= Solution
Measure energy from the single-particle ground state and write $z=e^{\mu/(k_BT)}\le1$. The <Bose-Einstein distribution> gives the number in excited levels as
$$
N_{\mathrm{ex}}=B\int_0^\infty\frac{\epsilon^{7/2}}{z^{-1}e^{\epsilon/(k_BT)}-1}d\epsilon
=B\Gamma(\tfrac92)(k_BT)^{9/2}\operatorname{Li}_{9/2}(z).
$$
Here $\operatorname{Li}_s(z)=\sum_{j\ge1}z^j/j^s$, obtained directly from the supplied integral identity. At the <Bose-Einstein condensation> threshold, $z\uparrow1$ and the excited states are just able to hold all $N$ particles. Thus
$$
\boxed{T_c=\frac1{k_B}\left[\frac{N}{B\Gamma(9/2)\zeta(9/2)}\right]^{2/9}.}
$$
The ground-state population is treated separately from the continuum <density of states>.
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