On the relationship between $\Re\operatorname{Li}_n(1+i)$ and $\operatorname{Li}_n(1/2)$ when $n\ge5$

Question

Motivation
$\newcommand{Li}{\operatorname{Li}}$ 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?
Thoughts
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)

Update
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.)

Answer 1

lets try to find it by Lerch Zeta function , firstly we have $$Re\left(\text{Li}_n(1+i)\right)=Re\left(\text{Li}_n(\sqrt{2}e^{\frac{\pi i}{4}})\right) $$ to use series we need $|x|<1$ So by using the identity of polylogarithm $$ \text{Li}_n(z)+(-1)^n\text{Li}_n\left(\frac{1}{z}\right)=-2\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} \eta(2k)\frac{\ln^{n-2k}(-z)}{(n-2k)!}, n\in N/\text{{1}} $$ So $$ \text{Li}_n(\sqrt{2}e^{\frac{\pi i}{4}})=(-1)^{n+1}\text{Li}_n\left(\frac{1}{\sqrt{2}}e^{\frac{-\pi i}{4}}\right)-2\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} \eta(2k) \frac{\ln^{n-2k}(-1-i)}{(n-2k)!} $$ now lets trying with new polylogarithm function which we can use its series : $$ Re\left(\text{Li}_n\left(\frac{1}{\sqrt{2}}e^{\frac{-\pi i}{4}}\right) \right)=Re\left(\sum_{k=1}^\infty \frac{\left( \sqrt{2} e^{\frac{\pi i}{4}} \right)^{-k}}{k^n}\right) =\sum_{k=1}^\infty \frac{2^{-\frac{k}{2}}}{k^n} cos\left(\frac{\pi k}{4}\right)$$ now using double summation which known as: $$ \sum_{k=1}^{\infty} f(k)=\sum_{k=1}^{\infty} \sum_{p=0}^{n-1} f(nk-p)=\sum_{p=0}^{n-1}\sum_{k=1}^{\infty} f(nk-p)$$ therefore putting n=4 $$ Re\left(\text{Li}_n\left(\frac{1}{\sqrt{2}}e^{\frac{-\pi i}{4}}\right) \right)=\sum_{p=0}^3 \sum_{k=1}^\infty \frac{2^{-\frac{4k-p}{2}}}{(4k-p)^n} cos\left(\frac{\pi (4k-p)}{4}\right) $$ the cosine function can be simplify $$cos\left(\frac{\pi (4k-p)}{4}\right)=cos\left(\pi k-\frac{\pi p}{4}\right)=(-1)^kcos\left(\frac{\pi p}{4}\right) $$ So, $$ Re\left(\text{Li}_n\left(\frac{1}{\sqrt{2}}e^{\frac{-\pi i}{4}}\right) \right)=\sum_{p=0}^3 2^{\frac{p}{2}}cos\left(\frac{\pi p}{4}\right) \sum_{k=1}^\infty \frac{(-4)^{-k}}{(4k-p)^n}$$ then using Lerch Zeta function we get $$Re\left(\text{Li}_n\left(\frac{1}{\sqrt{2}}e^{\frac{-\pi i}{4}}\right) \right)=-4^{-n-1}\sum_{p=0}^3 2^{\frac{p}{2}}cos\left(\frac{\pi p}{4}\right)\Phi \left(-\frac{1}{4},n,1-\frac{p}{4}\right) $$ $$ =\frac{1}{4^{n+1}} \left(4 \text{Li}_n \left(\frac{-1}{4} \right)-\Phi \left(-\frac{1}{4},n,\frac{3}{4}\right)+2\Phi \left(-\frac{1}{4},n,\frac{1}{4}\right) \right)$$ finally we have now for $n=2,3,4,5,...$ $$ \boxed{Re\left(\text{Li}_n(1+i)\right)=\left(\frac{-1}{4}\right)^{n+1}\left(4 \text{Li}_n \left(\frac{-1}{4} \right)-\Phi \left(-\frac{1}{4},n,\frac{3}{4}\right)+2\Phi \left(-\frac{1}{4},n,\frac{1}{4}\right) \right)-\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} \frac{2\eta(2k)}{(n-2k)!}Re\left[\ln^{n-2k}(-1-i)\right]}$$ to get it simply lets use $J_n$,$A_n$ and $B_n$ $$ Re\left(\text{Li}_n(1+i)\right)=J_n=\left(\frac{-1}{4}\right)^{n+1}A_n-B_n$$ where $B_n$ is defined by $$ B_n=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} \frac{2\eta(2k)}{(n-2k)!}Re\left[\left(\frac{\ln2}{2}-\frac{3\pi}{4}i\right)^{n-2k}\right] $$ So $$ B_2=\frac{\ln^2{2}}{8}-\frac{11\pi^2}{96} ,\quad B_3=\frac{\ln^3{2}}{48}-\frac{11 \pi^2}{192} \ln2 ,\quad B_4=-\frac{1313}{92160}\pi^4+\frac{\ln^42}{384}-\frac{11\pi^2}{768} \ln^22 $$ $$ B_5=\frac{\ln^52}{3840}-\frac{11\pi^2}{4608}\ln^32-\frac{1313\pi^4}{184320}\ln2 \quad , B_6=-\frac{90941\pi^6}{61931520}-\frac{1313\pi^4\ln^22}{737280}-\frac{11\pi^2\ln^42}{36864}+\frac{\ln^62}{46080}$$ And $$ A_n=4 \text{Li}_n \left(\frac{-1}{4} \right)-\Phi \left(-\frac{1}{4},n,\frac{3}{4}\right)+2\Phi \left(-\frac{1}{4},n,\frac{1}{4}\right) $$ which has an integral representation $$ A_n=\frac{(-4)^{n+1}}{(n-1)!} \int_0^1 \frac{x}{x^2+1} \left(\ln(1-x) \right)^{n-1}dx$$ from your results for $J_2$ , $J_3$ and $J_4$ we can see : $$ A_2=\frac{10\pi^2}{3}-8\ln^22 , \quad A_3=140\zeta(3)+\frac{16}{3}\ln^32-\frac{20\pi^2}{3} \ln2 $$ $$ A_4=320 \text{Li}_4\left(\frac{1}{2}\right)+\frac{343}{90}\pi^4+\frac{32}{3}\ln^42-\frac{20\pi^2}{3}\ln^22 $$ and we have now for $J_5$ and $J_6$ $$J_5=\frac{1}{4096} \left(4\text{Li}_5 \left(\frac{-1}{4} \right)-\Phi \left(-\frac{1}{4},5,\frac{3}{4}\right)+2\Phi \left(-\frac{1}{4},5,\frac{1}{4}\right) \right)-\frac{\ln^52}{3840}+\frac{11\pi^2}{4608}\ln^32+\frac{1313\pi^4}{184320}\ln2$$ $$J_6=\frac{-1}{16384} \left(4\text{Li}_6 \left(\frac{-1}{4} \right)-\Phi \left(-\frac{1}{4},6,\frac{3}{4}\right)+2\Phi \left(-\frac{1}{4},6,\frac{1}{4}\right) \right)+\frac{90941\pi^6}{61931520}+\frac{1313\pi^4\ln^22}{737280}+\frac{11\pi^2\ln^42}{36864}-\frac{\ln^62}{46080}$$