Integral $\int_0^1 \frac{dx}{\prod_{n=1}^\infty (1+x^n)}$

The following integral appeared this summer on AoPS. However it received no answer until today. $$I=\lim_{n\to \infty } \int_0^1\frac{dx}{(1+x)(1+x^2)\dots(1+x^n)}=\int_0^1 \frac{dx}{\prod_{n=1}^\infty (1+x^n)}$$ I have learnt recently from here that: $$\frac{1}{\prod_{n=1}^\infty (1+x^n)}=\prod_{n=1}^\infty\left(1-x^{2n-1}\right)\Rightarrow I=\int_0^1\prod_{n=1}^\infty\left(1-x^{2n-1}\right)dx$$ I suspect this has a closed form since a similar integral to the last equality appeared here on MSE before; however this one is a bit different since the product goes only on odd powers and I don't see how to make a connection between the two of them, so I will appreciate some help with that.

• The series expansion of the integrand is here. An alternate form of the problem is $$I = \int_0^1 (q;q^2)_\infty \,dq,$$ where $(a;q)_\infty$ is the $q$-Pochhammer symbol. – user153012 Dec 16 '18 at 12:11
• With a lot of help from Mathematica I can only obtain a (conjectural) infinite series form for the integral $$I=8 \pi\sqrt{3} \sum _{n=0}^{\infty } \frac{p(n) \sinh \left(\frac{1}{6} \pi \sqrt{48 n+23}\right)}{\sqrt{48 n+23} \left(2 \cosh \left(\frac{1}{3} \pi \sqrt{48 n+23}\right)-1\right)}$$ where $p(n)$ is the number of unrestricted partitions of the integer n. This leads me to suspect that $I$ does not have a closed form. – James Arathoon Dec 19 '18 at 0:12
• By numeric computation, this integral is about $0.42888151226615025372$. If you put it in WolframAlpha or RIES, you will see that this does not look like anything with a closed form. – ablmf Jan 14 '19 at 19:52
• $I=\int\limits_0^1 \dfrac2{(-1;x)_\infty}\,\mathrm dx\approx 0.428882.$ – Yuri Negometyanov Jan 15 '19 at 17:20

$$\color{brown}{\textbf{Analysis of the production.}}$$

Let us consider the production $$p(x)=\prod\limits_{n=0}^{\infty}\dfrac1{1+x^n},\quad x\in(0,1).$$

First, $$\prod\limits_{k=0}^{\infty}(1-x^{2k+1})\cdot\prod\limits_{k=0}^{\infty}\log(1-x^{2k}) = \prod\limits_{k=0}^{\infty}\log(1-x^k),$$ so $$p(x)=\prod\limits_{n=0}^{\infty}\dfrac1{1+x^n} = \prod\limits_{k=0}^{\infty}(1-x^{2k+1}),\quad x\in(0,1).\tag1$$

At the second, looks right the prove $$\sum\limits_{k=0}^{\infty}\ln(1-x^{2k+1}) = -\sum\limits_{m=1}^{\infty}\sum\limits_{k=0}^{\infty}\dfrac{x^{(2k+1)m}}{m} = -\sum\limits_{m=1}^{\infty}\dfrac{x^m}{m(1-x^{2m})}\\ = -\sum\limits_{m=1}^{\infty}\dfrac1{2m}\left(\dfrac1{1-x^m} + \dfrac1{1+x^m}\right) = -\dfrac12\sum\limits_{m=1}^{\infty}\sum\limits_{k=0}^{\infty} \left(\dfrac{x^{km}}{m}+\dfrac{(-x)^{km}}{m}\right),$$ $$\ln p(x)= \dfrac12\sum\limits_{k=0}^{\infty}\left(\ln(1-x^k)+\ln(1+x^k)\right),$$ $$\ln p(x)= \dfrac13\sum\limits_{k=0}^{\infty}\ln(1-x^k),$$ $$p(x)=\sqrt[3]{\prod\limits_{k=0}^{\infty}(1-x^k)} = \sqrt[3]{(x;x)_\infty}, \tag{*}$$ where $$(x,x)_\infty$$ is q-Pochhammer symbol.

However, identity $$(*)$$ $$\color{red}{\textrm{is wrong}}$$ (see Wolfram Alpha counterexample).

$$\color{brown}{\textbf{Results.}}$$

Right identity is $$p(x)=\dfrac2{(-1;x)_\infty}\tag2$$ (see also Wolfram Alpha example).

There are not detalized information about $$q$$-Pochhammer symbols, so the value of integral is calculated numerically, wherein

$$\boxed{I=\int\limits_0^1 p(x)\,\mathrm dx \approx 0.428882.}$$

• @Zacky A possible error in the proof is a permutation of the members of a two-dimensional alternating signs sequence. – Yuri Negometyanov Jan 16 '19 at 10:44