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Here in my answer I got the real part for the 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 $A_n$ is defined as $$ A_n=4 \text{Li}_n \left(\frac{-1}{4} \right)-\Phi \left(-\frac{1}{4},n,\frac{3}{4}\right)+2\Phi \left(-\frac{1}{4},n,\frac{1}{4}\right) $$ 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 of this integral for $n=2,3,4$: precisely $$ \begin{split} A_2 &=\frac{10\pi^2}{3}-8\ln^22,\\ A_3 &=140\zeta(3)+\frac{16}{3}\ln^32-\frac{20\pi^2}{3} \ln2\\ A_4 &=320 \text{Li}_4\left(\frac{1}{2}\right)+\frac{343}{90}\pi^4+\frac{32}{3}\ln^42-\frac{20\pi^2}{3}\ln^22 \end{split}$$ Question: is it possible to find a closed form expression for $A_n$ for every natural $n$?

Note: by closed form I meant way to simplify the Lerch $\Phi$ function and use only the polylogarithm function (without complex numbers) or Clausen function, because some values of this functions are known as constants and it cant be express simply as constants just like $Li_4\left(\frac{1}{2}\right)$.

this question asked on MSE.

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  • $\begingroup$ What do you mean by "closed form"? Please give the definition of "closed form" $\endgroup$
    – GH from MO
    Commented Feb 14 at 19:27
  • $\begingroup$ Probably, they want to express $A_n$ in terms of $Li_k(1/2)$, $\pi$, $\ln 2$, $\zeta(k)$, where $k\le n$, or show that this is not possible. $\endgroup$
    – Nemo
    Commented Feb 14 at 19:49
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    $\begingroup$ @Faoler Note that the condition "without complex numbers" is not included in your question. The complex conjugate pairs stem from the cosine in my answer below. You might be able to rewrite $A_n$ using more and more complicated expressions with growing $n$, but what is the benefit? $\endgroup$
    – Fred Hucht
    Commented Feb 15 at 13:59
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    $\begingroup$ It is not clear what it means "use polylogarithm only". The verb "use" is not a mathematical term. Please explain precisely what kind of answer you expect, otherwise your question is not well-defined. $\endgroup$
    – GH from MO
    Commented Feb 15 at 17:56
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    $\begingroup$ In your last comment, the first expression is a function of $n$, while the second expression is a function of $N$ and $x$. So there is a mismatch of variables. $\endgroup$
    – GH from MO
    Commented Feb 15 at 19:42

1 Answer 1

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I’m not sure if this is what the OP wants, but…

Mathematica can integrate the indefinite integral for positive integer $n$. Inserting the upper limit $x=1$ gives zero for $n \in \mathbb N$, while the lower limit $x=0$ can be evaluated, leading to the "closed form" $$\tag{1}\label{eq:1} A_n = 2^{2n+1} \left[ \operatorname{Li}_n \left(\tfrac{1+i}{2} \right)+ \operatorname{Li}_n \left(\tfrac{1-i}{2} \right) \right] \,. $$ Note the relation to the Clausen function, see also here. Note also that $$\tag{2}\label{eq:2} A_n = 2^{2n+2} \sum _{k=1}^{\infty } \frac{ \cos \left( \tfrac{\pi k}{4}\right)}{2^{k/2} k^{n}}\,. $$ Sums over sine or cosine often lead to these conjugate sums, or real/imaginary parts, when expressed via the polylogarithm. The Clausen family of functions avoid these constructs.

Edit

The detour via indefinite integrals war not necessary, Mathematica can directly evaluate the definite integral for $n \in \mathbb N$.

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  • $\begingroup$ I am not understanding "Inserting the upper bound 𝑥=1 gives zero in all cases". (If you're talking about a definite integral, what is the lower bound?) And what "cases" are you referring to? $\endgroup$ Commented Feb 15 at 1:58
  • $\begingroup$ @DanielAsimov: I have clarified the answer, I mixed up "bound" and "limit". $\endgroup$
    – Fred Hucht
    Commented Feb 15 at 8:05
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    $\begingroup$ using complex numbers for $A_n$ does not solve the problem because as you see I need that to get result of $\Re(Li_n(1+i))$ and for using Clausen function we must to delete the $\sqrt{2}^{-k}$ to get a simplest result. $\endgroup$
    – Faoler
    Commented Feb 15 at 13:52

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