here in my answer I got real part for polylogarithm function at $1+i$ for natural $n$

$$ \Re\left(\text{Li}_n(1+i)\right)=\left(\frac{-1}{4}\right)^{n+1}A_n-B_n $$

where $$ B_n=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} \frac{2\eta(2k)}{(n-2k)!}\operatorname{Re}\left[\left(\frac{\ln2}{2}-\frac{3\pi}{4}i\right)^{n-2k}\right] $$ which is easy to evaluate the closed form by using Dirichlet eta function .

but for $A_n$ which defined $$ A_n=4 \text{Li}_n \left(\frac{-1}{4} \right)-\Phi \left(-\frac{1}{4},n,\frac{3}{4}\right)+2\Phi \left(-\frac{1}{4},n,\frac{1}{4}\right) $$ by using Lerch zeta function So it has an integral representation $$ A_n=\frac{(-4)^{n+1}}{(n-1)!} \int_0^1 \frac{x}{x^2+1} \left(\ln(1-x) \right)^{n-1}dx$$ I got closed form for $n=2,3,4$ where $$ A_2=\frac{10\pi^2}{3}-8\ln^22 , \quad A_3=140\zeta(3)+\frac{16}{3}\ln^32-\frac{20\pi^2}{3} \ln2 $$ $$ A_4=320 \text{Li}_4\left(\frac{1}{2}\right)+\frac{343}{90}\pi^4+\frac{32}{3}\ln^42-\frac{20\pi^2}{3}\ln^22 $$ my Question is how to find closed form for $A_n$ for natural $n$ ?

this question asked on math over flow .

  • 1
    $\begingroup$ I think Feynman trick on $\log(1-ax)$ results in a mess of partial fractions but could be solvable every time $\endgroup$
    – Zima
    Commented Feb 13 at 9:56


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