# An integration of product $(1-x^n)$

Prove $$\int_0^1\prod_{n=1}^\infty(1-x^n)dx=\frac{4\pi\sqrt3}{\sqrt{23}}\frac{\sinh\frac{\pi\sqrt{23}}3}{\cosh\frac{\pi\sqrt{23}}2}.$$ In fact, product $(1-x^n)$ is difficult to compute, I hope you can show me some ideas, thank you!

## 2 Answers

We may use the pentagonal number theorem and get $$\int_{0}^{1}\prod_{n\geq1}\left(1-x^{n}\right)dx=\sum_{k\in\mathbb{Z}}\left(-1\right)^{k}\int_{0}^{1}x^{k\left(3k-1\right)/2}dx=\sum_{k\in\mathbb{Z}}\left(-1\right)^{k}\frac{2}{3k^{2}-k+2}$$ and now we can use the well known summation formula $$\sum_{n\in\mathbb{Z}}\left(-1\right)^{n}f\left(n\right)=-\sum\left\{ \textrm{residues of }\pi\csc\left(\pi z\right)f(z)\textrm{ at }f\left(z\right)\textrm{'s poles}\right\}$$ and since we have poles at $z=\frac{1}{6}\left(1\pm i\sqrt{23}\right)$ we can conclude.

• Efficient! (+1) – Olivier Oloa Aug 5 '16 at 7:57
• Good job! Thanks! – Eufisky Aug 5 '16 at 8:09

I didn't manage to find Marco's answer I seemed to recall, so I will just re-do the proof.

By Euler pentagonal number theorem we have $$\prod_{k=1}^{+\infty}(1-x^k) = \sum_{m=-\infty}^{+\infty}(-1)^m x^{\frac{3m^2-m}{2}} \tag{1}$$ hence by termwise integration the initial integral equals $$\sum_{m=-\infty}^{+\infty}\frac{2(-1)^m}{3m^2-m+2} \tag{2}$$ and the last series can be computed through the following identity, that comes from the logarithmic derivative of the Weierstrass product for the $\cosh$ function:

$$\tanh(x) = \sum_{n\geq 0}\frac{8x}{4x^2+(2n+1)^2\pi^2} \tag{3}$$ or through $$\sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b} \tag{4}$$ and the dilogarithm reflection and duplication formulas. Putting all together, $$\boxed{\int_{0}^{1}\prod_{k\geq 1}(1-x^k)\,dx = \color{red}{\frac{4\pi\sqrt{3}\sinh{\frac{\pi\sqrt{23}}{3}}}{{\sqrt{23}\cosh{\frac{\pi\sqrt{23}}{2}}}} }}$$ follows.

• you must really be fond of special functions – user159517 Aug 19 '16 at 23:01
• +1 for the use of Euler pentagonal theorem. I was trying to use the link between eta function and elliptic integrals but did not succeed as it led to very cryptic results. – Paramanand Singh Aug 20 '16 at 8:54