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? Way(?) harder: "Shew that" 2 (d592) 4 li (- sqrt(1 - x ) + %i x + 1) 2 2 2 - li (2 %i x sqrt(1 - x ) + 2 x ) 2 is pure imaginary for -1<=x<=1.