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Double integrals for Euler’s constant and $$\ln \frac 4\pi$$ and an analog of Hadjicostas’s formula. (English) Zbl 1138.11356
From the introduction: Euler’s constant $$\gamma$$ is defined as the limit
$\gamma= \lim_{N\to\infty} \bigl(1+ \tfrac12+ \tfrac13+\cdots+ \tfrac1N-\ln N\bigr). \tag{1}$
In this note we prove the formulas
\begin{aligned} \gamma&= \sum_{n=1}^\infty \biggl(\frac1n-\ln \frac{n+1}{n}\biggr)= \iint_{[0,1]^2} \frac{1-x}{(1-xy)(-\ln xy)}\,dx\,dy, \tag{2}\\ \ln \frac 4\pi&= \sum_{n=1}^\infty (-1)^{n-1}\biggl(\frac1n-\ln \frac{n+1}{n}\biggr)= \iint_{[0,1]^2} \frac{1-x}{(1+xy)(-\ln xy)}\,dx\,dy. \tag{3} \end{aligned}
In view of series (2), which is due to Euler, series (3) reveals $$\ln(4/\pi)$$ to be an “alternating Euler constant”.

##### MSC:
 11Y60 Evaluation of number-theoretic constants
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