)$ uses the real natural logarithm for $t>0$. The Gamma function recurrence gives its meromorphic continuation to the complex plane, with simple poles at $0,-1,-2,\dots$ and no zeros. See Gamma function residue at a nonpositive integer and Gamma function has no zeros for proof outlines.^[ref]

For nonnegative integers $n$, the factorial identity has a shift of one:Thus $\Gamma (n)=(n-1)!$ for positive integers $n$. Half-integer arguments supply a different familiar constant:The first value follows from the Gaussian integral; the second follows from the Gamma function recurrence.^[ref]

The plot shows only positive real arguments; the Gamma function also accepts complex numbers away from its poles. Matching the factorials alone does not select a unique interpolation, even if the Gamma function recurrence is imposed: $f(x)=\Gamma (x)\exp (\sin (2\pi x))$ has the same positive-integer values and the same recurrence relation. The Bohr-Mollerup theorem supplies the extra condition that selects the Gamma function.

The useful idea is to replace a discrete product of numbers by an integral whose exponent varies continuously. The Gamma integral then evaluates whole families of integrals at once, provides the normalizing constant for the Gamma distribution in probability theory, and appears in physics, for example in the Bose integral. Its large-positive-argument growth reconnects this continuous description to factorials through Stirling's formula.^[ref]

Historically, Leonhard Euler studied the interpolation of factorials in his paper E19, written in 1729 and published in 1738. His integral over $(0,1)$ becomes the Gamma integral after a change of variables, with the same one-step shift in the argument.^[ref]

The Gamma function obeys the recurrence relationInitially, for $\mathrm{Re}z>0$, integration by parts in the Gamma integral gives the short proofThe boundary term vanishes at zero because $|t^z|=t^{\mathrm{Re}z}\to 0$, and at infinity because of exponential decay. Together with $\Gamma (1)={\int }_0^{\infty }{e}^{-t}dt=1$, the Gamma function recurrence immediately gives $\Gamma (n+1)=n!$ for every nonnegative integer $n$.^[ref]

Read backwards, the Gamma function recurrence defineswhere $m$ is a positive integer. These overlapping formulas agree by the Gamma function recurrence, giving the meromorphic continuation. Since $\Gamma (1)=1$, they give a simple pole at zero and then at every negative integer; Gamma function residue at a nonpositive integer records the residues. The recurrence relation therefore holds throughout the continuation as an identity of meromorphic functions.

For real $a>0$ and $\mathrm{Re}s>0$, the change of variables $u=at$ in the defining improper integral for the Gamma function givesThe improper integral converges absolutely: near zero its absolute integrand behaves like $t^{\mathrm{Re}s-1}$, and at infinity the exponential function overcomes every fixed power.^[ref]

For real $s=1,2,4$ and $a=1$, the integrands are $e^{-t}$, $t{e}^{-t}$, and $t^3{e}^{-t}$, with total areas $\Gamma (1)=1$, $\Gamma (2)=1$, and $\Gamma (4)=6$. The third integrand is taller and has six times the area of the first, although the exponential function makes each tail decay.

For real $s>0$, the kernel $t^{s-1}{e}^{-t}$ becomes the probability density function of a Gamma distribution with shape $s$ and unit rate only after division by its normalizing constant $\Gamma (s)$. In particular, the shape-four probability density function is $t^3{e}^{-t}/6$. More generally, the Gamma integral verifies that $a^s{t}^{s-1}{e}^{-at}/\Gamma (s)$ integrates to one for shape $s>0$ and rate $a>0$.

For the half-integer evaluation, the change of variables $t=u^2$ reduces the Gamma integral to the Gaussian integral:Consequently, for real $u>0$,

Repeated integration by parts gives the Gamma integralThe logarithm of its integrand is $n\log t-t$, whose unique maximum is at $t=n$. Write $t=n+\sqrt{n}s$. For fixed $s$ and large $n$, a Taylor expansion givesThe Big O notation here is for a fixed $s$, or uniformly for $s$ in a fixed bounded interval. It does not claim a uniform approximation over the entire growing integration domain. Consequently the peak, divided by its height, approaches $e^{-s^2/2}$.

The exact substitution givesReplacing the integral by the Gaussian integral ${\int }_{\mathbb{R}}{e}^{-s^2/2}ds=\sqrt{2\pi }$ produces $S_n$. Thus the estimate is peak height $n^n{e}^{-n}$ times width $\sqrt{n}$ times the area of a standard Gaussian peak. This is Laplace's method in a concrete form; controlling the integral's tails justifies the replacement. The next terms, including their products after exponentiation, produce the Stirling correction by Gaussian moments.

