I'm interested in evaluating the following integral
$ \DeclareMathOperator{\Li}{Li}$
$$ \mathcal{A} = -\int_0^1 \frac{\ln(1-x)}{1+x} \Li_2(x) \, \mathrm{d}x $$
My most successful attempt thus far went like this:
First, converting the dilogarithm to its integral form yields
$$ \mathcal{A} = \int_0^1 \int_0^1 \frac{\ln(1-x) \ln(1-xt) }{t(1+x)} \, \mathrm{d}t \, \mathrm{d}x $$
Interchanging the bounds of integration yields
$$ \mathcal{A} = \int_0^1 \int_0^1 \frac{\ln(1-x) \ln(1-xt) }{t(1+x)} \, \mathrm{d}x \,\mathrm{d}t $$
For the inner integral, we have
$$ \mathfrak{J}(t) = \int_0^1 \frac{ \ln(1-x)\ln(1-xt) }{(1+x)} \, \mathrm{d}x $$
Differentiating under the integral with respect to $t$ and then applying partial fractions yields
$$ \mathfrak{J}'(t) = \frac{1}{1+t} \int_{0}^{1} \frac{\ln(1-x) }{tx-1} \, \mathrm{d}x +\frac{1}{1+t} \int_0^1 \frac{\ln(1-x)}{1+x} \, \mathrm{d}x $$
This evaluates (not) very nicely to
$$ \mathfrak{J}'(t) = \frac{1}{t(1+t) } \Li_2\left(\frac{t}{1-t} \right) +\frac{1}{1+t} \left( \frac{\ln^2(2)}{2} -\frac{\pi^2}{12} \right) $$
This means that our original integral is equivalent to solving
$$ \mathcal{A} = \int_0^1 \int_0^t \frac{1}{at(1+a)} \Li_2 \left(\frac{a}{1-a} \right) \, \mathrm{d}a \, \mathrm{d}t + \left(\frac{\ln^2(2)}{2}-\frac{\pi^2}{12} \right) \int_0^1 \int_0^t \frac{1}{t(1+a)} \, \mathrm{d}a \, \mathrm{d}t $$
The second part of the integral above is trivial, what's giving me trouble is the first part. Any help whatsoever is much appreciated!
Best Answer
The integral is immediately derived by combining the integral result at the point $i)$, Sect. $1.27$, page $17$, from the book (Almost) Impossible Integrals, Sums, and Series and Landen's identity, and we get $$\int_0^1 \frac{\ln(1-x)}{1+x} \operatorname{Li}_2(x)\textrm{d}x=3\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{1}{4}\log^2(2)\zeta(2)-\frac{29}{16}\zeta(4)+\frac{1}{8}\log^4(2).$$ The other resulting integral is trivial, that is $\displaystyle \int_0^1 \frac{\log^3(1-x)}{1+x}\textrm{d}x=-6\operatorname{Li}_4\left(\frac{1}{2}\right).$
End of story (also subtler ways are possible)
Additional information: If also interested in the following very similar integral, $\displaystyle \int_0^1 \frac{\log(1-x)}{1+x} \operatorname{Li}_2(-x)\textrm{d}x$, one may find it calculated here.