Lattice Gaussian Riemann limits and the singular Bose endpoint

For $\mu <0$, the Bose denominator is bounded below by $e^{\beta (-\mu )}-1>0$ on compact sets; for finite $\mu$, the Fermi denominator is at least $1$. At large $\mathrm{\mid }p\mathrm{\mid }$, each occupation is bounded by $C{e}^{-c\mathrm{\mid }p{\mathrm{\mid }}^2}$ for positive constants $C,c$ depending on a compact parameter neighborhood. A polynomial factor $\mathrm{\mid }p{\mathrm{\mid }}^s$ is absorbed by a slower Gaussian because $r^s{e}^{-c{r}^2\mathrm{/}2}$ is bounded on $[0,\mathrm{\infty })$. Thus the continuum integrands and their polynomially weighted tails are integrable by comparison with products of one dimensional Gaussians and Tonelli [F3,F1]. For each fixed $h>0$, the max-norm shell $\mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}=n$ has at most $C_d{n}^{d-1}$ points, and its summand is bounded by a fixed polynomial in $n$ times $e^{-c{h}^2{n}^2}$; the ratio of successive polynomial-Gaussian terms is eventually below a constant less than one, proving every lattice sum is finite.

At $\mu =0$ and $p\mathrm{\ne }0$, $e^x-1\ge x$ gives $F_B(p)\le (\beta \alpha \mathrm{\mid }p{\mathrm{\mid }}^2{)}^{-1}$. If $\mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}=n$, the number of integer lattice points in that shell is $(2n+1{)}^d-(2n-1{)}^d\le C_d{n}^{d-1}$, while $\mathrm{\mid }k\mathrm{\mid }\ge n$. Hence for $0<h\le \delta$, $h^d{\sum }_{0<\mathrm{\mid }hk\mathrm{\mid }\le \delta }{F}_B(hk)\le C{h}^{d-2}{\sum }_{n\le \delta \mathrm{/}h}{n}^{d-3}\le C^{\mathrm{\prime }}{\delta }^{d-2}$ when $d>2$; the last estimate follows by comparison on each unit interval with the integral of $x^{d-3}$ (and for $d=3$ the sum is bounded by a constant times $\delta \mathrm{/}h$). The continuum small-ball integral has the same bound by decomposing into dyadic annuli, each with volume at most a constant times its radius to power $d$.

On a fixed compact cube, uniform continuity of $\mathrm{\mid }p{\mathrm{\mid }}^{s}F(p)$ implies that replacing its value on each mesh cube by its value at the lattice corner changes the integral by at most the cube volume times the common modulus of continuity; summing finitely many cubes makes this error tend to zero. The measure of each mesh cube is $h^d$ by [F5]. This proves convergence on compact cubes.

To control tails uniformly for $0<h\le 1$, let $Q_{k,h}=hk+[-h\mathrm{/}2,h\mathrm{/}2{]}^d$. For $p\in Q_{k,h}$, $\mathrm{\mid }hk{\mathrm{\mid }}^2\ge \mathrm{\mid }p{\mathrm{\mid }}^2\mathrm{/}2-d{h}^2\mathrm{/}4$. The Gaussian bound from step 1.1 therefore bounds each weighted lattice summand times $h^d$ by a fixed constant times the integral of a slower Gaussian over $Q_{k,h}$; the polynomial weight is absorbed by reducing its exponent once more. Cubes whose centers satisfy $\mathrm{\mid }hk\mathrm{\mid }>R$ lie outside the ball of radius $R-\sqrt{d}\mathrm{/}2$. The integral of the slower Gaussian on that exterior tends to zero as $R\to \mathrm{\infty }$, by [F1,F3]. Compact convergence from step 2.1 and this uniform tail bound prove the stated full-space limit and finiteness of every lattice sum.

Outside any fixed ball around zero, $F_B$ is continuous and has Gaussian tails, so the compact-annulus and tail argument in steps 2.1 and 3.1 applies. Split both the sum and integral into $0<\mathrm{\mid }p\mathrm{\mid }\le \delta$, $\delta <\mathrm{\mid }p\mathrm{\mid }\le R$, and $\mathrm{\mid }p\mathrm{\mid }>R$. First take $h\downarrow 0$ on the annulus, then $R\to \mathrm{\infty }$ using Gaussian tails, and finally $\delta \downarrow 0$ using step 1.2. This proves convergence of the missing-origin Bose sums without assigning an origin value.

For $p\mathrm{\ne }0$, the positive geometric series gives $F_B(p;0)={\sum }_{\mathrm{\ell }\ge 1}{e}^{-\mathrm{\ell }\beta \alpha \mathrm{\mid }p{\mathrm{\mid }}^2}$. Its partial sums increase pointwise, so [F4] and Tonelli [F3] give the equality as extended nonnegative integrals $\int F_B={\sum }_{\mathrm{\ell }\ge 1}\int e^{-\mathrm{\ell }\beta \alpha \mathrm{\mid }p{\mathrm{\mid }}^2}dp$. The product Gaussian integral [F1] evaluates each term as $(\pi \mathrm{/}(\mathrm{\ell }\beta \alpha ){)}^{d\mathrm{/}2}$. The resulting positive series converges exactly when $d\mathrm{/}2>1$, by comparison of $\sum {\mathrm{\ell }}^{-d\mathrm{/}2}$ with ${\int }_1^{\mathrm{\infty }}{x}^{-d\mathrm{/}2}dx$, and diverges otherwise. For $\mu <0$, $F_B(p;\mu )$ increases pointwise to $F_B(p;0)$ as $\mu \uparrow 0$, so [F4] also gives convergence of the corresponding integrals to this endpoint value. This proves the extended endpoint formula, its divergence for $d\le 2$, and the integral limit used by the Bose condensation theorem.