I managed to convert the integral $\large\int_0^1 \frac{\operatorname{Li}_3^2(-x)}{x^2}\ dx$ to $\int_0^1 \ln^2x\ln(1+x)\operatorname{Li}_3\left(\frac1x\right)\ dx$ in hope of evaluating it in a different way. I checked both numerically and I got $0.89767$ for the first integral and $0.89767 – 1.11049\ i$ for the second one.
My question is why I got the extra imaginary part? Did I do any mistake in my following solution:
using $\frac{\operatorname{Li}_3(x)}{x}=\frac12\int_0^1\frac{\ln^2y}{1-xy}\ dy$
we write
$$\int_0^1 \left(\frac{\operatorname{Li}_3(-x)}{-x}\right)^2\ dx=\frac14\int_0^1\int_0^1\int_0^1\frac{\ln^2y\ln^2z}{(1+x)(1+xy)(1+xz)}\ dxdydz$$
$$=\frac14\int_0^1\int_0^1\ln^2y\ln^2z\left(\color{blue}{\int_0^1\frac{dx}{(1+xy)(1+xz)}}\right)\ dydz$$
$$=\int_0^1\int_0^1\ln^2y\ln^2z\left(\color{blue}{\frac{\ln(1+y)-\ln(1+z)}{y-z}}\right)\ dy dz$$
$$=\frac14\int_0^1\frac{\int_0^1\ln^2y\ln^2z\ln(1+y)}{y-z}\ dy dz-\frac14\underbrace{\int_0^1\int_0^1\frac{\ln^2y\ln^2z\ln(1+z)}{y-z}\ dy dz}_{\text{interchange}\ y \ \text{and}\ z}$$
$$=\frac12\int_0^1\int_0^1\frac{\ln^2y\ln^2z\ln(1+y)}{y-z}\ dy dz$$
$$=\int_0^1\ln^2y\ln(1+y)\left(\color{red}{\frac12\int_0^1\frac{\ln^2z}{y-z}\ dz}\right)\ dy$$
$$=\int_0^1\ln^2y\ln(1+y)\color{red}{\operatorname{Li}_3\left(\frac1y\right)}\ dy$$
Thanks in advance.
Best Answer
Good to know that
which, of course, may be generalized. The integrals have to be understood in a Cauchy principal value sense.
Proof to $i)$
Since $\displaystyle \int \frac{x\log(y)}{1-x y}\textrm{d}y=-\log(y)\log(1-x y)-\operatorname{Li}_2(x y)+C$, by the Cauchy principal value all reduces to calculating $\displaystyle\lim_{\epsilon \to0^{+}} \left(\int_0^{1/x-\epsilon} \frac{x\log(y)}{1-x y}\textrm{d}y+\int_{1/x+\epsilon}^1 \frac{x\log(y)}{1-x y}\textrm{d}y\right)$, which leads immediately to the desired value, that is $\displaystyle \Re\{\operatorname{Li}_2(x)\}=-\operatorname{P.V.} \int_0^1 \frac{x\log(y)}{1-x y}\textrm{d}y, \ x>1$.
https://en.wikipedia.org/wiki/Cauchy_principal_value