I had a question regarding the Pochhammer symbol, specifically integrating $[x;x]_{\infty} = \prod_{j=0}^{\infty} (1-x^{j+1})$. Firstly, is there a closed form for the $n^{th}$ partial product of this, or $[x;x]_{n+1}$? I tried computing the first 3 partial products and couldn't find a pattern.
Secondly, is there a closed form for the area under this from $0$ to $1$? I was thinking about trying to find a closed form for $[x,x]_n$ as a series, then you can integrate the series and get a series representation of the integral, but if I don't know a closed form for $[x,x]_n$, this isn't useful. Can someone help?
This is just for personal interest, as I just learned about the Pochhammer symbol.
Best Answer
I do not know a closed form for $$I_n=\int_{-1}^{+1} (x;x)_{n+1}\,dx$$
For even values of $n$, this generates the sequence $$\left\{2,\frac{152}{105},\frac{4672}{3465},\frac{333280768}{253161909},\frac{141841 28217088}{10883356008525},\frac{257964528813440303104}{198899418641149766925}\right\}$$ For odd values of $n$ $$\left\{\frac{4}{3},\frac{128}{99},\frac{328448}{255255},\frac{34963456}{27188525}, \frac{530068713377923072}{412106981947605945}\right\}$$ which both decrese when $n$ increase. As you wrote, I do not see any clear pattern. I did not find these terms in $OEIS$.
Concerning
$$I_\infty=\int_{-1}^{+1} (x;x)_\infty\,dx$$ it is just a number $\color{red}{\text{(this is WRONG})}$ $$I_\infty=1.28830088867392123018090\cdots$$ which is not recognized by inverse symbolic calculators.
Edit (after @Brian Constantinescu's comment)
In Quora, using residues, @Brian Constantinescu provided a very elegant closed form of the result.
Simplifying his last result $$I_\infty=\int_{-1}^{+1} (x;x)_\infty\,dx=4\pi \sqrt{\frac{6}{23}}\,\, \frac{\sinh \left(\frac{\sqrt{23} \pi }{4}\right)+\sqrt{2} \sinh \left(\frac{\sqrt{23} \pi }{6}\right)}{2 \cosh \left(\frac{\sqrt{23} \pi }{3}\right)-1}$$
Using the same steps and notations as @Brian Constantinescu, computing the sums in terms of polylogarithms and simplifying, we have $$A=\sum_{-\infty}^{\infty}\frac{(-1)^n}{3 n^2-n+2}$$ $$A=-\frac{i \pi \left(\tan \left(\frac{1}{12} \left(5+i \sqrt{23}\right) \pi \right)+\cot \left(\frac{1}{12} \left(5+i \sqrt{23}\right) \pi \right)-2 \csc \left(\frac{1}{6} \left(1+i \sqrt{23}\right) \pi \right)\right)}{2 \sqrt{23}}$$ Expanding the complex numbers $$\color{blue}{A=4 \pi \sqrt{\frac{3}{23}}\,\,\frac{\sinh \left(\frac{\sqrt{23} \pi }{6}\right)}{2 \cosh \left(\frac{\sqrt{23} \pi}{3}\right)-1}} $$ $$B_x=\sum_{-\infty}^{\infty}\frac{(-1)^x}{12 x^2-2 x+2}$$ $$B_x=\frac{i \pi \left(\csc \left(\frac{1}{12} \left(1+i \sqrt{23}\right) \pi \right)-\text{sech}\left(\frac{1}{12} \left(\sqrt{23}-5 i\right) \pi \right)\right)}{2\sqrt{23}}$$ $$\color{blue}{B_x=-\pi\sqrt{\frac{2}{23}} \left(\sqrt{3}+1\right)\,\,\frac{\sinh \left(\frac{\sqrt{23} \pi }{12}\right)}{\sqrt{3}-2 \cosh \left(\frac{\sqrt{23} \pi }{6}\right)}}$$ $$B_y=\sum_{-\infty}^{\infty}\frac{(-1)^y}{12 y^2+10 y+4}$$ $$B_y=\frac{i \pi \left(\csc \left(\frac{1}{12} \left(5+i \sqrt{23}\right) \pi \right)-\text{sech}\left(\frac{1}{12} \left(\sqrt{23}-i\right) \pi \right)\right)}{2 \sqrt{23}}$$ $$\color{blue}{B_y=\pi\sqrt{\frac{2}{23}} \left(\sqrt{3}-1\right)\,\,\frac{\sinh \left(\frac{\sqrt{23} \pi }{12}\right)}{\sqrt{3}+2 \cosh \left(\frac{\sqrt{23} \pi }{6}\right)}}$$ $$\color{red}{I_\infty=2(A+B_x+B_y)=4\pi \sqrt{\frac{6}{23}}\,\, \frac{\sinh \left(\frac{\sqrt{23} \pi }{4}\right)+\sqrt{2} \sinh \left(\frac{\sqrt{23} \pi }{6}\right)}{2 \cosh \left(\frac{\sqrt{23} \pi }{3}\right)-1}}$$