Let $\displaystyle f(x)=\sum_{n=1}^{\infty} \dfrac{x^n}{n^2} , \; x \in (0, 1)$. Evaluate $f(1/2)$ without using the known formulae of the dilogarithm or the equation it satisfies.
May I have some hints on how to begin? I tried Laplace Transformation and a link to a theorem but it did not work. I also tried the Taylor series of $\ln(1+x)$ and $\ln(1-x)$ but I cannot a see a pattern.
We may exploit the identity: $$ \int_{0}^{1} x^n\log x\,dx = -\frac{1}{(n+1)^2}\tag{1} $$ leading to (just multiply both terms of $(1)$ by $\frac{1}{2^{n+1}}$ then sum over $n$): $$\begin{eqnarray*} \operatorname{Li}_2\left(\frac{1}{2}\right)&=&-\frac{1}{2}\int_{0}^{1}\frac{\log x}{1-x/2}\,dx\\&=&-\frac{\log 2}{2}\int_{0}^{1}\frac{dx}{1-x/2}-\frac{1}{2}\int_{0}^{1}\frac{\log(x/2)}{1-x/2}\,dx \\ &=&-\log^2 2-\int_{0}^{1/2}\frac{\log x}{1-x}\,dx\\&=&-\log^2 2-\int_{0}^{1}\frac{\log x}{1-x}\,dx+\int_{1/2}^{1}\frac{\log x}{1-x}\,dx\\&=&\zeta(2)-\log^2 2+\int_{1}^{2}\frac{\log x}{x(1-x)}\,dx\\&=&\zeta(2)-\frac{\log^2 2}{2}+\int_{1}^{2}\frac{\log x}{1-x}\,dx\tag{2}\end{eqnarray*}$$ where: $$\begin{eqnarray*} \int_{1}^{2}\frac{\log x}{1-x}\,dx&=&-\int_{0}^{1}\frac{\log(x+1)}{x}\,dx\\&=&\sum_{n\geq 1}\frac{(-1)^{n}}{n^2}=-\frac{\zeta(2)}{2}\tag{3}\end{eqnarray*}$$ from which: $$\operatorname{Li}_2\left(\frac{1}{2}\right)=\color{red}{\frac{\zeta(2)-\log^2 2}{2}}.$$