How to derive the Dilogarithm reflection $\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\frac{\pi^2}{6}-\ln(x)\ln(1-x)$?

Asked May 29 '19 at 23:15

Active May 29 '19 at 23:15

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While exploring some interesting sums involving the harmonic numbers, I stumbled onto a useful identity for the Dilogarithm. $$\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\frac{\pi^2}{6}-\ln(x)\ln(1-x)$$

I tried using the sum formula for the Dilogarithm and combining $\operatorname{Li}_2(x)$ and $\operatorname{Li}_2(1-x)$ but had no success. I would be very grateful if anybody could show me how this reflection formula is derived.

asked May 29 '19 at 23:15

aleden

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Differentiate & verify this is true, then integrate to obtain the original formula. – Donald Splutterwit May 29 '19 at 23:18

How to derive the Dilogarithm reflection $\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\frac{\pi^2}{6}-\ln(x)\ln(1-x)$?

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