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Sherlock
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- Jan 28, 2012
- 59
Inspired by this thread: By considering the integral $\displaystyle \int_{0}^{\infty} \frac{\ln(1+x^2)}{1+x^2}\;{dx}$ or otherwise,
show that $\displaystyle \sum_{k \ge 0}~\sum_{0 \le j \le k}\frac{(-1)^k}{(2k+3)(j+1)} = \frac{1}{2}\pi\ln(2)-G$ where $G$ is Catalan's constant.
show that $\displaystyle \sum_{k \ge 0}~\sum_{0 \le j \le k}\frac{(-1)^k}{(2k+3)(j+1)} = \frac{1}{2}\pi\ln(2)-G$ where $G$ is Catalan's constant.