OFFSET
1,1
COMMENTS
-dilog(phi) = -polylog(2, 1-phi) = -Sum_{k>=1} ((1-phi)^k/k^2) = Pi^2/15 - (log(phi-1)^2)/2. See the Abramowitz-Stegun link, p. 1004, eqs. 27.7.3 - 27.7.6 with x = phi-1, solving for -dilog(x+1) = -f(1+x), using log(2-phi) = 2*log(phi-1).
This solution for -Sum)k>=1} (-2*sin(Pi/10)^k/k^2) should also have been mentioned in the Jolley reference pp. 66-69 under (360).
REFERENCES
L. B. W. Jolley, Summation of Series, Dover, 1961.
LINKS
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
FORMULA
-dilog(phi) = -polylog(2, 1-phi) = -sum(((1-phi)^k)/k^2, k=1..infinity) = Pi^2/15 - (log(phi-1)^2)/2 = 0.542191216450693..., with the golden section phi = (1 + sqrt(5))/2.
MATHEMATICA
RealDigits[PolyLog[2, 1 - GoldenRatio], 10, 120][[1]] (* Amiram Eldar, May 30 2023 *)
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Wolfdieter Lang, Jun 16 2014
STATUS
approved