The Mellin transform is an integral transform suited to changes of scale: replacing $f(x)$ by $f(ax)$ multiplies its transform by $a^{-s}$ for $a>0$. For a locally integrable function $f:(0,\infty )\to \mathbb{C}$, definewhere the improper integral converges. The complex number $x^{s-1}=\exp ((s-1)\log x)$ uses the real natural logarithm because $x>0$. Its absolute value is $x^{\sigma -1}$, so absolute convergence depends on the real part $\sigma$ rather than the imaginary part $\tau$. The Mellin fundamental strip records the open range of such real parts.^[ref]

A familiar example is the exponential function $f(x)=e^{-x}$:This is exactly the Gamma integral. Near zero $e^{-x}$ tends to one, so absolute convergence requires $\sigma >0$; at infinity $x^r{e}^{-x}\to 0$ for every real $r$, so there is no upper boundary. The Gamma function has a meromorphic continuation with simple poles at $0,-1,-2,\dots$, but a finite continued value elsewhere in $\sigma \le 0$ does not make the defining improper integral converge.^[ref]

The Mellin transform of an interval indicator gives an elementary example with absolute convergence for every complex number $s$. The Mellin transform in logarithmic coordinates explains why changes of scale become translations, connecting the scaling property of the Mellin transform to the Fourier transform.

Use $\Xi (u)={\pi }^{-u/2}\Gamma (u/2)\zeta (u)$ for the completed Riemann zeta function, without the polynomial factor used to define the entire Riemann xi function. For complex $\nu$ putThis is the Modified Bessel function of the second kind. The substitution $t=e^v$ writes it as $\frac{1}{2}{\int }_{\mathbb{R}}{e}^{-Y\cosh v+\nu v}dv$, proving that it is entire in $\nu$, that $K_{\nu }=K_{-\nu }$, and that it decays exponentially as $Y\to \infty$, uniformly for $\nu$ in a compact set.

For $\Re s>1$ all the lattice sums and the integrals below converge absolutely. The $m=0$ terms in the completed nonholomorphic Eisenstein series contribute $\Xi (2s)y^s$. For $m\ne 0$, use the Gamma integral and then Poisson summation in $n$:Here the Poisson identity is ${\sum }_n{e}^{-\pi u(n+mx{)}^2}=u^{-1/2}{\sum }_h{e}^{-\pi h^2/u}{e}^{2\pi ihmx}$. At $h=0$, evaluating the Gamma integral and summing over $m$ gives $\Xi (2s-1)y^{1-s}$. At $h\ne 0$, the substitution $u=|h|t/(|m|y)$ and the integral representation of the modified Bessel function of the second kind giveCollect the frequency $r=mh\ne 0$. There are two signed choices of $m$ for each positive divisor of $|r|$. Thus, with ${\sigma }_a(n)={\sum }_{d\mid n}{d}^a$, the full Fourier expansion isThe factor $2$ is important: the lattice sum includes both signs, while the defining factor $1/2$ was already used in its Mellin integral. This expansion uses exactly the completion in the question.

We can deduce both of the offered conclusions. First recall, with a short proof, the needed functional equation of the Riemann zeta function. Put $\Theta (t)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}t}$. Poisson summation gives $\Theta (t)=t^{-1/2}\Theta (1/t)$. Splitting the Mellin integral of $\Theta -1$ at one yieldsThe integral is entire and invariant under $u\mapsto 1-u$. Hence $\Xi (u)=\Xi (1-u)$, with only simple poles at $u=0,1$, of residues $-1,1$. This is the pole-bearing completion, not the entire xi function.

For fixed $y>0$, the nonconstant part of the Eisenstein expansion is entire in $s$ and locally normally convergent: the divisor sums grow at most polynomially on a compact set of $s$, whereas $K_{s-1/2}(2\pi |r|y)$ decays exponentially in $|r|$. The same argument gives local uniformity for $z$ in compact subsets of the complex upper half-plane. Thus only the constant terms can have poles. Their apparent poles at $s=1/2$ cancel: $\Xi (2s)$ has principal part $1/(2(s-1/2))$, while $\Xi (2s-1)$ has its negative, and both multiply $y^{1/2}$ there. The remaining poles are simple, at $s=0,1$, with residues $-1/2,1/2$.

Under $s\mapsto 1-s$, the two constant terms interchange by the zeta functional equation. The nonconstant terms are unchanged because $K_{\nu }=K_{-\nu }$ andwhich follows by replacing a divisor $d$ by $|r|/d$. Therefore

For the Kronecker limit formula, evaluate the finite Laurent coefficient at $s=1$. Write ${\gamma }_E$ for the Euler--Mascheroni constant. From $\zeta (u)=1/(u-1)+{\gamma }_E+O(u-1)$ and ${\Gamma }^{\prime }(1/2)/\Gamma (1/2)=-{\gamma }_E-2\log 2$ we obtainThe gamma derivative follows by differentiating the Gamma duplication formula and using ${\Gamma }^{\prime }(1)=-{\gamma }_E$. Multiplication by $y^{1-s}$ adds $-\frac{1}{2}\log y$ to the constant term; the other constant term contributes $\Xi (2)y=\pi y/6$.

