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Multiple logarithmic integral

Let \zeta denote the Riemann zeta function. Evaluate the integral

    \[\mathcal{J} = \int_{0}^{1} \int_{0}^{1} \cdots \int_{0}^{1} \ln \prod_{k=1}^{n} x_k \ln \left ( 1 - \prod_{k=1}^{n} x_k \right ) \, \mathrm{d} \left ( x_1, x_2, \dots, x_n \right )\]

Solution

Based on symmetries,

    \begin{align*} \mathcal{J} &=n\int_{0}^{1} \int_{0}^{1} \cdots \int_{0}^{1} \ln x_{1}\ln \left ( 1 - \prod_{k=1}^{n} x_k \right ) \, \mathrm{d} \left ( x_1, x_2, \dots, x_n \right ) \\ &=-n\int_{0}^{1} \int_{0}^{1} \cdots \int_{0}^{1}\ln x_{1}\sum_{i=1}^{ \infty}\frac{(x_{1}\dots x_{n})^i}{i}\, \mathrm{d} \left ( x_1, x_2, \dots, x_n \right )\\ &=-n \sum_{i=1}^{\infty}\frac{1}{i(i+1)^{n-1}}\int_{0}^{1}x_{1}^i\ln x_{1}dx_{1} \\ &=n \sum_{i=1}^{\infty}\frac{1}{i(i+1)^{n+1}} \end{align*}

Let \displaystyle \mathcal{S}_n = \sum_{i=1}^{\infty}\frac{1}{i(i+1)^{n+1}}. It follows that

    \begin{align*} \mathcal{S}_n &= \sum_{i=1}^{\infty} \frac{1}{(i+1)^n} \left ( \frac{1}{i} - \frac{1}{i+1} \right ) \\ &= \mathcal{S}_{n-1} - \sum_{i=2}^{\infty} \frac{1}{i^{n+1}} \\ &= \mathcal{S}_{n-1} + 1 - \zeta (n +1) \end{align*}

Using the recursion we get that

    \[\mathcal{S}_n = n + 1 - \sum_{k=2}^{n+1} \zeta(k)\]

Thus,

    \[\mathcal{J} = n \left( n +1 - \sum_{k=2}^{n+1} \zeta(k) \right)\]

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