Calculate a $\operatorname{Li}_{2}(-1)$ using Integral Representation

Question

$\newcommand{LogI}{\operatorname{Li}}$ I know that $\LogI_{2}(-1)=-\frac{\pi^2}{12}$, but I have never seen a proof of this result without using a functional identity of the Dilogarithm or a series expansion. I would like to know how calculate $\LogI_{2}(-1)$ using the integral representation of the Dilogarithm function: $\LogI_{2}(z)=\int_{z}^{0}\frac{\ln{(1-t)}}{t}dt$ or $\LogI_{2}(z)=-\int_{0}^{1}\frac{\ln{(1-zt)}}{t}dt$. This amounts to calculating either of the following integrals: $$\LogI_{2}(-1)=\int_{-1}^{0}\frac{\ln{(1-t)}}{t}dt$$ $$\LogI_{2}(-1)=-\int_{0}^{1}\frac{\ln{(1+t)}}{t}dt$$ I am interested in seeing how to evaluate these integrals without using series expansion or functional identities of the Dilogarithm. Methods involving the use of Complex Analysis are welcome with some explanation.

Answer 1

Enforce the substitution $x\to x^2$. Then,

$$\begin{align} \int_0^1\frac{\log(1-x)}{x}\,dx&=2\int_0^1\frac{\log(1-x^2)}{x}\,dx\\\\ &=2\int_0^1 \frac{\log(1-x)}{x}\,dx+2\int_0^1\frac{\log(1+x)}{x}\,dx\tag1 \end{align}$$

Hence, we see from $(1)$ that

$$\begin{align} -\int_0^1\frac{\log(1+x)}{x}\,dx&=\frac12\int_0^1\frac{\log(1-x)}{x}\,dx\tag2\\\\ &=-\frac12 \int_0^1 \int_0^1 \frac{1}{1-xy}\,dx\,dy\tag3 \end{align}$$

Next, in THIS ANSWER, I evaluated the integral in $(3)$ by using the transformation of coordinates $x=s+t$ and $y=s-t$ and using elementary integral analysis. The result as given in that answer was

$$\int_0^1 \int_0^1 \frac{1}{1-xy}\,dx\,dy=\frac{\pi^2}{6}\tag 4$$

Using $(4)$ in $(3)$ reveals

$$-\int_0^1\frac{\log(1+x)}{x}\,dx=-\frac{\pi^2}{12}$$

as expected! And we are done!