The half-order Modified Bessel function of the second kind is $K_{1/2}(Y)=\sqrt{\pi /(2Y)}{e}^{-Y}$. For example, in its even integral representation set $v/2$ as variable and then $u=\sinh (v/2)$: $\cosh v=1+2u^2$ and the $\cosh (v/2)$ factor converts the integral to a Gaussian. At $s=1$ the nonconstant Eisenstein coefficients are consequently ${\sigma }_{-1}(|r|)e^{-2\pi |r|y}$. On the other hand, the absolutely convergent logarithmic series givesAdding this identity to its conjugate identifies the nonconstant sum as $-2\log |{\prod }_{n\ge 1}(1-q^n)|$. With the analytic Dedekind eta function product $\eta (z)=q^{1/24}{\prod }_{n\ge 1}(1-q^n)$, the resulting answer isThis derivation of the Kronecker limit formula uses the analytic eta product only; it does not assume its modular transformation. The lattice series is invariant for $\Re s>1$ by an integral change of the pair $(m,n)$ and $\Im (\gamma z)=y/|cz+d{|}^2$. Analytic continuation preserves this invariance, so its finite Laurent coefficient is invariant too. This supplies the independent weight-two transformation argument used in Question 2.

Mix the conditional Poisson distribution over the rate-$\nu$ exponential distribution intensity. The Gamma integral giveswhere $p=\nu /(1+\nu )$ and $q=1/(1+\nu )$. Thus the annual claim count has the zero-based geometric distribution, with probabilities $\mathbb{P}(N=0)=p$, $\mathbb{P}(N=1)=pq$ and $\mathbb{P}(N\ge 2)=q^2$.

For the intended Markov chain calculation, take these annual counts to be independent. Order the states by discounts $0,\alpha ,\beta$. From either of the first two states, a positive count returns the policy to zero, while a zero count raises it by one level. From the top, two or more claims send it to zero, exactly one sends it to $\alpha$, and no claim keeps it at $\beta$. Hence the transition matrix isWrite its stationary distribution as $({\pi }_0,{\pi }_{\alpha },{\pi }_{\beta })$. The final two balance equations give $q{\pi }_{\beta }=p{\pi }_{\alpha }$ and ${\pi }_{\alpha }=p{\pi }_0+pq{\pi }_{\beta }$. It follows that ${\pi }_{\alpha }=p{\pi }_0/[q(1+p)]$ and ${\pi }_{\beta }=p^2{\pi }_0/[q^2(1+p)]$. Normalizing yields the three-level discount equilibrium with geometric annual counts:The Markov chain is irreducible for $0<p,q<1$ and has a self-loop, so it is aperiodic and this equilibrium is the limiting distribution. Weighting the three premiums $c,c(1-\alpha ),c(1-\beta )$ givesThere is a necessary qualification to the temporal model: the one-year mixture alone does not specify independence between years. A persistent claim intensity can destroy the discount Markov property. If one fixed $\Lambda$ is drawn for the policyholder and annual counts are conditionally independent given it, then histories $0,\alpha ,\beta ,\beta$ and $0,0,\alpha ,\beta$ both end at the top after three years but give different next-year no-claim probabilities. The first history is three zero counts, so its posterior intensity has exponential rate $\nu +3$ and next no-claim probability $(\nu +3)/(\nu +4)$. The second is a positive count followed by two zero counts; its posterior density is proportional to $(1-e^{-\lambda })e^{-(\nu +2)\lambda }$, giving next no-claim probability $(\nu +2)/(\nu +4)$. Thus the displayed matrix and premium require independent annual marginal sampling, for example a fresh intensity each year. They are not valid for a persistent unobserved policy intensity without conditioning on that intensity.

Under the persistent-intensity interpretation, one can instead condition on $\lambda$. Put $a_0=e^{-\lambda }$ and $b_1=\lambda e^{-\lambda }$. The conditional transition matrix has rows $(1-a_0,a_0,0)$, $(1-a_0,0,a_0)$ and $(1-a_0-b_1,b_1,a_0)$. Its stationary distribution isThe corresponding population limiting expected value of the premium is $c{\int }_0^{\infty }[1-\alpha {\pi }_{\alpha }(\lambda )-\beta {\pi }_{\beta }(\lambda )]\nu e^{-\nu \lambda }d\lambda$. This gives the correct alternative when the intensity persists, and makes the extra assumption behind the requested marginal matrix explicit.

