Consider the series $$ s(x)=\sum_{j=1}^{\infty} \frac{x^j+(1-x)^j}{j^2} . $$ Evaluate $s(x)$ for $0<x<1$. (You may use that $\sum_1^{\infty} \frac{1}{j^2}=\frac{\pi^2}{6}$.)
I have proved that $s(x)$ is continues and continuously differentiable in $(0,1)$ as well as uniform convergent by M-Test. But how can I derive the answer which is
$s(x)=\frac{\pi^2}{6}-(\ln x)(\ln (1-x))$
I have also done the following according to the comment
$$ S(x)=\sum_{j=1}^{\infty} \frac{x^j+(1-x)^j.}{j^2}=\sum_{j=1}^{\infty} \frac{\frac{1}{1-x}+\frac{1}{1-(1-x)}}{j^2} $$ $S(x)$ converge unitormly so $\int s(x) d x=\sum_{j=1}^{\infty} \frac{1}{j^2} \int \frac{1}{(1-x) x} d x$ $$ =\frac{\pi^2}{6}(\operatorname{In}(x)-\operatorname{In}(1-x))+c $$ Am I doing something wrong here? Thanks in advance! :D
Consider the derivative of $\ln(x) \, \ln(1-x)$ which is $$ \frac{d}{dx} \, \ln(x) \, \ln(1-x) = \frac{\ln(1-x)}{x} - \frac{\ln(x)}{1-x}.$$ Now integrate both sides which gives $$ \text{Li}_{2}(x) + \text{Li}_{2}(1-x) = c_{0} - \ln(x) \, \ln(1-x),$$ where $\text{Li}_{2}(x)$ is the dilogarithm function. Setting $x = 0$, or $x=1$ leads to $c_{0} = \text{Li}_{2}(1) = \zeta(2)$ and $$ \text{Li}_{2}(x) + \text{Li}_{2}(1-x) = \zeta(2) - \ln(x) \, \ln(1-x). $$
Since $$\text{Li}_{2}(x) = \sum_{n=1}^{\infty} \frac{x^n}{n^2}$$ then $$ \sum_{n=1}^{\infty} \frac{x^n + (1-x)^n}{n^2} = \zeta(2) - \ln(x) \, \ln(1-x). $$
Process without using any named functions:
By using \begin{align} \frac{d}{dx} \, \ln(x) \, \ln(1-x) &= \frac{\ln(1-x)}{x} - \frac{\ln(x)}{1-x} \\ &= \frac{\ln(1-x)}{x} - \frac{\ln(1 - (1-x))}{1-x} \\ &= - \sum_{n=1}^{\infty} \frac{x^{n-1}}{n} + \sum_{n=1}^{\infty} \frac{(1-x)^{n-1}}{n} \end{align} and then integrating both sides gives $$ \sum_{n=1}^{\infty} \frac{x^n + (1-x)^n}{n^2} = c_{0} - \ln(x) \, \ln(1-x). $$ Setting $x = 0$, or $x=1$, it will be shown that $c_{0} = \zeta(2)$ and leads to the series $$ \sum_{n=1}^{\infty} \frac{x^n + (1-x)^n}{n^2} = \zeta(2) - \ln(x) \, \ln(1-x). $$