Evaluate $\Im(\operatorname{Li}_3(2i) + \operatorname{Li}_3(\frac i2))$

Question

Applying the trilogarithm identity $$ \operatorname{Li}_{3}\left(z\right) - \operatorname{Li}_{3}\left(1 \over z\right) = -{1 \over 6}\ln^{3}\left(-z\right) - {\pi^{2} \over 6}\ln\left(-z\right)\tag{1} $$ along with $\Im\operatorname{Li}_{3}\left(z\right) =- \Im\operatorname{Li}_{3}\left(-z\right)$, I am able to obtain $$ \Im\operatorname{Li}_{3}\left(2{i}\right) + \Im\operatorname{Li}_{3}\left({{i} \over 2}\right) = {\pi^{3} \over 16} + {\pi \over 4}\ln^{2}2 $$ However, the identity (1) seems a blackbox to me, offering little intuition. I would like to see the expression evaluated independently, preferably, self-contained.

Answer 1

Special functions are always. They are for eternity. But to find a common definition mathematicians do very very hard. I would solve this with a formula collection for example Abramovitch, Stegun nowadays in the Springer Verlag available.

The first equation is following Wolfram Alpha, which is available online, a series expansion for the left-hand side difference. This is polylog with n=3 common name is trilog or trilogarithm. The pod from WolframAlpha is:

This is slightly different to the given one. This is called the Puiseux series expansion. This can be found in the wikipedia.org. For Polylog Polylogarithm in wikipedia.org there is the identity:

This evaluates to

-(1/6) ln[-z] (π^2 + ln[-z]^2)

and there is Your result. The section is named relationships to other functions. Here to the Riemann zeta function. s==3 in this case. The evaluation can be done too with Wolfram Alpha. This is proven by Jonquiere in 1889 in research on the functional equation for the Hurwitz zeta function.

Using the first equation with the identity gives the left-hand side in general for the imaginary part. So the calculation has to be done on the right-hand side of equation (1) at z==2i.

-(1/6) ln[-z] (\[Pi]^2 + ln[-z]^2) /. z -> 2 I

-(1/6) (-((I 𝜋)/2) + 
   ln[2]) (𝜋^2 + (-((I 𝜋)/2) + ln[2])^2)

The imaginary part of this is:

𝜋 (𝜋^2/16 + 𝜋/4 ln[2]^2

That is the confirmation that Your identity is correct. A simple verification as simplification and symbolic representation of the result.

The above difference is probably just formally.