How do you get inf ==== 2 i + 1 \ (sqrt(2) - 1) li (sqrt(2) - 1) - li (1 - sqrt(2)) = 2 > -------------------- = 2 2 / 2 ==== (2 i + 1) i = 0
sqrt(2) - 1 / 2 2 [ atanh(x) %pi log (sqrt(2) + 1) 2 I -------- dx = ---- - ----------------- ] x 8 2 / 0
with the usual dilog relations?
Here's another: 2 2 1 1 %pi 3 log (%phi) li (-----) - li (- -----) = ---- - ------------. (%phi := golden ratio.) 2 3 2 3 12 2 %phi %phi (Yes, Henry, ln phi = asinh 1/2 .-) It's equivalent to 2 %phi 1 %phi 2 2 %pi li (----) + li (- - ----) = 2 log (%phi) - log (2) + ----. 2 2 2 2 2 12 How do you prove these?? They're obscuring drastic simplifications of huge expressions. GACK, look at this one! %phi 1 1 8 li (----) + 2 li (- 8 %phi - 4) - 12 li (-----) - li (-----) = 2 2 2 2 3 2 6 %phi %phi 2 2 2 %pi 35 log (%phi) - 40 log(2) log(%phi) - 4 log (2) + ---- 6 We're obviously missing a piece of technology to uncreate these things. Just now I got a twelve term relation with six different trilogs! Help. --rwg