Re: An exact simplification challenge - 42 (dilogarithms) - Go and surpass all the CASs!

First, reminder that dilog(z) == PolyLog[2,1-z]

Using

http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/PolyLog2/17/01/03/0001/

and subtracting two such identities (each being exact zero)
for z=I*(Sqrt[2]+1) and z=I*(Sqrt[2]-1) we get

In[32]:= With[{z = I (1 + Sqrt[2])},
PolyLog[2, z] - PolyLog[2, -z] + PolyLog[2, (1 - z)/(1 + z)] -
PolyLog[2, -((1 - z)/(1 + z))] - (Pi^2/4 +
Log[z] Log[(1 + z)/(1 - z)])] -
With[{z = I (-1 + Sqrt[2])},
PolyLog[2, z] - PolyLog[2, -z] + PolyLog[2, (1 - z)/(1 + z)] -
PolyLog[2, -((1 - z)/(1 + z))] - (Pi^2/4 +
Log[z] Log[(1 + z)/(1 - z)])] // FullSimplify

Out[32]= -I \[Pi] ArcSinh[1] + PolyLog[2, -I (-1 + Sqrt[2])] -
PolyLog[2, I (-1 + Sqrt[2])] - PolyLog[2, -I (1 + Sqrt[2])] +
PolyLog[2, I (1 + Sqrt[2])]

Therefore the answer is I*Pi*ArcSinh[1] == I Pi Log[1+Sqrt[2]].

Oleksandr Pavlyk
.