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It seems that $$\int_0^1\prod_{k=1}^\infty (1-x^k)\, dx=\frac{8\sqrt{69}\pi \sinh (\sqrt{23}\pi/6)}{46\cosh (\sqrt{23}\pi /3)-23}.$$

Below I present a proof, but there is one problem with that "proof".

First we use the pentagonal number theorem: $$I=\int_0^1\prod_{k=1}^\infty (1-x^k)\, dx=\int_0^1 \sum_{k=-\infty}^\infty (-1)^k x^{\frac{k(3k-1)}{2}}\, dx,$$ then interchange integral and series to obtain $$I=2\sum_{k=-\infty}^\infty \frac{(-1)^k}{3k^2-k+2}.$$ Then $$I=-1+2\sum_{k=0}^\infty \frac{(-1)^k}{3k^2-k+2}+2\sum_{k=0}^\infty \frac{(-1)^k}{3k^2+k+2}$$ and we use partial fraction decomposition: $$I=-1+\frac{2i}{\sqrt{23}}\sum_{k=0}^\infty \left(\frac{(-1)^k}{k-\frac{1}{6}+i\frac{\sqrt{23}}{6}}-\frac{(-1)^k}{k-\frac{1}{6}-i\frac{\sqrt{23}}{6}}\right)+\frac{2i}{\sqrt{23}}\sum_{k=0}^\infty \left(\frac{(-1)^k}{k+\frac{1}{6}+i\frac{\sqrt{23}}{6}}-\frac{(-1)^k}{k+\frac{1}{6}-i\frac{\sqrt{23}}{6}}\right).$$

Now $$\Phi (z,s,a)=\sum_{k=0}^\infty \frac{z^k}{(a+k)^s},\quad \Re s\gt 1, |z|\ge 1.$$ If we suppose that $s$ is a positive integer then $a\notin -\mathbb{N}$. If we suppose that $z\notin [1,\infty)$, then the Lerch transcendent has the following integral representation: $$\Phi (z,s,a)=\frac{1}{\Gamma (s)}\int_0^\infty \frac{x^{s-1}e^{-ax}}{1-ze^{-x}}\, dx,\quad \Re s\gt 0,\Re a\gt 0.$$

Here comes the important part: If we ignore that $\Re s\gt 1$ in the series definition of $\Phi$ and if we ignore that $\Re a\gt 0$ in the integral representation, then we can write $I$ as $$\begin{align}I&=-1+\frac{2i}{\sqrt{23}}\left(\Phi \left(-1,1,-\frac{1}{6}+i\frac{\sqrt{23}}{6}\right)-\Phi\left(-1,1,-\frac{1}{6}-i\frac{\sqrt{23}}{6}\right)+\Phi\left(-1,1,\frac{1}{6}+i\frac{\sqrt{23}}{6}\right)-\Phi\left(-1,1,\frac{1}{6}-i\frac{\sqrt{23}}{6}\right)\right)\\&=-1+\frac{2i}{\sqrt{23}}\int_0^\infty \frac{1}{1+e^{-x}}\left(e^{\frac{1}{6}x-i\frac{\sqrt{23}}{6}x}-e^{\frac{1}{6}x+i\frac{\sqrt{23}}{6}x}+e^{-\frac{1}{6}x-i\frac{\sqrt{23}}{6}x}-e^{-\frac{1}{6}x+i\frac{\sqrt{23}}{6}x}\right)\, dx\\&=-1+\frac{8}{\sqrt{23}}\int_0^\infty \frac{\sin\frac{\sqrt{23}x}{6}\cosh\frac{x}{6}}{1+e^{-x}}\, dx.\end{align}$$ By utilizing Laplace transforms, this can be rewritten to $$I=\frac{4}{\sqrt{23}}\int_{-\infty}^\infty \frac{\sin\frac{\sqrt{23}x}{6}}{1+e^{-x}}e^{\frac{x}{6}}\, dx.$$ Then a routine application of the residue theorem yields $$I=\frac{4\pi}{\sqrt{23}\sinh \sqrt{23}\pi}\sum_{k=0}^5 \cosh\left(\frac{\sqrt{23}}{6}(2k+1)\pi-\sqrt{23}\pi\right)e^{(2k+1)\frac{i\pi}{6}},$$ which, after using Euler's formula, gives $$I=\frac{8\sqrt{69}\pi \sinh (\sqrt{23}\pi/6)}{46\cosh (\sqrt{23}\pi /3)-23}.$$

Question

How to make the proof rigorous? In other words: notice how we ignored two conditions when using the Lerch transcendent. How can this be patched?

As a reference for the Lerch transcendent, I used https://dlmf.nist.gov/25.14

Nomas2
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1 Answers1

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firstly you don't need to use this function here its more easy to use the identity $$ \sum_{k=-\infty}^{\infty} (-1)^k f(k)=\sum_{k=-\infty}^{\infty} \left(2f(2k)-f(k)\right)$$ then use partial fraction and the known identity which proved here $$ \pi \cot(\pi z)=\sum_{k=-\infty}^{\infty} \frac{1}{k+z}$$ and for your question about ignore the conditions , you did it in a correct way you can sometimes Take some values outside the conditions of the formula(series or integral) but you need to make your formula converge , for example we have $$ \Phi(-1,1,a)-\Phi(-1,1,b)=\int_0^\infty \frac{e^{-ax}-e^{-bx}}{1+e^{-x}}dx$$ now if $a=-A+Bi , b=-A-Bi $ for positive $A,B$ we get $$ \Phi(-1,1,-A+Bi)-\Phi(-1,1,-A-Bi)=-2i\int_0^\infty \frac{e^{Ax} \sin(Bx)}{1+e^{-x}}dx$$ which can be converge for some values of $A>0$

In short, the result of the integration of the subtraction of two functions can be converge, but the integration of each function separately is divergent and this can be found also in the series .

Faoler
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