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An arccot sum

Evaluate the sum

    \[\mathcal{S} = \sum_{n=1}^{\infty} \left( n \arccot n -1 \right)\]

Solution

Well, first of all we note that

    \begin{align*} n \arccot n - 1 &= n \arctan \left ( \frac{1}{n} \right ) -1 \\ &=n \int_{0}^{1/n} \frac{{\rm d}x}{1+x^2} -1 \\ &= -n \int_{0}^{1/n} \frac{x^2}{1+x^2}\, {\rm d}x \end{align*}

We also recall the Fourier series expansion of \pi \coth \pi z which is no other than

(1)   \begin{equation*} \pi \coth \pi z = \frac{1}{z} + \sum_{n=1}^{\infty} \frac{2z}{n^2+z^2}  \end{equation*}

Hence

    \begin{align*} \mathcal{S} &= -\sum_{n=1}^{\infty} n \int_{0}^{1/n} \frac{x^2}{1+x^2} \, {\rm d}x\\ &=-\sum_{n=1}^{\infty} \int_{0}^{1} \frac{x^2}{x^2+n^2} \, {\rm d}x \\ &= \frac{1}{2} \int_{0}^{1} \left ( 1 - \pi x \coth \pi x \right ) \, {\rm d}x\\ &= \frac{1}{2} \int_{0}^{1} \left [ 1- \pi x \left ( 1 + \frac{2e^{-2\pi x}}{1-e^{-2\pi x}} \right ) \right ] \, {\rm d}x \\ &= \frac{1}{2} + \frac{17 \pi}{24} - \frac{1}{2} \log \left ( e^{2\pi} - 1 \right ) + \frac{1}{4\pi} {\rm Li}_2 \left ( e^{-2\pi} \right ) \end{align*}

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