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I'm trying to solve all the exercises from Tenenbaum's book but am unfortunately stuck on problem 9 of the very first ("tools") chapter. The problem is supposed to be an application of the Euler-Maclaurin formula in order to prove a classical approximation of the generating function for the partition function $p(n)$, but I can't see how the formula would help. I know how to prove this through other methods (residues plus Mellin transform) but I think they're not in the spirit of the problem. The precise exercise is as follows:

By applying the Euler-Maclaurin formula on the interval $[1,N]$ to the function $$ f(x):=\ln\left(\frac{x\theta}{1-e^{-x\theta}}\right), $$ and letting $N$ tend to infinity, show that, for any fixed integer $k\geq2$ we have, as $\theta$ approaches $0$ through positive values, $$ \prod_{n\geq1}\left(\frac{1}{1-e^{-n\theta}}\right)=\left\{1+O\left(\theta^k\right)\right\}e^{\pi^2/6\theta-\theta/24}\sqrt{\frac{\theta}{2\pi}}. $$

So I think the $O(\theta^k)$ part is a consequence of the function we are summing being Schwarz class and thus all intermediate Euler-Maclaurin terms being $0$, but I'm not exactly sure how to understand this. In general, I think it comes from the Fourier series expansion, with the error term actually being on the order of $e^{-C/\theta}$ (someone correct me on this). But overall, I'm not sure what's going on and how to get the approximation without significantly more advanced technology. Euler-Maclaurin here has me calculate the integral $$ \int_0^\infty B_1(x)f'(x)\mathrm{d}x=\int_0^\infty\left(\frac{\{x\}-1/2}{x}-\frac{(\{x\}-1/2)\theta e^{-x\theta}}{1-e^{-x\theta}}\right)\mathrm{d}x. $$ Any tips?

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  • $\begingroup$ After some more calculations, I noticed that $$ \frac{\pi^2}{6\theta}=\int_0^\infty\log\left(\frac{1}{1-e^{-x\theta}}\right)\mathrm{d}x. $$ This gives the main term; what about the fractional part ones? And I guess the above integral can be computed through Taylor series; it gives $$ \sum_n\frac{1}{n^2\theta}. $$ $\endgroup$
    – confused
    Commented Jul 21, 2023 at 4:44

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