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A dilogarithm and log Γ integral

Let {\rm Li}_2 denote the dilogarithm function. Prove that

    \[\zeta(3) = 2 \bigintsss_{0}^{1} \bigg( {\rm Li}_2 \left ( e^{-2 \pi i x} \right ) + {\rm Li}_2 \left ( e^{2\pi i x} \right ) \bigg) \log \Gamma (x) \, {\rm d}x\]

Solution

We begin from the very simple observation

    \[\sum_{n=1}^\infty\frac{\cos (2 \pi n x)}{n^2}=\frac{{\rm Li}_2(e^{-2\pi i x})+{\rm Li}_2(e^{2\pi i x})}{2}\]

which is an immediate consequence of the well known identity

    \[\sum_{n=1}^{\infty} \frac{\sin nx}{n} = \frac{\pi-x}{2} \quad , \quad x \in (0, 2\pi)\]

Thus letting \mathcal{J} denote the given integral ,

    \begin{align*} \mathcal{J} &= 2\int_{0}^{1} \log \Gamma(x) \sum_{n=1}^{\infty} \frac{\cos 2 \pi n x}{n^2} \, {\rm d}x\\ &= 2\sum_{n=1}^{\infty} \frac{1}{n^2} \int_{0}^{1}\cos 2 n \pi x \log \Gamma(x) \, {\rm d}x\\ &= 2 \sum_{n=1}^{\infty} \frac{1}{4n^3} \\ &= \frac{\zeta(3)}{2} \end{align*}

since it is known that

    \[\int_{0}^{1} \cos 2 n \pi x \log \Gamma(x) \, {\rm d}x = \frac{1}{4n}\]

due to the Fourier expansion of \log \Gamma.

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