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

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?

• @anomaly No, the conjecture is not true or no it is not possible to prove it true? – omegadot Jul 12 at 12:38
• Yes, I know numerical evidence is not sufficient to prove it. If the conjecture is not true I was suggesting that perhaps after so many decimal places this would be clearly obvious. – omegadot Jul 12 at 12:40
• @anomaly : OP suggested disproving the conjecture with numerical evidence, not proving it. – The_Sympathizer Jul 12 at 12:47
• @omegadot: Oh, jeez, this is why I shouldn't post before breakfast. Never mind. – anomaly Jul 12 at 13:49
• A complementary post about the imaginary part. – Tito Piezas III Jul 12 at 17:38

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 $$\;\displaystyle \int_0^{\frac{\pi}2} \left(\frac {\pi}2-t\right)\log^2\left(2\sin\frac t2\right)\,dt$$

• math.stackexchange.com/questions/984026/… math.stackexchange.com/questions/1080247/… Referring to these two questions, it is possible to deduce $$\operatorname{Re} \operatorname{Li}_4 (1 + i)$$ – user178256 Jul 13 at 18:37
• @user178256: this question and answer show only that you may replace an expression containing $\operatorname{Re} \operatorname{Li}_4 (1 + i)$ with another expression containing $\operatorname{Li}_4 \left (\frac{1}{2} \right )$ (and conversely). Both terms can disappear only if the expression contains $16\operatorname{Re} \operatorname{Li}_4 (1 + i) \stackrel{?}+5\operatorname{Li}_4 \left (\frac{1}{2} \right )$ or something equivalent (no 'simple' expression is known for $\operatorname{Li}_4 \left (\frac{1}{2} \right )$ and thus for the real part of $\operatorname{Li}_4 (1 + i)$). – Raymond Manzoni Jul 13 at 18:52
• (generalized log-sine integrals and Nielsen's generalized polylogarithms being considered more complex) – Raymond Manzoni Jul 13 at 18:59

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.