TITLE: A conjectured value for $\operatorname{Re} \operatorname{Li}_4 (1 + i)$
QUESTION [14 upvotes]: In evaluating the integral given here it would seem that:

$$\operatorname{Re} \operatorname{Li}_4 (1 + i) \stackrel{?}{=} -\frac{5}{16} \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{97}{9216} \pi^4 + \frac{\pi^2}{48} \ln^2 2 - \frac{5}{384} \ln^4 2$$

I arrived at a result involving the $\operatorname{Re} \operatorname{Li}_4 (1 + i)$ term for the value of the integral while the OP is convinced the integral in question has a simple, elementary answer. If both of us are right, then the conjecture holds. 
So my question is, is it possible to either (i) prove the conjecture true analytically or (ii) disprove the conjecture based on (very high precision) numerical evidence?

REPLY [5 votes]: 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.