1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

1
$\begingroup$

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)]

  ComplexExpand[Re[%]]

-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 $$\text{Li}_3(\exp (2 \pi i x))=-\frac{1}{2} \int_0^1 \log ^2(t) \log (1-t \exp (2 \pi i x)) \, dt $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.