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This question has some relationship to this integral: Let $\mathrm{Li}_2$ be the dilogarithm. Then, numerically, $$ \int_0^1 \Bigl(\mathrm{Li}_2\bigl(\frac{x-1}{x}\bigr)\Bigr)^2 \mathrm{d}x = \frac{17}{180}\pi^4, $$ as can be checked by Mathematica code like

NIntegrate[ PolyLog[2, (x-1)/x]^2, {x,0,1}, WorkingPrecision -> 100]

and

N[ 17/180 * Pi^4, 100]   

both of which output the same digits

9.199747486544674627886031420599927173585383 ...

Is this a known identity? Some related results appear in Freitas and Laurenzi, but without this specific argument in the dilogarithm.

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OK, this follows from Freitas' paper. In Table 3 there

$$\int_0^1 \mathrm{Li}_2^2(x) \mathrm{d}x = 6 - 2\zeta(2) - 4\zeta(3) + \frac{5}{2}\zeta(4),$$ and Formula (2.2.5) of Table of integrals and formulae for Feynman diagram calculations states $$\mathrm{Li}_2\Bigl(-\frac{y}{1-y}\Bigr) = - \mathrm{Li}_2(y) - \frac{1}{2}\ln^2(1-y),$$ and Mathematica can solve $$ \frac{1}{4}\int_0^1 \ln^4(1-x)\mathrm{d}x = 6 $$ as well as $$ \int_0^1 \mathrm{Li}_2(x) \ln^2(1-x) \mathrm{d}x = \frac{1}{15} \left(60 (\zeta (3)-3)+5 \pi ^2+\pi ^4\right) $$

Using these formulae, the claim follows.

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  • $\begingroup$ Impressive! I wanted to come back to this integral later, and I probably will - as the cancellation of $\zeta(3)$ hints that there should be also an elementary derivation. (+1) $\endgroup$ – Start wearing purple May 14 '13 at 16:05

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