A infinite product problem from physics. Background: During the process of computing the thermal entropy of massless scalar field in a box of volume $V_{box}$ at finite temperature $T$. I run into a difficult of calculating a product of infinite many terms. 
The entropy is:
$$S=T\partial_{T}\prod_{n_1,n_2,\dotsc,n_d=0}^{\infty}Z(n_1,n_2,...,n_d)$$($n_1,n_2,...,n_d$ can not be all $0$ at same time.)
$$Z(n_1,n_2,...,n_d)=\frac{1}{1-e^{-P_{n_1,n_2,...,n_d}/T}}$$
$$P_{n_1,n_2,...,n_d}=\sqrt{n_1^2+n_2^2+...+n_d^2}\frac{2\pi}{V_{box}^{1/d}}$$.
I do not know how to calculate the final expression for $S$ (Mathematica does not give me the answer).
The equation $(7)$ on page $6$ of arxiv:1104.3712v1[hep-th] states the answer is 
$$S= \frac{d+1}{\pi^{\frac{d+1}{2}}}\Gamma(\frac{d+1}{2})\zeta(d+1)T^{d}V_{box}$$ but this is just an outcome  without any calculation.
I wish someone can help me out.
 A: As often in statistical physics, the "exact" result is calculated by making an approximation as the density of states becomes continuous (whatever that means...). In particular, let's make the most blasé approximation we can think of, and replace the sum by an integral, replacing the $n_i$ with $x_i$, varying continuously over all space (I think you need to be considering the $n_i$ negative as well). Then taking logs, we have
$$ \log{Z} = -\sum_{n_i=0}^{\infty} \log{(1-e^{-c\sqrt{n_i n_i}})} \approx \int_{\mathbb{R}^d} -\log{(1-e^{-cr})} \, dx, $$
where I have just put $c=2\pi/V^{1/d}T$ to consolidate the constants. Doing the new integral:
$$ \int_{\mathbb{R}^d} -\log{(1-e^{-cr})} \, dx = -S_{d-1} \int_0^{\infty} r^{d-1} \log{(1-e^{-cr})} \, dr; $$
$ S_{d-1} = 2\pi^{d/2}/\Gamma(d/2) $, changing variables to $y=cr$ gives
$$ -\frac{2\pi^{d/2}}{\Gamma(d/2)} c^{-d} \int_0^{\infty} y^{d-1}\log{(1-e^{-y})} \, dy. $$
Integrating by parts,
$$ -\int_0^{\infty} y^{d-1}\log{(1-e^{-y})} \, dy = 0 + \frac{1}{d}\int_0^{\infty} \frac{y^d e^{-y}}{1-e^{-y}} \, dy = \frac{\Gamma(1+d)\zeta(d+1)}{d} = \Gamma(d)\zeta(d+1). $$
Therefore the whole lot is
$$ \log{Z} = \frac{2\pi^{d/2}\Gamma(d)\zeta(d+1)}{\Gamma(d/2)} c^d = \frac{2^d\pi^{(d-1)/2}\Gamma((d+1)/2)\zeta(d+1)}{(2\pi)^{d}} T^{d} V = \frac{\Gamma((d+1)/2)\zeta(d+1)}{\pi^{(d+1)/2}} T^{d} V $$
The entropy is given by
$$ S = \partial_T (T\log{Z}) = (d+1)\frac{\Gamma((d+1)/2)\zeta(d+1)}{\pi^{(d+1)/2}} T^{d} V, $$
as in the paper you cite.
