I am trying to calculate how the chemical potential of a non-interacting Bose gas above the condensation temperature, which boils down to obtaining the leading order expansion of the function $g(z) = \sum_{n=1}^\infty \frac{z^n}{n^{3/2}}$, for $z = 1-\epsilon$, as a function of $\epsilon >0$.
Here is a mathematica plot of the function.
I tried naively expanding the binomial series of $(1-\epsilon)^n$, but all the coefficient of the Taylor series are divergent.
For example, $\sum_{n=1}^\infty \frac{(1-\epsilon)^n}{n^{3/2}} = \sum_{n=1}^\infty \frac{(1-n\epsilon+\cdots)}{n^{3/2}} = \zeta(3/2) – \epsilon \sum_{n=1}^\infty\frac{1}{n^{1/2}} + \cdots$.
Here $\sum_{n=1}^\infty\frac{1}{n^{1/2}}$ diverges.
From numerics, $\left(g(1-\epsilon) – g(1)\right) \sim -\sqrt{\epsilon}$ upto leading order
(which implies, the chemical potential decreases quadratically with temperature beyond the condensation temperature).
How to demonstrate this square root dependence using analytic methods?
Best Answer
Note that your function is the polylogarithm of order $\frac{3}{2}$:
$$ \operatorname{Li}_{3/2}(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^{3/2}} $$
Invoking the gamma integral $\frac{\Gamma(\alpha)}{s^{\alpha}} = \int_{0}^{\infty} t^{\alpha-1} e^{-st} \, \mathrm{d}t$, valid for $\operatorname{Re}(\alpha) > 0$ and $\operatorname{Re}(s) > 0$, it is easy to obtain the following integral representation,
$$ \operatorname{Li}_{3/2}(z) = \frac{1}{\Gamma(\frac{3}{2})} \int_{0}^{\infty} \frac{z \sqrt{t}}{e^t - z} \, \mathrm{d}t, $$
valid for all $z \in \mathbb{C}\setminus[1,\infty)$.
Now, consider $z = 1 - \varepsilon$ for sufficiently small $\varepsilon > 0$. Then
\begin{align*} \operatorname{Li}_{3/2}(1-\varepsilon) - \operatorname{Li}_{3/2}(1) &= - \frac{1}{\Gamma(\frac{3}{2})} \int_{0}^{\infty} \frac{\varepsilon \sqrt{t} e^t}{(e^t - 1)(e^t - 1 + \varepsilon)} \, \mathrm{d}t \\ &= - \frac{\sqrt{\varepsilon}}{\Gamma(\frac{3}{2})} \int_{0}^{\infty} \frac{\varepsilon^2 \sqrt{s} e^{\varepsilon s}}{(e^{\varepsilon s} - 1)(e^{\varepsilon s} - 1 + \varepsilon)} \, \mathrm{d}s, \tag{1} \end{align*}
where the substitution $t = \varepsilon s$ is used in the last step. In order to make use of the last line, we remark that $e^{x} - 1 \geq x e^{x/2}$ holds for all $x \geq 0$. (This is easily proved by invoking Jensen's inequality.) From this, we get
$$ \frac{\varepsilon^2 \sqrt{s} e^{\varepsilon s}}{(e^{\varepsilon s} - 1)(e^{\varepsilon s} - 1 + \varepsilon)} \leq \frac{1}{\sqrt{s}(s + e^{-\varepsilon s/2})}, $$
This bound shows that dominated convergence theorem is applicable, yielding
\begin{align*} \lim_{\varepsilon \to 0^+} \frac{\operatorname{Li}_{3/2}(1-\varepsilon) - \operatorname{Li}_{3/2}(1)}{\sqrt{\varepsilon}} &= - \frac{1}{\Gamma(\frac{3}{2})} \int_{0}^{\infty} \lim_{\varepsilon \to 0^+} \frac{\varepsilon^2 \sqrt{s} e^{\varepsilon s}}{(e^{\varepsilon s} - 1)(e^{\varepsilon s} - 1 + \varepsilon)} \, \mathrm{d}s \\ &= - \frac{1}{\Gamma(\frac{3}{2})} \int_{0}^{\infty} \frac{1}{\sqrt{s}(s+1)} \, \mathrm{d}s \\ &= -2\sqrt{\pi}. \end{align*}
Therefore
$$ \operatorname{Li}_{3/2}(1-\varepsilon) = \operatorname{Li}_{3/2}(1) - 2\sqrt{\pi \varepsilon} + o(\sqrt{\varepsilon}) $$
as $\varepsilon \to 0^+$.
Addendum. Further terms in the asymptotic expansion of $\operatorname{Li}_{3/2}(1-\varepsilon)$ can be found by expanding the integrand of $\text{(1)}$ as a power series in $\varepsilon$,
\begin{align*} &\frac{\varepsilon^2 \sqrt{s} e^{\varepsilon s}}{(e^{\varepsilon s} - 1)(e^{\varepsilon s} - 1 + \varepsilon)} \\ &\qquad = \frac{1}{\sqrt{s}(s+1)} + \frac{\sqrt{s}}{2(s+1)^2} \varepsilon - \frac{s^{3/2} (s^2+3s-1)}{12 (s+1)^3} \varepsilon^2 + \cdots, \end{align*}
and integrating term-by-term, although rigorously justifying this would require much more works.