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Dec 24, 2018 at 14:50 history edited Greg Martin CC BY-SA 4.0
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Dec 23, 2018 at 3:50 vote accept user521337
Dec 22, 2018 at 7:11 comment added Bjørn Kjos-Hanssen Nice. WolframAlpha says $$\mu(\mathbb N)-\mu(\{1\})= \sum_{n=2}^\infty \int_1^\infty \frac{n^{-x}}{\zeta(x)}\,dx=\int_1^\infty \frac{\sum_{n=2}^\infty n^{-x}}{\zeta(x)}\,dx=\int_1^\infty1-\frac{1}{\zeta(x)}\,dx = 1.1239120333264 $$
Dec 22, 2018 at 6:38 history answered Greg Martin CC BY-SA 4.0