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An analytic logarithmic sum

Evaluate the sum

    \[\mathcal{S} = \sum_{n=-\infty}^{\infty} \frac{\log \left | n + \frac{1}{4} \right |}{n + \frac{1}{4}}\]

(Seraphim Tsipelis)

Solution [pprime]

We have successively:

\begin{aligned} \sum_{n=-\infty}^{\infty} \frac{\log \left | n + \frac{1}{4} \right |}{n + \frac{1}{4}} &= 4 \log \frac{1}{4} + \sum_{n=1}^{\infty} \left [ \frac{\log \left ( n + \frac{1}{4} \right )}{n + \frac{1}{4}} - \frac{\log \left ( n - \frac{1}{4} \right )}{n - \frac{1}{4}} \right ] \\ &= - 8 \log 2 + 8 \log 2 \sum_{n=1}^{\infty} \left [ \frac{1}{4n-1} - \frac{1}{4n+1} \right ] +\\ &\quad \quad +4 \sum_{n=1}^{\infty} \left [ \frac{\log(4n+1)}{4n+1} - \frac{\log(4n-1)}{4n-1} \right ] \\ &=-8 \log 2 + 8 \log 2 \left ( 1 - \frac{\pi}{4} \right ) + \\ &\quad \quad \quad +4\sum_{n=1}^{\infty} \frac{(-1)^n \log (2n+1)}{2n+1} \\ &= - 2 \pi \log 2 + 4 \sum_{n=1}^{\infty} \frac{\log n}{n} \sin \frac{n \pi}{2} \end{aligned}

We are invoking Kummer’s formula for the evaluation of the last sum. Evaluating the Fourier series that appear for z=\frac{1}{4} we get that

    \[\mathcal{S}= - \pi \cdot \left( {4 \cdot \log 2 - 4 \cdot \log \Gamma \left( {\dfrac{1}{4}} \right) + 3 \cdot \log \pi + \gamma } \right)\]

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