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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.

Moni145
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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 $$\color{blue}{A=4 \pi \sqrt{\frac{3}{23}}\,\,\frac{\sinh \left(\frac{\sqrt{23} \pi }{6}\right)}{2 \cosh \left(\frac{\sqrt{23} \pi}{3}\right)-1}} $$ $$B_x=\sum_{-\infty}^{\infty}\frac{(-1)^x}{12 x^2-2 x+2}$$ $$B_x=\frac{i \pi \left(\csc \left(\frac{1}{12} \left(1+i \sqrt{23}\right) \pi \right)-\text{sech}\left(\frac{1}{12} \left(\sqrt{23}-5 i\right) \pi \right)\right)}{2\sqrt{23}}$$ $$\color{blue}{B_x=-\pi\sqrt{\frac{2}{23}} \left(\sqrt{3}+1\right)\,\,\frac{\sinh \left(\frac{\sqrt{23} \pi }{12}\right)}{\sqrt{3}-2 \cosh \left(\frac{\sqrt{23} \pi }{6}\right)}}$$ $$B_y=\sum_{-\infty}^{\infty}\frac{(-1)^y}{12 y^2+10 y+4}$$ $$B_y=\frac{i \pi \left(\csc \left(\frac{1}{12} \left(5+i \sqrt{23}\right) \pi \right)-\text{sech}\left(\frac{1}{12} \left(\sqrt{23}-i\right) \pi \right)\right)}{2 \sqrt{23}}$$ $$\color{blue}{B_y=\pi\sqrt{\frac{2}{23}} \left(\sqrt{3}-1\right)\,\,\frac{\sinh \left(\frac{\sqrt{23} \pi }{12}\right)}{\sqrt{3}+2 \cosh \left(\frac{\sqrt{23} \pi }{6}\right)}}$$ $$\color{red}{I_\infty=2(A+B_x+B_y)=4\pi \sqrt{\frac{6}{23}}\,\, \frac{\sinh \left(\frac{\sqrt{23} \pi }{4}\right)+\sqrt{2} \sinh \left(\frac{\sqrt{23} \pi }{6}\right)}{2 \cosh \left(\frac{\sqrt{23} \pi }{3}\right)-1}}$$

Claude Leibovici
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    I believe the integral does have a closed-form due to Euler's Pentagonal Number Theorem for the case where $n\to\infty$ – Brian Constantinescu May 09 '22 at 02:59
  • @BrianConstantinescu. Your solution in Quora is more than elegant and beautiful. Tell me if you plan to add the solution here as an answer. If you do not, I shall edit and, quoting you for sure, I shall add the end result. Cheers and thanks. – Claude Leibovici May 09 '22 at 04:22
  • Certainly would be happy if you adapted the work here. My rambling post doesn't blend well with the brevity and precision found on StackExchange. – Brian Constantinescu May 09 '22 at 04:38
  • You can find a complete closed-form here: http://oeis.org/A258232/a258232_2.pdf – Brian Constantinescu May 09 '22 at 05:19
  • @BrianConstantinescu. In Quora, I think that there is a typo : the last $(\sqrt 3+1)$ is probably $(\sqrt 3-1)$ – Claude Leibovici May 09 '22 at 05:55
  • @BrianConstantinescu. I made your result more compact. – Claude Leibovici May 09 '22 at 06:18
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    @BrianConstantinescu. You inspired me a lot ! Look at my edit (without residues). Cheers :-) – Claude Leibovici May 09 '22 at 07:51
  • Awesome work! I'm glad to see this nice compact solution. All the best! – Brian Constantinescu May 09 '22 at 07:55
  • Thank you all for your help with this! Claude, how did you get this expression for A? You mention the use of the polylog, which I am familiar with, but don't readily see how to get it. Thanks. – Moni145 May 10 '22 at 10:23
  • @Luna145. Have a look at https://www.wolframalpha.com/input?i=Sum%5B%28-1%29%5En%2F%282+-+2n+%2B+3n%5E2%29%2C%7Bn%2C-Infinity%2CInfinity%7D%5D and simplify (a lot). Otherwise, use partial fraction decomposition and use $$\sum_{-\infty}^\infty\frac {(-1)^n}{n-\gamma}=-\pi \csc (\pi \gamma )$$ – Claude Leibovici May 10 '22 at 11:14