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I'd like to evaluate $\zeta(4)$ using iterated integrals. We already know the numerical answer, so it remains to set up the integral and do some of the steps. From the recipe of Ihara-Kaneko-Zagier one has: $$ \zeta(4) = \int_{1 > t_1 > t_2 > t_3 > t_4 > 0} \omega(t_1) \omega(t_2)\omega(t_3) \omega(t_4) = \int_{1 > t_1 > t_2 > t_3 > t_4 > 0} \frac{dt_1}{t_1} \frac{dt_2}{t_2} \frac{dt_3}{t_3} \frac{dt_4}{(1-t_4)}$$ Does this look correct? The domain of integration is a simplex (a 5-cell). What are some of the intermediate steps to evaluating this integral?

cactus314
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1 Answers1

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In pages $8-9$ of my notes it is proved that by evaluating $$ \int_{0}^{1}\frac{\log(x)\log^2(1-x)}{x}\,dx $$ (which, of course, can be unfolded into a multiple integral) in two different ways we have $$ \zeta(4) = \frac{2}{5}\zeta(2)^2 $$ even without knowing the identity $\zeta(2)=\frac{\pi^2}{6}$, which, on its turn, can be proved by essentially squaring the Taylor series of the $\arctan$ or $\arcsin$ functions. Many equivalent approaches are outlined in this historical thread.

Jack D'Aurizio
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  • your notes are fascinating!! This paper of Kaneko-Ihara-Zagier starts you off with a specific multiple integral. So I'm asking to do the computation from there... I guess the exercise is to show that RHS is indeed $\zeta(4)$. I"m not even sure they demonstrate $\zeta(4) \in \pi^4 \mathbb{Q}$. – cactus314 Apr 17 '18 at 17:20
  • @cactus314: by my initial integral should be equivalent to IKZ's one by writing $\log(x)$ as $\int_{1}^{x}\frac{dt_1}{t_1}$ etc, then performing simple substitutions. – Jack D'Aurizio Apr 17 '18 at 17:23