∫(0,∞)1/(1+x^6)dx=2π/3は間違いのようである。

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∫(0,∞)1/(1+x^2)dx=π/2{sinπ/2}=π/2

∫(0,∞)1/(1+x^4)dx=π/4{sinπ/4+sin3π/4}=π/4・√2

∫(0,∞)1/(1+x^6)dx=π/6{sinπ/6+sin3π/6+sin5π/6}=π/2 ???

table of integrals,series and products

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