A conjectured value for $\operatorname{Re} \operatorname{Li}_4 (1 + i)$

Solution 1:

Integrations by parts of the recursive definition of $\;\operatorname{Li}_n(x)$ : $\;\displaystyle\operatorname{Li}_{n+1}(x)=\int\frac {\operatorname{Li}_{n}(x)}x\,dx\;$ allowed Lewin (in his $1981$ reference book "Polylogaritms and associated functions") to write :

$$\tag{7.62}\operatorname{Li}_4(x)=\log(x)\operatorname{Li}_3(x)-\frac 12\log^2(x)\operatorname{Li}_2(x)-\frac 16\log^3(x)\log(1-x)-\frac 16\int_0^x\frac{\log^3(y)}{1-y}\,dy$$

for $\; x:=1-e^{it}\;$ this becomes $$\tag{7.66}\operatorname{Li}_4(1-e^{it})=\log(1-e^{it})\operatorname{Li}_3(1-i)-\frac 12\log^2(1-i)\operatorname{Li}_2(1-e^{it})-\frac {it}6\log^3(1-e^{it})\\+\frac i6\int_0^{t}\log^3(1-e^{iv})\,dv$$

We may rewrite the last integral as $\;\displaystyle\int_0^{t}\left(\frac i2(v-\pi)+\log\left(2\sin\frac v2\right)\right)^3\,dv\;$ to expand it using binomials in terms of generalized log-sine integrals $\;\displaystyle\operatorname{Ls}_j^{(k)}(t):=-\int_0^t v^k\,\left(\log\left(2\sin\frac v2\right)\right)^{j-k-1}\,dv$.

After quite some rewriting and reduction Lewin obtained his equation $(7.68)$ for the real part : \begin{align} &\Re\operatorname{Li}_4\left(1-e^{it}\right)=\frac 14\operatorname{Ls}_4^{(1)}\left(t\right)-\frac t4\operatorname{Ls}_3\left(t\right)+\frac {t^2}8\log^2\left(2\sin\frac t2\right)+\frac{\operatorname{Li}_3(1)-\operatorname{Cl}_3(t)}2\log\left(2\sin\frac t2\right)-\frac{t^4}{192}\\ &\text{giving for $t=\frac {\pi}2\;$ since $\;\displaystyle\operatorname{Cl}_3\left(\frac {\pi}2\right)=-\frac{3}{32}\zeta(3)$ :}\\ \tag{1}&\Re\operatorname{Li}_4\left(1-i\right)=\frac 14\operatorname{Ls}_4^{(1)}\left(\frac {\pi}2\right)-\frac {\pi}8\operatorname{Ls}_3\left(\frac {\pi}2\right)+\frac{\pi^2}{32}\log^2\left(\sqrt{2}\right)+\frac {35}{64}\zeta(3)\log\left(\sqrt{2}\right)-\frac{\pi^4}{3072}\\ \end{align} But the two log-sine terms disappear using the first of the $(A.14)$ relations : $$\operatorname{Ls}_{4}^{(1)}\left(\tfrac{\pi}{2}\right)-\tfrac{\pi}{2} \operatorname{Ls}_{3}\left(\tfrac{\pi}{2}\right) = -\tfrac{5}{96} \tag{2}\log^4(2) + \tfrac{5}{16} \zeta(2) \log^2(2) - \tfrac{35}{32} \zeta(3) \log(2) + \tfrac{125}{32} \zeta(4) - \tfrac{5}{4} \operatorname{Li}_{4}\left(\tfrac{1}{2}\right)\\ $$ as provided by Davydychev and Kalmykov in the appendix of their paper "New results for the epsilon-expansion of certain one-, two- and three-loop Feynman diagrams" (from this SE answer detailing the notations used here)

