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replaced latex.mathoverflow with mathjax http://meta.mathoverflow.net/a/385/3948
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Willie Wong
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Qiaochu, Parseval does apply, it just doesn't yield anything elementary. :) Still, it is easy to deduce (for example) the identity

\int\sb {0}^{2\pi}x(2\pi-x)\log(1-\cos{x})dx=8\pi \zeta(3)-\frac{4}{3}\pi^3\log{2} http://latex.mathoverflow.net/png?%5Cint%5F%7B0%7D%5E%7B2%5Cpi%7Dx%282%5Cpi%2Dx%29%5Clog%281%2D%5Ccos%7Bx%7D%29dx%3D8%5Cpi%20%5Czeta%283%29%2D%5Cfrac%7B4%7D%7B3%7D%5Cpi%5E3%5Clog%7B2%7D

but$$\int_{0}^{2\pi}x(2\pi-x)\log(1-\cos{x})dx=8\pi \zeta(3)-\frac{4}{3}\pi^3\log{2}$$ but its not clear what good such an identity is, relative to the original problem.

Qiaochu, Parseval does apply, it just doesn't yield anything elementary. :) Still, it is easy to deduce (for example) the identity

\int\sb {0}^{2\pi}x(2\pi-x)\log(1-\cos{x})dx=8\pi \zeta(3)-\frac{4}{3}\pi^3\log{2} http://latex.mathoverflow.net/png?%5Cint%5F%7B0%7D%5E%7B2%5Cpi%7Dx%282%5Cpi%2Dx%29%5Clog%281%2D%5Ccos%7Bx%7D%29dx%3D8%5Cpi%20%5Czeta%283%29%2D%5Cfrac%7B4%7D%7B3%7D%5Cpi%5E3%5Clog%7B2%7D

but its not clear what good such an identity is, relative to the original problem.

Qiaochu, Parseval does apply, it just doesn't yield anything elementary. :) Still, it is easy to deduce (for example) the identity

$$\int_{0}^{2\pi}x(2\pi-x)\log(1-\cos{x})dx=8\pi \zeta(3)-\frac{4}{3}\pi^3\log{2}$$ but its not clear what good such an identity is, relative to the original problem.

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David Hansen
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Qiaochu, Parseval does apply, it just doesn't yield anything elementary. :) Still, it is easy to deduce (for example) the identity

\int\sb {0}^{2\pi}x(2\pi-x)\log(1-\cos{x})dx=8\pi \zeta(3)-\frac{4}{3}\pi^3\log{2} http://latex.mathoverflow.net/png?%5Cint%5F%7B0%7D%5E%7B2%5Cpi%7Dx%282%5Cpi%2Dx%29%5Clog%281%2D%5Ccos%7Bx%7D%29dx%3D8%5Cpi%20%5Czeta%283%29%2D%5Cfrac%7B4%7D%7B3%7D%5Cpi%5E3%5Clog%7B2%7D

but its not clear what good such an identity is, relative to the original problem.