Reduction. As @Raffaele mentioned, by expanding dilog, integrating term by term (Beta integral) and summing up, one reduces the integral to a generalized hypergeometric function, i.e.
$$\int^1_0 x^a (1-x)^b\operatorname{Li}_2 (x)\, \mathrm dx=B(a+2,b+1) \, _4F_3(1,1,1,a+2;2,2,a+b+3;1)$$
Claim. This hypergeometric function 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:
- $a=b$
- $a+b\in \mathbb Z$
- $\text{At least one of}\ a+2b, b+2a \in \mathbb Z$
Proof. For proof of the MZV part (mainly by Fourier Legendre expansion), see Theorem $1, 2$ of this paper (there seems to be no general formula though). For case $1$ of polygamma part, recall reflection formula of dilog one have a general formula that
- $\int^1_0 x^{a-1} (1-x)^{a-1}\operatorname{Li}_2 (x)\, \mathrm dx=\frac12\int^1_0 x^{a-1} (1-x)^{a-1}(\operatorname{Li}_2 (x)+\operatorname{Li}_2 (1-x))\, \mathrm dx$
$=\frac12\int^1_0 x^{a-1} (1-x)^{a-1}(\zeta(2)-\log(x)\log(1-x))\, \mathrm dx=\frac12(\zeta(2)B(a,a)-\partial^{(1,1)}B(a,a))$
$=\frac{1}{12} \left(-6 (\psi ^{(0)}(a)-\psi ^{(0)}(2 a))^2+6 \psi ^{(1)}(2 a)+\pi ^2\right) B(a,a)$
For case $2$ of polygamma part, see Theorem $3$ of paper linked above. For case $3$, due to second entry of this page the integral is evaluable whenever $b+2a=-2$. As @pisco mentioned, by repeated IBP and modulo polygamma integrals $\int_0^1 x^* (1-x)^* \log^*(1-x) \mathrm dx$ one may evaluate all $I(a+m,b+n)$ once $I=I(a,b)$ is known; this completes the proof of case $b+2a\in\mathbb Z$. The case $a+2b\in \mathbb Z$ is direct by reflection again (see case $1$).
Examples. Case $1$ of MZV part:
- $a=\frac12, b=\frac32: I=\frac{\pi ^3}{96}-\frac{17 \pi }{576}-\frac{1}{8} \pi \log ^2(2)+\frac{1}{48} \pi \log (2)$
Case $2$ of MZV part:
- $a=\frac54, b=\frac34: \\ I=\frac{\pi \left(-480 C+35 \pi ^2-178 \pi -134-540 \log ^2(2)+180 \pi \log (2)+1068 \log (2)\right)}{1536 \sqrt{2}}$
- $a=-1, b=\frac14: \\ I=-32 C-48 \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)-14 \zeta (3)+\frac{7 \pi ^3}{8}-4 \pi ^2+64+\frac{3}{2} \pi \log ^2(2)+3 \pi ^2 \log (2)$
Case $3$ of MZV part:
- $a=b=-\frac54: I= \frac{\left(-5 \pi ^2+24 \pi +12 \log ^2(2)-12 \pi \log (2)-48 \log (2)\right) \Gamma \left(\frac{3}{4}\right)^2}{3 \sqrt{\pi }}$
Polygamma part:
- $a=b=-\frac56: I=-\frac{\sqrt{\pi } \Gamma \left(\frac{7}{6}\right) \left(\pi ^2+24 \log ^2(2)+8 \pi \sqrt{3} \log (2)-6 \psi ^{(1)}\left(\frac{1}{3}\right)\right)}{\sqrt[3]{2} \Gamma \left(\frac{2}{3}\right)}$
Notes.
- In MZV Case $1$ example I omit the trivial case $a, b\in \mathbb Z$ since they are direct via IBP.
- MZV Case $2$ examples are more complicated; for reason of appearance of certain constants, see paper linked above.
- MZV Case $3$ example can be established by either Fourier Legendre expansion or method of "polygamma part", with the former generalizable to trilog case (and above) but the latter not.
- Method of polygamma part, case $1,3$ are ad-hoc and not generalizable to trilog. Case $2$ of polygamma part is generalizable since only partial fractions and Beta derivatives are involved.
- I only give a polygamma Case $1$ example; further examples are left to readers.
- This post is the trilog counterpart of this problem; some more difficult series are evaluated there. For more information, see paper linked above. The theoretical basis of evaluating these integrals and series are given here.