Use $q=e^{2\pi i\tau }$, so $g(iy)={\sum }_{n\ge 1}{b}_n{e}^{-2\pi ny}$. Put $u=\mathrm{Re}s$. For $u>\max \{0,\sigma +1\}$,The Gamma integral and absolute convergence therefore justify exchanging summation and integration. For every positive integer $n$, substitution $t=2\pi ny$ givesSumming these identities proves the Mellin transform formula for the Dirichlet series:The power of the positive real variable $y$ uses the real logarithm; the initial half-plane stated above is a sufficient one even when $\sigma$ is negative.

Normalizing the true density givesa gamma distribution of shape $1/2$ and rate $\beta$. Its mean is $1/(2\beta )=\mu$, and the Gamma integral givesPut $x=R/\beta$. Since $\mu \beta =1/2$, the positive adjustment coefficient satisfiesBoth sides are positive in this domain. Squaring, multiplying by $1-x$, and dividing out the zero root $x=0$ yieldsThe discriminant is $(1+\rho {)}^2[(\rho -3{)}^2+16\rho ]=(1+\rho {)}^2(\rho +9)(\rho +1)$. The negative constant term makes one root positive and the other negative. The polynomial is negative at zero and has value four at one, so the positive root lies in $(0,1)$ and is not an extraneous solution introduced by squaring. Therefore the adjustment coefficient for shape-one-half gamma claims is

To compare it with the exponential calculation, write $R_I=2\beta \rho /(1+\rho )$. Their difference issince $9(1+\rho {)}^2-(\rho +9)(\rho +1)=8\rho (1+\rho )>0$. Thus the two proposed Lundberg inequality bounds are ordered asOnly ${\psi }_{\mathrm{true}}(u)\le e^{-Ru}$ is justified for the true claim law. The exponential calculation gives ${\psi }_{\mathrm{exponential}}(u)\le e^{-R_{I}u}$ for a different model, not a valid guaranteed bound on the true ruin probability.

This exponential misspecification of shape-one-half gamma claims is optimistic: the true variance is $1/(2{\beta }^2)=2{\mu }^2$, whereas the assumed exponential variance is ${\mu }^2$. Its tail also decays at rate $\beta$, more slowly than the assumed rate $1/\mu =2\beta$. Equivalently, for $0<r<\beta$, the true transform $(1-2\mu r{)}^{-1/2}$ is larger than $(1-\mu r{)}^{-1}$, since $(1-\mu r{)}^2>1-2\mu r$. This stronger transform growth makes the true adjustment root smaller. Comparing upper bounds alone does not by itself prove an ordering of the two actual ruin probabilities. At $u=0$ both displayed bounds equal one.

In the nondegenerate regime, $\mu$ lies well inside the gap, with $\mu \gg kT$ and $E_g-\mu \gg kT$. Then conduction occupations are approximately $e^{(\mu -E)/(kT)}$, while the hole probability in a valence state is approximately $e^{(E-\mu )/(kT)}$. Using the near-edge density of states and the supplied Gamma integral givesThereforeThis assumes the edge approximations are valid over the thermally populated ranges. For an intrinsic semiconductor $n=p$, so each carrier density is proportional to $T^{3/2}{e}^{-E_g/(2kT)}$. Both mobile conduction Electrons and holes contribute to conductivity, $\sigma =|e|(n{b}_e+p{b}_h)$ with mobilities $b_e,b_h$. The exponentially varying carrier densities explain strong temperature dependence even when the mobilities vary only algebraically.

Donors supply Electrons and acceptors remove Electrons to create holes. With complete ionization and negligible minority carriers, the majority densities are $n\simeq N_d$ on the n side and $p\simeq N_a$ on the p side. The corresponding isolated chemical potentials areCarrier diffusion at contact leaves positively charged donors and negatively charged acceptors in a depletion layer. At equilibrium the electrochemical potential is constant; the Electron energy shift is $-|e|\varphi$. Hence the n-minus-p electric potential isAll $N$ quantities must have a consistent normalization: if $N_c,N_v$ are state densities per volume, $N_d,N_a$ are dopant densities rather than bare atom counts. The formula neglects incomplete ionization, degeneracy and changes in band structure.

At a common temperature the two equilibrium junction/contact effects do not drive a perpetual current. At different junction temperatures, Electrons and holes transport entropy differently, giving different Seebeck coefficients on the two sides. The loop has a thermoelectric electromotive force, oriented as ${\int }_{T_1}^{T_2}(S_n-S_p)dT$, and therefore generally carries current when closed. This is the Seebeck effect: energy is supplied by heat flow from hot to cold. The temperature dependence of the junction barrier is useful intuition, but an exact terminal voltage is not obtained simply by subtracting two equilibrium built-in potentials; it also depends on bulk transport and contact conditions. The given equilibrium density-of-states data do not determine the numerical thermopowers.

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.