Evaluate $\int_0^1 x^{a-1}(1-x)^{b-1}\operatorname{Li}_3(x) \, dx$
Solution 1:
Solution of OP's problem.
$1$. To compute $f(n,1/2)$, using $x^{n-1}=((1-x)-1)^{n-1} (-1)^{n-1}$ and binomial theorem the problem boils down to evaluation of $\int_0^1 (1-x)^{k+1/2} \text{Li}_3(x)dx\ (k\in\mathbb{Z},-1\leq k<n)$. Now, it's a fact that FL (Fourier-Legendre) expansion of $(1-x)^{k+1/2}$ above can be deduced from that of $(1-x)^{-1/2}$ by repeated integration; moreover by induction its FL coefficients is of form $\sum_{i=1}^{m_k} \frac{A_{i,k}}{2n+i}$. Applying Parseval and using FL expansion of $\text{Li}_3$ (see paper 1 for all FL related things), modulo trivial rational sums all $\int_0^1 (1-x)^{k+1/2} \text{Li}_3(x)dx$ reduce to $\int_0^1 \frac{\log(x)}{1+x} \sum_n \sum_{i=1}^{m_k} \frac{A_{i,k} (-x)^n}{2n+i}dx$ due to the repesentation of $\widetilde{H_n^{(2)}}$ via $\int_0^1 \frac{\log(x)}{1+x} (-x)^ndx$. Thus by reindexing and substitution $x\to x^2$ all resulting $\int_0^1 \frac{\log(x)}{1+x} \sum_n \sum_{i=1}^{m_k} \frac{A_{i,k} (-x)^n}{2n+i}dx$ are reduced to NQLIs (nonhomogeneous quadratic log integrals) with $W\leq3$ since denominators $2n+i$ only produce quadratic $\tan^{-1}, \tanh^{-1}$ terms (here $W$ denotes the weight of certain integral). According to paper 2 all such NQLIs are computed by repeated IBP, therefore all $f(n,1/2)$ hence (by OP's observation) $f(n,m+1/2)$ are solvable. Definitions of NQLI/NPLI(see below)/etc are also due to paper 2.
$2$. To compute $f(1/2,n)$, let $x\to x^2$ one arrive at $\int_0^1 (1-x^2)^{n-1}\text{Li}_3(x^2)dx$. By binomial and $\text{Li}_3(x^2)\to \text{Li}_3(\pm x)$ these reduce to several NPLIs with $W\leq 3$ i.e. $\int_0^1 x^{2k}\ \text{Li}_3(\pm x)dx$, all of which are dealt by the paper 2 either (IBP). Thus all $f(1/2,n)$ hence (by OP's observation) $f(1/2+m,n)$ are solvable.
$3$. Dual results. Moreover, using FL expansion and computing NQLIs one confirms OP's conjecture on solvability of $f(1/4,1/4)$ i.e.:
- $\small \int_0^1 \frac{\text{Li}_3(x)}{(x (1-x))^{3/4}} \, dx=\frac{\Gamma \left(\frac{1}{4}\right)^2 \, _5F_4\left(1,1,1,1,\frac{5}{4};\frac{3}{2},2,2,2;1\right)}{2 \sqrt{\pi }}=\frac{\Gamma \left(\frac{1}{4}\right)^2 }{\sqrt{\pi }}\left(2 \pi C+4 C \log (2)-8 \Im(\text{Li}_3(1+i))-\frac{21 \zeta (3)}{8}+\frac{3 \pi ^3}{16}+\frac{\log ^3(2)}{6}+\frac{3}{4} \pi \log ^2(2)-\frac{5}{24} \pi ^2 \log (2)\right)$
Details and variations are also given by paper 2.
Generalizations.
