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Square logarithmic integral

Prove that

    \[\int_{0}^{\pi/2} \ln^2 \sin x \, \mathrm{d}x = \frac{\pi^3}{24} + \frac{\pi \ln^2 2}{2}\]

Solution

Lemma 1: Let n, m \in \mathbb{N}. It holds that

    \[\int_{0}^{\pi/2}\cos nx \cos mx \, \mathrm{d}x = \left\{\begin{matrix} 0 & , & n \neq m \\ \frac{\pi}{2} & , & n=m \end{matrix}\right.\]

Lemma 2: Let x \in (0, 2\pi). It holds that

    \[\ln 2\sin \frac{x}{2} = - \sum_{n=1}^{\infty} \frac{\cos nx}{n}\]

We begin by squaring the identity of lemma 2. Hence,

    \[\ln^2 \left ( 2 \sin \frac{x}{2} \right ) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{\cos nx \cos mx}{nm}\]

Integrating the last equation we get,

    \[\int_{0}^{\pi}\ln^2 \left ( 2 \sin \frac{x}{2} \right ) \, \mathrm{d}x = \int_{0}^{\pi}\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{\cos nx \cos mx}{nm} \, \mathrm{d}x = \frac{\pi^3}{12}\]

Expanding the LHS we get that

    \begin{align*} \frac{\pi^3}{12} &= \int_{0}^{\pi} \ln^2 \left ( 2 \sin \frac{x}{2} \right ) \, \mathrm{d}x \\ &=\int_0^\pi \left ( \ln 2 + \ln \sin \frac{x}{2} \right )^2 \, \mathrm{d}x \\ &= \int_{0}^{\pi} \ln^2 2 \, \mathrm{d}x + 2 \ln 2 \int_{0}^{\pi} \ln \sin \frac{x}{2} \, \mathrm{d}x + \\ &\quad \quad + \int_{0}^{\pi} \ln^2 \sin \frac{x}{2} \, \mathrm{d}x \\ &= \pi \ln^2 2 + 4 \ln 2 \int_{0}^{\pi/2} \ln \sin x \, \mathrm{d}x + 2\int_{0}^{\pi/2} \ln^2 \sin x \,\mathrm{d}x \\ &= \pi \ln^2 2 - 2 \pi \ln^2 2 + 2\int_{0}^{\pi/2} \ln^2 \sin x \, \mathrm{d}x \\ &= -\pi \ln^2 2 + 2\int_{0}^{\pi/2} \ln^2 \sin x \, \mathrm{d}x \end{align*}

Finally,

    \[\mathbf{\int_{0}^{\pi/2} \ln^2 \sin x \, \mathrm{d}x = \frac{\pi \ln^2 2}{2} + \frac{\pi^3}{24}}\]

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