Show that $\int_0^1 \prod_{n\geq 1} (1-x^n) \, dx = \frac{4\pi\sqrt{3}\sinh(\frac{\pi}{3}\sqrt{23})}{\sqrt{23}\cosh(\frac{\pi}{2}\sqrt{23})}$

Question

$$\int_0^1 \prod_{n\geq 1} (1-x^n) \, dx = \frac{4\pi\sqrt{3}\sinh(\frac{\pi}{3}\sqrt{23})}{\sqrt{23}\cosh(\frac{\pi}{2}\sqrt{23})}$$

This monstrous expression is from Tolaso Network (tolaso.com.gr). I have no idea how to approach it - converting the product to a sum of logarithms does no good, and one cannot switch the order of product/integral either. The product in itself doesn't converge to anything nice either.

I am interested in seeing the proof of the above identity, as well as an explanation of how exactly $\sqrt{23}$ becomes involved in such a deceptive integral. Both real and complex analytic solutions are welcome. A proof without the pentagonal number theorem would be nice as well, since that somewhat trivializes the problem.

Use the pentagonal number theorem, then switch order of sum and integration, the resulting series is easy to evaluate using contour integration. — pisco, Nov 30 '18 at 07:01
Related: https://math.stackexchange.com/questions/1541601/closed-form-of-the-integral-large-int-0-infty-e-x-prod-n-1-infty-left/1541666#1541666 — Olivier Oloa, Nov 30 '18 at 08:46
This is (4) is Glasser, Some Integrals of the Dedekind $\eta$-Function, specialized to $y = 23/24$. — Eric Towers, Jun 02 '21 at 14:05

score 10 · Answer 1 · answered Nov 30 '18 at 07:16

10

Note that $$\prod_{n=1}^\infty(1-x^n)=\sum_{m=-\infty}^\infty(-1)^mx^{m(3m+1)/2} $$ (Euler's pentagonal number formula), so the integral equals \begin{align} I&=\sum_{m=-\infty}^\infty(-1)^m\frac2{3m^2+m+2}\\ &=\frac4{\sqrt{23}} \sum_{m=-\infty}^\infty(-1)^m\left(\frac{1}{m+(1-\sqrt{23})/6} -\frac{1}{m+(1+\sqrt{23})/6}\right). \end{align} Now you can attack this with identities for the digamma function.

answered Nov 30 '18 at 07:16

Angina Seng

158,341

Those digamma identities are...? – clathratus Nov 30 '18 at 16:58
The Wolfram MathWorld page has all the ones you'll need: http://mathworld.wolfram.com/DigammaFunction.html – Isaac Browne Dec 02 '18 at 16:43

score 7 · Answer 2 · answered Nov 30 '18 at 07:15

As suggested by Pisco, use the Pentagonal number theorem, for $|x|<1$, $$\prod\limits_{n=1}^{+\infty}(1-x^n)=\sum\limits_{m=-\infty}^{+\infty}(-1)^m x^{\frac{3m^2-m}{2}}.$$ Then by integrating each term (here some details are needed), we get $$2\sum\limits_{m=-\infty}^{+\infty}\frac{(-1)^m }{3m^2-m+2}.$$ Now note that the discriminant of the quadratic polynomial at the denominator is the number $-23$.

Can you take it from here?

Show that $\int_0^1 \prod_{n\geq 1} (1-x^n) \, dx = \frac{4\pi\sqrt{3}\sinh(\frac{\pi}{3}\sqrt{23})}{\sqrt{23}\cosh(\frac{\pi}{2}\sqrt{23})}$

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