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I encountered the following problem:

Show that $$\zeta(2n+1)=\frac{(-1)^{n+1}(2\pi)^{2n+1}}{2(2n+1)!}\int_0^{1}B_{2n+1}(x)\cot({\pi}x)dx$$

where $B_{2n+1}(x)$ is the Bernoulli polynomial.

This problem can be found here.

Roger209
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    What have you tried, where are you stuck? What do you know about the Bernoulli polynomials and the relations they satisfy? – Steven Stadnicki Oct 24 '13 at 15:25
  • This is just a guess, but this identity may be helpful $\frac{B_{2\mu}}{(2\mu)!} = \frac{2(-1)^{\mu+1}}{(2\pi)^{2\mu}}\zeta(2\mu)$. This identity is derived in concrete mathematics (I believe), which is surprising (the source is surprising). Also, this problem may help/interest you: http://math.stackexchange.com/questions/855740/analytic-continuation-of-zeta-function-using-bernoulli-numbers – Clyde Jan 10 '15 at 06:54

1 Answers1

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We have that, if $0<x\leq1$ , the Bernoulli polynomial can be written as $$B_{2n+1}\left(x\right)=\left(-1\right)^{n+1}\frac{2\left(2n+1\right)!}{\left(2\pi\right)^{2n+1}}\sum_{k\geq1}\frac{\sin\left(2\pi kx\right)}{k^{2n+1}}.$$ You can find a proof on Apostol, "Introduction to analytic number theory", page 267. It remain to note that, if $k$ is an integer$$\int_{0}^{1}\sin\left(2\pi kx\right)\cot\left(\pi x\right)dx=1.$$

Marco Cantarini
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  • Thank you very much. It's kind of you to answer my question even a year later. – Roger209 Mar 17 '15 at 13:11
  • @Roger209 You're welcome. Maybe you've solved by yourself, but I hope is still useful ;) – Marco Cantarini Mar 17 '15 at 13:24
  • Typically, here I would like to recommend a paper titled "Numerical calculation of the Riemann zeta function at odd-integer arguments: a direct formula method"(http://download.springer.com/static/pdf/993/art%253A10.1007%252Fs40096-015-0146-9.pdf?auth66=1426606253_162ab734e12cb01cbedae548f1f82ee9&ext=.pdf). In this paper, equation (5) and (14) are the solution too. – Roger209 Mar 17 '15 at 15:20