What can we say about $\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^k\log (2k+1)}{2k+1}$? Clearly it takes a value of negative real number, and seems to be an analogy of $\displaystyle \sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}=\frac{\pi}{4}$, but I don't know the way of evaluating $\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^k\log (2k+1)}{2k+1}$.
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Isn't it $L'(1,\chi)$ for a suitable L-function? – Angina Seng Oct 14 '17 at 07:00
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It is $L'(1,\chi_4)= \beta'(1)$. Not convinced by the given closed-form – reuns Oct 14 '17 at 07:30
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You may want to have a look at this: https://math.stackexchange.com/q/1284941/72031 – Paramanand Singh Oct 14 '17 at 09:39
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As shown here (https://math.stackexchange.com/a/2472198/44121), the question boils down to evaluating the integral $$\int_{0}^{+\infty}\frac{\log x}{\cosh x},dx.$$ – Jack D'Aurizio Oct 14 '17 at 20:39