$\newcommand{Li}{\operatorname{Li}}$ It is already known that:
$$\Re\Li_2(1+i)=\frac{\pi^2}{16}$$ $$\Re\Li_3(1+i)=\frac{\pi^2\ln2}{32}+\frac{35}{64}\zeta(3)$$

And by this question, assuming $\Li(1/2)$ is a simpler form than $\Re\Li(1+i)$, we are able to simplify $\Re\Li_4(1+i)$. $$\Re\Li_4(1+i)=-\frac5{16} \Li_4\left(\frac{1}{2}\right)+\frac{97 \pi ^4}{9216}-\frac{5}{384} \ln^42+\frac{1}{48} \pi ^2 \ln^22$$

Going further, I tried to find some numerical evidence to simpify $\Re\Li_5(1+i)$, and I conjectured that

$$\Re\Li_5(1+i)\stackrel?=\frac{5}{32} \text{Li}_5\left(\frac{1}{2}\right)+\frac{2139 \zeta (5)}{4096}-\frac{1}{768} \ln^52+\frac{1}{288} \pi ^2 \ln^32+\frac{97 \pi ^4 \ln2}{18432}$$

As I verified it up to 1000 digits, I believe it holds. How can we prove it?
Well, I honestly does not have much thoughts on it but I think turning it into an integral or Euler sum is a possible pathway. Therefore, I turned it into an integral: $$\int_0^1\frac{54x^2-150x+59}{2x^3-6x^2+5x-2}\ln^4xdx\stackrel?=\ln^52-\frac83\pi^2\ln^32-\frac{97}{24}\pi^4\ln2-\frac{6417}{16}\zeta(5)$$ Further Question
Can $\Re\Li_n(1+i)$ be simplified? Or at least can it be turned into a combination of $\Li(1/2)$ and some other simpler constants? (Assuming Multiple Zeta Values are simpler)

Another numerical experiment suggests $$\Re\operatorname{Li}_6(1+i)\stackrel?=-\frac{5 \text{Li}_6\left(\frac{1}{2}\right)}{64}+\frac{69 S}{512}+\frac{207 \zeta (3)^2}{4096}+\frac{2139 \zeta (5) \log (2)}{8192}+\frac{18779 \pi ^6}{30965760}-\frac{\log ^6(2)}{9216}+\frac{\pi ^2 \log ^4(2)}{2304}+\frac{97 \pi ^4 \log ^2(2)}{73728},$$where $S=\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{n^5}$.

Further numerical experiments suggest that $$J_n=\int_0^1\left(\frac{1-x}{x^2-2 x+2}+\frac{5 x}{x^2-2}\right) \ln^nxdx$$ may be turned into the same form of the result of "logarithm integral", mentioned in the paper in the comment, containing $1−x,x,x+1$ only, of weight $n+1$. If we can find a substitution that can directly convert it into "log integral", the whole question will be solved by evaluating this integral.
Another Update
An interesting phenomenon discovered by observing the coefficient between adjacent terms: $$J_{n-1}=\frac4n\left(\frac{\ln2}2\right)^n\sum_{k=0}^na_kP_k(n)$$ where $$P_k(n)=n(n-1)(n-2)\cdots(n-(k-1))(4\ln2)^{-k}$$ and $$\begin{array}{|c|c|c|c|c|c|c|c|}\hline k&0&1&2&3&4&5&6\\\hline a_k&-1&0&5\zeta(2)&0&-343\zeta(4)&4278\zeta(5)&8832S+3312\zeta^2(3)-53383\zeta(6)\\\hline\end{array}$$ Therefore, it may be true that all $a_k$ is a linear combination of alternating multiple zeta function with integer coefficients of weight $k$. (i)
Also, there might be a formula to calculate $a_k$ directly. (ii) $$\begin{array}{|c|c|}\hline k&7\\\hline a_k&-304707\zeta(7)-2208(7\zeta(3)\zeta(4)+6\zeta(2)\zeta(5)+16\zeta(-5,1,1)+8\zeta(-4,2,1))\\\hline k&8\\\hline a_k&\frac1{51}\left(-\frac{199260125\zeta(8)}{144}+280600\zeta(3)\zeta(5)-261088\zeta(-7,1)-34827\zeta(6,2)+2944(3\zeta^2(3)\zeta(2)+7\zeta(-3,3,2)+4\zeta(-3,3,1,1))\right)\\\hline\end{array}$$ This supports the conjecture (i). (Note that $S$ can also be regarded as Multiple Zeta Value.)

  • 3
    $\begingroup$ Great question (+1) $\endgroup$ Dec 10, 2019 at 18:05
  • 1
    $\begingroup$ Can you even briefly elaborate on how you managed to conjecture that? $\endgroup$ Apr 5, 2020 at 16:41
  • $\begingroup$ @stokes-line I used Mathematica's FindIntegerNullVector function to test if there is possibility to connect $\Re\operatorname{Li}_5(1+i)$ with $\operatorname{Li}_5(1/2)$. Then, I got coefficient that might be right. Finally, I run the numerical test of the equality. $\endgroup$ Apr 5, 2020 at 16:45
  • 2
    $\begingroup$ @Displayname I used command N[Re[PolyLog[5, 1 + I]] - 5/32 PolyLog[5, 1/2] - 2139/4096 Zeta[5] + Log[2]^5/768 - \[Pi]^2 Log[2]^3/288 - 97/18432 \[Pi]^4 Log[2], 1000] with Mathematica version The output is $0.\times10^{-1049}$. $\endgroup$ Apr 6, 2020 at 5:01


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