# How can I ask Mathematica to give the explicit real form of the given function?

In part of my calculations, I obtain this expression which contains the imaginary unit I but I expect that this expression might be real (from the comments below, now we are sure that it is real). How can I ask Mathematica to simplify this expression and give its explicit real form?

exp=1/96 (\[Pi]^2 +
12 I ArcCos[-(5/4)] ArcTanh[
1/3 Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] +
20 \[Pi]^2 Cosh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])] +
6 I ArcCos[-(5/4)] Log[2] -
12 ArcTanh[
1/3 Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Log[
2] + 3 Log[2]^2 -
12 Cosh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])] Log[
2]^2 - 12 I \[Pi] Cosh[
1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])] Log[4] -
6 I ArcCos[-(5/4)] Log[9] +
12 I ArcCos[-(5/4)] Cosh[
1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])] Log[16] +
12 ArcTanh[
3 Coth[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Log[
1 - I Sqrt[15]] -
48 I ArcCos[
I Sqrt[2/3]
Sinh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Cosh[
1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])] Log[
1 - I Sqrt[15]] -
12 ArcTanh[
3 Coth[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Log[
1 + I Sqrt[15]] +
48 I ArcCos[
I Sqrt[2/3]
Sinh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Cosh[
1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])] Log[
1 + I Sqrt[15]] +
6 I ArcCos[-(5/4)] Log[I/(
2 Sqrt[(I + Sqrt[15])/(3 I - 3 Sqrt[15])])] +
12 ArcTanh[
3 Coth[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Log[I/(
2 Sqrt[(I + Sqrt[15])/(3 I - 3 Sqrt[15])])] -
12 ArcTanh[
1/3 Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Log[
I/(2 Sqrt[(I + Sqrt[15])/(3 I - 3 Sqrt[15])])] +
6 I ArcCos[-(5/4)] Log[I/(
2 Sqrt[(I - Sqrt[15])/(3 I + 3 Sqrt[15])])] -
12 ArcTanh[
3 Coth[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Log[I/(
2 Sqrt[(I - Sqrt[15])/(3 I + 3 Sqrt[15])])] +
12 ArcTanh[
1/3 Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Log[
I/(2 Sqrt[(I - Sqrt[15])/(3 I + 3 Sqrt[15])])] -
6 I ArcCos[-(5/4)] Log[(-1 +
Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])/(-3 +
Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])] +
12 ArcTanh[
1/3 Tanh[
1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Log[(-1 +
Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])/(-3 +
Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])] -
6 I ArcCos[-(5/4)] Log[(
1 + Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])/(-3 +
Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])] -
12 ArcTanh[
1/3 Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Log[(
1 + Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])/(-3 +
Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])] +
6 Log[2] Log[(
3 + Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])/(-3 +
Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])] +
3 Log[(3 +
Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])/(-3 +
Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])]^2 -
48 Cosh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])] PolyLog[
2, -(1/2)] +
6 PolyLog[
2, -((-3 +
Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])/(
2 (3 + Tanh[
1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])))] +
6 PolyLog[
2, -((3 +
Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])/(
2 (-3 + Tanh[
1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])))]) Sech[
1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])];
• N@exp gives 0.0889796 - 7.40149*10^-17 I and you can Chop the I part. So it looks Real.
– Syed
Apr 11 at 14:01
• @Syed Thanks. I do not want to use the numerics N; if possible, I like to have it in its explicit real form. Apr 11 at 14:07
• First, use FullSimplify[exp ∈ Reals] to get $\text{Li}_2\left(\frac{i \sqrt{15}}{8}-\frac{1}{8}\right)+\text{Li}_2\left(-\frac{i \sqrt{15}}{8}-\frac{1}{8}\right)$, then apply the mirror symmetry to see that the result is indeed real. It doesn't look like Mathematica knows about the mirror symmetry, though ... Apr 11 at 14:17
• FullSimplify[Im[exp]] returns zero and that doesn't use N.
– Bill
Apr 11 at 15:28
• @Bill Thanks, yes, it works. And is there any hope to find the explicit real form of the function? Apr 11 at 15:40