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Integral with infinite product

Prove that

\displaystyle \int_{0}^{1}\prod_{k=1}^{\infty}(1-x^k)\;{\rm d}x=\frac{4\pi\sqrt{3}}{\sqrt{23}}\frac{\sinh \frac{\pi\sqrt{23}}{3}}{\cosh \frac{\pi\sqrt{23}}{2}}


Using the pentagonal number theorem we have the following:

\begin{aligned} \int_{0}^{1}\prod_{n=1}^{\infty} \left ( 1-x^n \right ) \, {\rm d}x &= \int_{0}^{1}\sum_{m=-\infty}^{\infty} (-1)^m x^{(3m^2-m)/2} \, {\rm d}x\\ &=\sum_{m=-\infty}^{\infty} (-1)^m \int_{0}^{1} x^{(3m^2-m)/2} \, {\rm d}x\\ &= 2 \sum_{m=-\infty}^{\infty} \frac{(-1)^m}{3m^2-m+2} \\ &= \frac{4 \pi \sqrt{3} \sinh \frac{\pi \sqrt{23}}{3}}{\sqrt{23} \cosh \frac{\pi \sqrt{23}}{2}} \end{aligned}

The sum \displaystyle \sum_{m=-\infty}^{\infty} \frac{(-1)^m}{3m^2-m+2} is dealt by the classic approach by summing the residues of the function

\displaystyle f(z) =\frac{\pi \csc \pi z}{3z^2-z+2}

at the poles of the denominator. Hence the result.

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1 Comment

  1. Let us see a derivation of the series. Consider the function

        \begin{align*} f(z) &=\frac{\pi \csc \pi z}{3z^2-z+2}  \end{align*}

    as well as a circle contour of radius R. Then

        \begin{align*} \frac{1}{2\pi i}\oint \limits_{\left | z \right |=R} f(z) \, {\rm d}z &= \sum {\rm Res} \left ( f(z) \right ) \\  &= \sum_{m=-\infty}^{\infty} \frac{(-1)^m}{3m^2-m+2} + \\  & \quad \quad \quad \quad +\sum {\rm Res}\left ( f ; z =\frac{1 \pm i \sqrt{23}}{6}  \right ) \end{align*}

    hence sending R \rightarrow +\infty the contour vanishes and thus

        \begin{align*} \sum_{m=-\infty}^{\infty} \frac{(-1)^m}{3m^2-m+2} &= - \sum {\rm Res}\left ( f ; z =\frac{1 \pm i \sqrt{23}}{6}  \right )   \end{align*}

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