In this book (M. Abramowitz, I. Stegun (Eds.), Handbook of Mathematical Functions with Formulas, Graphs and Mathematical Tables, Dover Publications, New York, 1972), page 807, Equation 23.2.17, it is stated that it is possible to represent the odd values of the Riemann zeta function as this integral:

$$\zeta\left(2n+1\right) = \frac{\left(-1\right)^{n+1}\left(2\pi\right)^{2n+1}}{2\left(2n+1\right)!}\int_{0}^{1}B_{2n+1}\left(x\right)cot\left(\pi x\right)dx$$

I just don't understand how to prove this, I've looked everywhere and haven't found decent proof of this. How can this be correct if $x^ncot(\pi x)$ can not be evaluated at 1?

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    $\begingroup$ Formula (84) at Wolfram Mathworld: Riemann Zeta Function) indicates $\cot\left(\frac{\pi x}{2}\right)$ instead of $\cot(\pi x)$, but perhaps the two integrals are equivalent. $\endgroup$ Apr 2 at 23:01
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    $\begingroup$ Your objection just means that this is an improper integral, like $\int_0^1x^{-1/2}\,dx$. $\endgroup$ Apr 2 at 23:02
  • $\begingroup$ @StevenClark in fact, plugging that 1/2 into the cotangent makes the interval integrable. Perhaps the equation in the book is correct and is equivalent as you said, however, the one you mentioned makes much more sense. Thanks. If you turned this comment into an answer, I would select it as correct answer. $\endgroup$ Apr 3 at 5:30


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