$(1)$ then becomes : \begin{align} \Re\operatorname{Li}_4\left(1-i\right)&=\frac 14\left[-\tfrac{5}{96} \log^4(2) + \tfrac{5}{16} \zeta(2) \log^2(2) - \tfrac{35}{32} \zeta(3) \log(2) + \tfrac{125}{32} \zeta(4) - \tfrac{5}{4} \operatorname{Li}_{4}\left(\tfrac{1}{2}\right)\right]+\frac{\pi^2}{32}\log^2\left(\sqrt{2}\right)+\frac {35}{64}\zeta(3)\log\left(\sqrt{2}\right)-\frac{\pi^4}{3072}\\ &=- \frac{5}{16} \operatorname{Li}_{4}\left(\tfrac{1}{2}\right)-\frac{5}{384} \log^4(2) + \frac{5}{64} \zeta(2) \log^2(2) + \frac{125}{128} \zeta(4) +\frac{\pi^2}{128}\log^2\left(2\right)-\frac{\pi^4}{3072}\\ \tag{3}\Re\operatorname{Li}_4\left(1-i\right)&=- \frac{5}{16} \operatorname{Li}_{4}\left(\tfrac{1}{2}\right)-\frac{5}{384} \log^4(2)+ \frac{97}{9216}\pi^4+ \frac{\pi^2}{48}\log^2\left(2\right)\\ \end{align} Which is exactly your result ($\Re\operatorname{Li}_4\left(1+i\right)=\Re\operatorname{Li}_4\left(1-i\right)$ of course).
The only problem is that the relations $(A.14)$ from the paper of Davydychev and Kalmykov is followed by the words (much sweeter for physicists than for mathematicians...) :

$\qquad$"All relations $(A.9)–(A.14)$ have been obtained using the PSLQ procedure".

So that $(A.14)$ may or not have been proved since $2001$ (the paper was updated in $2017$). Anyway the remaining problem could be to prove the expression $(2)$ with the LHS given by $\;\displaystyle \int_0^{\frac{\pi}2} \left(\frac {\pi}2-t\right)\log^2\left(2\sin\frac t2\right)\,dt$

Solution 2:

NOT AN ANSWER TILL NOW, TOO LONG FOR A COMMENT (only a possible pathway)

Conjecture and its Motivation.
$$\Re\operatorname{Li}_4(1+i)=-\frac{5}{64} \, _5F_4\left(1,1,1,1,1;\frac{3}{2},2,2,2;1\right)+\frac{13 \pi ^4}{1536}+\frac{3}{64} \pi ^2 \ln ^22$$ This is very similar to this answer. As @Cleo gave us similar representations, I have confidence proving this conjecture by this pathway. I have numerically confirmed it to 1000 digits.

We are able to evaluate the $_5F_4(1)$ part.

Statement. $$H={}_5F_4\left(\{1\}^{5};\frac{3}{2},\{2\}^{3};1\right)=4 \text{Li}_4\left(\frac{1}{2}\right)-\frac{19 \pi ^4}{720}+\frac{\ln^42}{6}+\frac{1}{3} \pi ^2 \ln^22$$

