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

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.

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?