$4$. FL expansion is in fact applicable for higher weights; for instance $\small\int_0^1 \frac{\text{Li}_n(x)}{\sqrt[4]{x (1-x)}} dx$ ($n=4,5$) is computable in terms of MZV closed-forms using a very complicated FL argument (the following hypergeometric series correspond to case $n=4,5$ respectively, via expanding polylogarithm & Beta integral):
-
$\small \, _6F_5\left(\{1\}_5,\frac{7}{4};\{2\}_4,\frac52;1\right)=-\frac{5 \pi ^2 C}{3}-8 \pi C+4 C \log ^2(2)-4 \pi C \log (2)+16 C \log (2)+32 \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)-32 \Im\left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)-22 \text{Li}_4\left(\frac{1}{2}\right)+\frac{21 \zeta (3)}{2}+7 \pi \zeta (3)-14 \zeta (3) \log (2)+\frac{\zeta \left(4,\frac{1}{4}\right)}{16}-\frac{\zeta \left(4,\frac{3}{4}\right)}{16}+\frac{277 \pi ^4}{960}-\pi ^3+\frac{5 \pi ^2}{3}+8 \pi -32-\log ^4(2)-\frac{2 \log ^3(2)}{3}+\frac{9}{8} \pi ^2 \log ^2(2)-4 \log ^2(2)-\frac{1}{2} \pi ^3 \log (2)+\frac{5}{6} \pi ^2 \log (2)+4 \pi \log (2)-16 \log (2)$
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$\small \, _7F_6\left(\{1\}_6,\frac{7}{4};\{2\}_5,\frac52;1\right)=36 \Im(\text{CMZV}(4,\{4,1\},\{i,1\}))-20 \Im(\text{CMZV}(4,\{4,1\},\{i,-1\}))-64 \Re(\text{CMZV}(4,\{3,1,1\},\{1,1,i\}))+64 C \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+21 C \zeta (3)-2 \pi ^3 C+\frac{10 \pi ^2 C}{3}-8 \pi C^2+16 \pi C-\frac{4}{3} C \log ^3(2)-8 C \log ^2(2)+16 C^2 \log (2)+\frac{5}{3} \pi ^2 C \log (2)+8 \pi C \log (2)-32 C \log (2)-64 \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+64 \Im\left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)-64 \Im\left(\text{Li}_5\left(\frac{1}{2}+\frac{i}{2}\right)\right)+44 \text{Li}_4\left(\frac{1}{2}\right)-2 \text{Li}_5\left(\frac{1}{2}\right)+20 \text{Li}_4\left(\frac{1}{2}\right) \log (2)-\frac{37 \pi ^2 \zeta (3)}{16}-14 \pi \zeta (3)-21 \zeta (3)-\frac{457 \zeta (5)}{64}+7 \zeta (3) \log ^2(2)-7 \pi \zeta (3) \log (2)+28 \zeta (3) \log (2)-\frac{\zeta \left(4,\frac{1}{4}\right)}{8}+\frac{1}{16} \pi \zeta \left(4,\frac{1}{4}\right)+\frac{\zeta \left(4,\frac{3}{4}\right)}{8}-\frac{1}{16} \pi \zeta \left(4,\frac{3}{4}\right)-\frac{7}{32} \zeta \left(4,\frac{1}{4}\right) \log (2)+\frac{7}{32} \zeta \left(4,\frac{3}{4}\right) \log (2)+\frac{95 \pi ^5}{384}+2 \pi ^3-\frac{10 \pi ^2}{3}-\frac{277 \pi ^4}{480}-16 \pi +64+\frac{13 \log ^5(2)}{15}+2 \log ^4(2)-\frac{67}{72} \pi ^2 \log ^3(2)+\frac{4 \log ^3(2)}{3}+\frac{1}{4} \pi ^3 \log ^2(2)-\frac{9}{4} \pi ^2 \log ^2(2)+8 \log ^2(2)-\frac{97}{960} \pi ^4 \log (2)+\pi ^3 \log (2)-\frac{5}{3} \pi ^2 \log (2)-8 \pi \log (2)+32 \log (2)$
Here $\text{CMZV}$ denotes level $4$ colored Multiple Zeta Values. Proofs are recorded in paper 3, in which the final steps (i.e. computing $4$-admissible integrals) depends on the algorithm given by paper 4. By combining these one have that for all $n\in\mathbb{N}, p=1/\frac12, q=\frac14/\frac34$, the integral $\int_0^1 \frac{\text{Li}_n(x^p)}{(x(1-x))^q}dx$ is computable via MZVs.
Summary.
$5.$ Due to arguments above and paper 3 one conclude that OP's integral can be expressed by Multiple Zeta Values of level $1,2,4$ (up to Gamma factors) whenever:
- $2a\in \mathbb Z, 2b\in \mathbb Z$
- $4a\in \mathbb Z, 4b\in \mathbb Z, \text{and at least one of}\ a, b, a+b\in \mathbb Z$
- $4a\in \mathbb Z, 4b\in \mathbb Z, a=b$
And can be expressed via polygamma function whenever:
- $a+b\in \mathbb Z, a,b\in\mathbb R$
One may notice the similarity and difference of this answer (trilog case) and answer to this post (dilog case). More specifically, only $4$ out of $6$ classes of integrals there are generalizable to arbitrary weights (like here) due to loss of symmetry of dilogarithm. Paper 3 also conjectures that the integral is expressible via level $4$ MZVs whenever $4a\in \mathbb Z, 4b\in \mathbb Z$ (without further restriction of $a,b$) but currently I have no proof.