We know that the series expansion of $\frac{\log(1+x)}{x}$ is
$$\frac{\log(1+x)}{x}=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}….$$
It is also known that the value of dilogarithm at 1 is
$$\int_{-1}^{1} \frac{\log(1+x)}{x}=2\sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}=\frac{3}{2}\zeta(2)=\frac{3}{2}Li_2(1)$$
For this to be true we must show that the RHS of the first equation that is $1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}….$ is termwise integrable. To prove that this series is termwise integral we must show that it's uniformly convergent in $(-1,1)$. Although it is simple to prove that the series converges by the ratio test.
How to show that $1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}….$ is uniformly convergent in $(-1,1)$?
Best Answer
For $x \in [-1+\delta,1-\delta]$ where $0 < \delta < 1$, we have
$$\left|(-1)^k \frac{x^k}{k+1}\right|\leqslant \frac{(1-\delta)^k}{k+1} \leqslant (1-\delta)^k$$
As the geometric series $\sum_{k \geqslant 0} (1-\delta)^k$ converges, it follows by the Weierstrass M-test that we have uniform convergence on $[-1+\delta,1+\delta]$ of the series
$$\frac{\log (1+x)}{x} = \sum_{k=0}^\infty(-1)^k \frac{x^k}{k+1}$$
Now the series may be integrated termwise over $[-1+\delta,1-\delta]$ to obtain
$$\begin{align} \int_{-1+\delta}^{1-\delta}\frac{\log(1+x)}{x} \, dx &= \sum_{k=0}^\infty (-1)^k \frac{(1-\delta)^{k+1} - (-1+\delta)^{k+1}}{(k+1)^2}\\ &= \sum_{k=0}^\infty (-1)^k \frac{(1-\delta)^{k+1} - (-1)^{k+1}(1-\delta)^{k+1}}{(k+1)^2} \\ &= \sum_{k=0}^\infty \frac{(1-\delta)^{k+1} + (-1)^{k}(1-\delta)^{k+1}}{(k+1)^2}\\&= 2\sum_{j=0}^\infty \frac{(1-\delta)^{2j+1} }{(2j+1)^2}\end{align}$$
The series on the RHS is uniformly convergent for $\delta \in [0,1]$, and, consequently we may interchange the limit as $\delta \to 0+$ with the sum to obtain
$$ \int_{-1}^{1}\frac{\log(1+x)}{x} \, dx = \lim_{\delta \to 0+}\int_{-1+\delta}^{1-\delta}\frac{\log(1+x)}{x} \, dx = 2\sum_{k=0}^\infty \frac{1 }{(2j+1)^2}$$