Proof.
Exploiting $$_5F_4\left(\{1\}^{4},a;\frac{3}{2},\{2\}^{2},b;1\right)=\frac{1}{B(a,b)}\int_0^1{}_4F_3\left(\{1\}^{4};\frac{3}{2},\{2\}^{2};x\right)x^{a-1}(1-x)^{b-1}dx,$$ (this can be proved by the Taylor expansion of $_4F_3$), $H$ can be represented by $$\int_0^1{}_4F_3\left(\{1\}^{4};\frac{3}{2},\{2\}^{2};x\right)dx$$ But we know (according to Wolfram) that the integrand equals $$\frac1x\left(-2 \arcsin\left(\sqrt{x}\right) \Im\left(\text{Li}_2\left(1-2 x-2 i \sqrt{(1-x) x}\right)\right)+\Re\left(\text{Li}_3\left(1-2 x-2 i \sqrt{(1-x) x}\right)\right)+\ln(4 x) \arcsin\left(\sqrt{x}\right)^2-\zeta (3)\right),$$ substitute $x=\sin^2t$, we get $$H=\int_0^{\pi/2}2 \cot (t) \left(-2 t \Im\left(\text{Li}_2\left(e^{-2it}\right)\right)+\Re\left(\text{Li}_3\left(e^{-2it}\right)-\zeta(3)\right)+2t^2\ln(2\sin (t))\right)dt\\ =:-4I_2+2I_3+4I_1$$ Evaluation of $I_1$: (Result of this post is used, I'm sure it's not a circular argument because CAS cannot use the formula we want to prove) $$I_1=\ln2\int_0^{\pi/2}t^2\cot tdt-\int_0^{\pi/2}t\ln^2(\sin t)dt\text{ (IBP)}\\ =\frac{1}{4} \pi ^2 \ln^22-\frac{7}{8} \zeta (3) \ln2-\left(\operatorname{Li}_4\left(\frac{1}{2}\right)+\frac{1}{24}\ln^42+\frac{\pi^2}{12}\ln^22-\frac{{19}\pi^4}{2880}\right)$$ Evaluation of $I_2$: $$\begin{align} I_2&=\int_0^{\pi/2}-t\cot t\Im\operatorname{Li}_2(e^{2it})dt \\\\ &=\frac14\Re\int_1^{-1}\ln u\frac{1+u}{u(1-u)}\operatorname{Li}_2(u)du\qquad\text{(contour is in the upper half plane)} \\\\ &=\bigg\{{\small\frac{\text{Li}_2(t)^2}{4}+ \text{Li}_4(1-t) -\text{Li}_4\bigg(\frac{t}{t-1}\bigg) -\frac34\text{Li}_4(t) -\frac12\text{Li}_2\bigg(\frac{t}{t-1}\bigg)\log^2\bigg(\frac{t}{1-t}\bigg)} \\ &\qquad{\small+\text{Li}_3\bigg(\frac{t}{t-1}\bigg)\log\bigg(\frac{t}{1-t}\bigg) +\text{Li}_3(t)\log\bigg(\frac{t}{1-t}\bigg) -\text{Li}_3(1-t)\bigg[\log(t)-\log\bigg(\frac{t}{1-t}\bigg)\bigg]} \\ &\qquad{\small+\text{Li}_3(t)\log(1-t) -\frac14\text{Li}_3(t)\log(t) +\frac{1}{24}\log^4\bigg(\frac{t}{1-t}\bigg) +\frac{1}{24}\log^4(t)} \\ &\qquad{\small-\frac16\log\bigg(\frac{1}{1-t}\bigg)\log^3\bigg(\frac{t}{1-t}\bigg) -\frac16\log(t)\log^3\bigg(\frac{t}{1-t}\bigg) -\frac16\log^3(t)\log\bigg(\frac{t}{1-t}\bigg)} \\ &\qquad{\small+\frac13\log(1-t)\log^3(t) +\frac14\log^2(t)\log^2\bigg(\frac{t}{1-t}\bigg) -\frac12\log(1-t)\log^2(t)\log\bigg(\frac{t}{1-t}\bigg)} \\ &\qquad{\small-\frac14\log^2(1-t)\log^2(t) +\frac12\text{Li}_2(1-t)\bigg[\log(t)-\log\bigg(\frac{t}{1-t}\bigg)\bigg]^2} \\ &\qquad{\small+\frac12\text{Li}_2(t)\bigg[-2\log\bigg(\frac{t}{1-t}\bigg)-\log(1-t)+\log(t)\bigg]\log(t)} \bigg\}\bigg|_{1}^{-1} \\\\ &=2\text{Li}_4\left(\frac{1}{2}\right)+\frac{7}{4}\zeta(3)\ln2-\frac{19\pi^4}{1440}+\frac{\ln^42}{12}-\frac{1}{12}\pi^2\ln^22 \end{align}$$ Evaluation of $I_3$: $I_3$ have a simple antiderivative that can be deduced easily by integrating by parts repeatedly. $$I_3=\Re\left(-\zeta (3) \ln\sin t+\frac{1}{2} \left(\text{Li}_2\left(e^{2 i t}\right){}^2-\text{Li}_4\left(e^{2 i t}\right)+2 \text{Li}_3\left(e^{2 i t}\right) \log \left(1-e^{2 i t}\right)\right)\right)\Bigg|_{0}^{\pi/2}\\ =-\frac74\zeta(3)\ln2$$ Combining these three result, the statement I mentioned above holds.