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By the way, $ fiets\int_{0}^{2\pi} \sin(x)dx = 0$, while $\sum_{n=1}^\infty n^{-1} = +\infty$ and

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\int_{0}^{2\pi} \sin^2(x)dx = \pi,\qquad \sum_{n=1}^\infty {1\over n^2 } = {\pi^2\over 6}.
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