Is this a valid way to find Apery's constant? Consider the power series generated by $\ln (1-x)$ Let $x= e^{ix}$. Integrate $f(e^{ix})$ two times. Then let $x=\pi$. By doing these manipulations do you get $\zeta(3)$ where $\zeta$ is the Riemann Zeta function?
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1If you find a closed form for $\zeta\left(, 3,\right)$ you'll be famous right away. – Felix Marin Nov 27 '14 at 21:41
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Let $x=e^{ix}$ is slightly unsettling. – snulty Nov 28 '14 at 02:49
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Dear @snulty, what I meant was to replace x with e^(ix). – Nov 28 '14 at 06:09
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If you integrate $\ln(1-e^{ix})$ twice, you get (up to a constant and an $x$ term) $\text{polylog}(3,e^{ix})$. Evaluate this at $x = \pi$, and according to Maple the result is $-3\; \zeta(3)/4$.
Robert Israel
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