Exact disk-model Laplace integral and heat capacity

Fix $\beta >0$ and set $a:=2\pi \sqrt{2C}$. Completing the square gives $-\beta x^2+ax=-\beta (x-a\mathrm{/}(2\beta ){)}^2+a^2\mathrm{/}(4\beta )$. The affine map $u=\sqrt{\beta }(x-a\mathrm{/}(2\beta ))$ is an increasing $C^1$ bijection $\mathbb{R}\to \mathbb{R}$; its proper change-of-variable hypotheses hold on every compact truncation. Thus [F3] and [F1] give $I(a):={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\beta x^2+ax}dx=e^{a^2\mathrm{/}(4\beta )}{\beta }^{-1\mathrm{/}2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-u^2}du=e^{a^2\mathrm{/}(4\beta )}\sqrt{\pi \mathrm{/}\beta }$. The same calculation gives $I(-a)=I(a)$. Since $e^{-\beta x^2}\cosh (ax)$ is even and $\cosh (ax)=(e^{ax}+e^{-ax})\mathrm{/}2$, its full-line integral is $I(a)$ and $F(a):={\int }_0^{\mathrm{\infty }}{e}^{-\beta x^2}\cosh (ax)dx=I(a)\mathrm{/}2=\frac{\sqrt{\pi }}{2\sqrt{\beta }}{e}^{a^2\mathrm{/}(4\beta )}$.

For each compact $J=[-A,A]\subset \mathbb{R}$, $f(x,t):=e^{-\beta x^2}\cosh (tx)$ and ${\mathrm{\partial }}_{t}f=x{e}^{-\beta x^2}\sinh (tx)$ are continuous on $(0,\mathrm{\infty })\times \mathbb{R}$. The inequality $Ax\le \beta x^2\mathrm{/}2+A^2\mathrm{/}(2\beta )$ bounds $\mathrm{\mid }{\mathrm{\partial }}_{t}f\mathrm{\mid }$ on $J$ by $e^{A^2\mathrm{/}(2\beta )}x{e}^{-\beta x^2\mathrm{/}2}$, whose integral is $e^{A^2\mathrm{/}(2\beta )}\mathrm{/}\beta <\mathrm{\infty }$; the base slice $f(x,0)=e^{-\beta x^2}$ is absolutely integrable by [F1] and [F3] with the increasing $C^1$ bijection $u=\sqrt{\beta }x:(0,\mathrm{\infty })\to (0,\mathrm{\infty })$, whose derivative is the positive constant $\sqrt{\beta }$. Thus [F2] applies and $F^{\mathrm{\prime }}(a)={\int }_0^{\mathrm{\infty }}x{e}^{-\beta x^2}\sinh (ax)dx$. Differentiating the closed form in step 1.1 yields $F^{\mathrm{\prime }}(a)=a\sqrt{\pi }{e}^{a^2\mathrm{/}(4\beta )}\mathrm{/}(4{\beta }^{3\mathrm{/}2})$.

Let $h(E):={\rho }_0(E)e^{-\beta E}$ and $\varphi (x):=x^2$ on $(0,\mathrm{\infty })$. The map $\varphi$ is an increasing $C^1$ bijection with ${\varphi }^{\mathrm{\prime }}(x)=2x>0$; on each compact truncation its proper change-of-variable hypotheses hold, and $h$ is locally continuous. By [A1], the transformed integrand is $2xh(x^2)=e^{S_0}C{\pi }^{-2}x{e}^{-\beta x^2}\sinh (ax)$, whose improper integral is finite by step 2.1. Thus [F3] gives $Z_0(\beta )={\int }_0^{\mathrm{\infty }}h(E)dE={\int }_0^{\mathrm{\infty }}h(x^2)2xdx=e^{S_0}C{\pi }^{-2}{F}^{\mathrm{\prime }}(a)$, where $a=2\pi \sqrt{2C}$. Substitution of step 2.1 gives the stated formula for $Z_0$. It is finite and strictly positive because $C,\beta >0$ and the displayed factors are finite and positive.

For $f(E,\beta ):={\rho }_0(E)e^{-\beta E}$ on $(0,\mathrm{\infty })\times (0,\mathrm{\infty })$, both $f$ and ${\mathrm{\partial }}_{\beta }f=-E{\rho }_0(E)e^{-\beta E}$ are continuous. On a compact $J\subset (0,\mathrm{\infty })$ let ${\beta }_0:=\min J>0$. Since $\sinh (a\sqrt{E})\le e^{a\sqrt{E}}\mathrm{/}2$ and $a\sqrt{E}\le {\beta }_{0}E\mathrm{/}2+a^2\mathrm{/}(2{\beta }_0)$, the derivative magnitude is bounded on $J$ by a constant times $E{e}^{-{\beta }_{0}E\mathrm{/}2}$, which has finite integral; the same estimate without the factor $E$ gives an integrable base slice. Apply [F2] on $(0,\mathrm{\infty })$ to differentiate $Z_0$: $Z_0^{\mathrm{\prime }}(\beta )=-{\int }_0^{\mathrm{\infty }}E{\rho }_0(E)e^{-\beta E}dE$. Hence $\bar{E}=-{\mathrm{\partial }}_{\beta }\log Z_0=3\mathrm{/}(2\beta )+2{\pi }^{2}C\mathrm{/}{\beta }^2$.

At fixed $C,S_0$, $T=1\mathrm{/}\beta$ gives $d\beta \mathrm{/}dT=-{\beta }^2$, so $C_V\mathrm{/}{k}_B=d\bar{E}\mathrm{/}dT=-{\beta }^2{\mathrm{\partial }}_{\beta }\bar{E}=3\mathrm{/}2+4{\pi }^{2}C\mathrm{/}\beta$. Since $C>0$ and $\beta >0$, this coefficient is strictly positive.