I tried to take the derivative of a long expression containing $\text{Re} \text{Li}_3(e^{2\pi ix})$, the trilogarithm while $x$ is real. Since the real part function is linear ,it should be like this $$ \frac{\partial }{\partial x}(F(x)+\text{Re}\text{Li}_3(e^{2\pi ix}))=F'(x)-2\pi\text{Im}\text{Li}_3(e^{2\pi ix})) $$ However, MMA's output is something strange like

Derivative[1][F][x] + 
 2 I \[Pi] PolyLog[2, E^(2 I \[Pi] x)] Derivative[1][Im][
   PolyLog[3, E^(2 I \[Pi] x)]]

or $$ F'(x)+2 i \pi \text{Li}_2\left(e^{2 i \pi x}\right) \text{Re}'\left(\text{Li}_3\left(e^{2 i \pi x}\right)\right) $$ I think the reason is that D didn't treat Re properly.

How can I let MMA know that Re is linear and output the correct result? Thanks in advance.

  • $\begingroup$ Mathematica can't figure what to do with or simplify Re[PolyLog[3, z]] nor with Im[PolyLog[3, z]] so it just returns D[Re[PolyLog]]' and D[Im[....]` when you apply the derivative operator on these. What should Re[PolyLog[3, ExpToTrig[E^(2 I Pi x)]] return? btw, try D[Im[Sin[x]], x] and see what you get. $\endgroup$
    – Nasser
    Mar 10 at 6:13

1 Answer 1


We start from the relation $\Re[g'[x]]=(\Re[g[x]])'$ for real values of $x$. Next, we apply the integral presentation of PolyLog (see Details in the documentation) $$\text{Li}_3(\exp (2 \pi i x))=\int_0^1 \log (t) \log (1-t \exp (2 \pi i x))/t \, dt $$ Now we differentiate under the integral sign, making use of Leibniz integral rule . We deal with an improper integral so we should consider the integral over $[\epsilon,1]$ and then we should pass to the limit when $\epsilon \to 0+$

  Assuming[x \[Element] Reals, D[Integrate[Log[t]*Log[1 - Exp[2*Pi*I*x]*t]/t, {t, 0, 1}], x]]

2 I \[Pi] PolyLog[2, E^(2 I \[Pi] x)]


-2 \[Pi] Im[PolyLog[2, E^(2 I \[Pi] x)]]

The above result is in accordance with numerical calculations , say for $x=\frac 1 3$.

Edit.The integral representation $$\text{Li}_3(\exp (2 \pi i x))= \int_0^1 \frac{\log (t) \log (1-t \exp (2 \pi i x))}{t} \, dt$$ instead of $$\text{Li}_3(\exp (2 \pi i x))=-\frac{1}{2} \int_0^1 \log ^2(t) \log (1-t \exp (2 \pi i x)) \, dt $$


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