A positive discrete Laplace asymptotic determines cumulative growth

For $R>0$, $F(R)\le eL(1\mathrm{/}R)<\mathrm{\infty }$, because every term with $E_n\le R$ obeys $e^{-E_n\mathrm{/}R}\ge e^{-1}$. Define the positive linear functional on bounded real functions $f:[0,1]\to \mathbb{R}$ by ${\mathrm{\Lambda }}_R(f):=R^{-\alpha }{\sum }_{n=0}^{\mathrm{\infty }}{a}_n{e}^{-E_n\mathrm{/}R}f(e^{-E_n\mathrm{/}R})\mathrm{.}$ The series is absolutely convergent since its absolute value is at most $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}{R}^{-\alpha }L(1\mathrm{/}R)$, and ${\mathrm{\Lambda }}_R(1)=R^{-\alpha }L(1\mathrm{/}R)\to 1$.

For every integer $k\ge 0$, the moment identity ${\mathrm{\Lambda }}_R(u^k)=R^{-\alpha }L((k+1)\mathrm{/}R)$ holds term by term. The assumed small-$s$ asymptotic therefore gives ${\mathrm{\Lambda }}_R(u^k)\to (k+1{)}^{-\alpha }$.

Define ${\mathrm{\Lambda }}_0(f):=\frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{\alpha -1}f(e^{-x})dx$ for bounded continuous $f$. Its absolute value is at most $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, because the positive density has integral $\mathrm{\Gamma }(\alpha )$. Substitution $t=(k+1)x$ in the Euler integral gives ${\mathrm{\Lambda }}_0(u^k)=\frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_0^{\mathrm{\infty }}{e}^{-(k+1)x}{x}^{\alpha -1}dx=(k+1{)}^{-\alpha },$ so the moments in step 2.1 converge to the moments of ${\mathrm{\Lambda }}_0$.

For any polynomial $p$, step 2.1 gives ${\mathrm{\Lambda }}_R(p)\to {\mathrm{\Lambda }}_0(p)$ by linearity and the finitely many monomial moments. For continuous $f:[0,1]\to \mathbb{R}$ and $\epsilon >0$, choose $p$ with $\mathrm{\parallel }f-p{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon$ by [F2]. Since ${\mathrm{\Lambda }}_R(1)\to 1$, the positive-functional bounds $\mathrm{\mid }{\mathrm{\Lambda }}_R(f-p)\mathrm{\mid }\le \epsilon {\mathrm{\Lambda }}_R(1),\mathrm{\mid }{\mathrm{\Lambda }}_0(f-p)\mathrm{\mid }\le \epsilon$ show ${\mathrm{\Lambda }}_R(f)\to {\mathrm{\Lambda }}_0(f)$.

Put $a=e^{-1}$. Let $g:[0,1]\to (0,\mathrm{\infty })$ equal $2\mathrm{/}a$ on $[0,a\mathrm{/}2]$ and $1\mathrm{/}u$ on $[a\mathrm{/}2,1]$; it is continuous. For $0<\delta <a\mathrm{/}2$, let ${\chi }_{\delta }^{-}$ be $0$ on $[0,a]$, linear from $0$ to $1$ on $[a,a+\delta ]$, and $1$ on $[a+\delta ,1]$. Let ${\chi }_{\delta }^{+}$ be $0$ on $[0,a-\delta ]$, linear from $0$ to $1$ on $[a-\delta ,a]$, and $1$ on $[a,1]$. Then $g{\chi }_{\delta }^{-}\le g{\mathbf{1}}_{[a,1]}\le g{\chi }_{\delta }^{+}$. By step 4.1 their ${\mathrm{\Lambda }}_R$-limits are their ${\mathrm{\Lambda }}_0$-values; the difference of these two limiting integrals tends to zero as $\delta \downarrow 0$, since the corresponding $x$-intervals shrink to $x=1$ and the density $e^{-x}{x}^{\alpha -1}\mathrm{/}\mathrm{\Gamma }(\alpha )$ is bounded in a neighborhood of $1$. Therefore ${\mathrm{\Lambda }}_R(g{\mathbf{1}}_{[a,1]})\to {\mathrm{\Lambda }}_0(g{\mathbf{1}}_{[a,1]})$.

In the series for ${\mathrm{\Lambda }}_R(g{\mathbf{1}}_{[a,1]})$, the indicator selects exactly $E_n\le R$, and $g(e^{-E_n\mathrm{/}R})=e^{E_n\mathrm{/}R}$ on that set. The exponential factors cancel, giving ${\mathrm{\Lambda }}_R(g{\mathbf{1}}_{[a,1]})=R^{-\alpha }F(R)\mathrm{.}$ The candidate limit is ${\mathrm{\Lambda }}_0(g{\mathbf{1}}_{[a,1]})=\frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_0^1{x}^{\alpha -1}dx=\frac{1}{\alpha \mathrm{\Gamma }(\alpha )}\mathrm{.}$ Integration by parts in the Euler integral gives $\mathrm{\Gamma }(\alpha +1)=\alpha \mathrm{\Gamma }(\alpha )$: the endpoint terms $t^{\alpha }{e}^{-t}$ vanish at $0$ because $\alpha >0$, and at $+\mathrm{\infty }$ by exponential decay. Thus the limit is $1\mathrm{/}\mathrm{\Gamma }(\alpha +1)$, proving the statement.