Evaluar $\int_0^1 \frac{\ln \left(1+x+x^2+\cdots +x^{n} \right)}{x}dx$

Question

Cómo evaluar $$\int_0^1 \frac{\ln \left(1+x+x^2+\cdots +x^{n} \right)}{x}dx$$ Mi intento: \begin{align*} \int_0^1 \frac{\ln \left(1+x+x^2+\cdots +x^{n} \right)}{x}dx &= \int_0^1 \frac{\ln \left(\dfrac{1-x^{n+1}}{1-x} \right)}{x}dx \\ &= \int_0^1 \frac{\ln \left({1-x^{n+1}} \right)}{x}dx -\int_0^1 \frac{\ln \left({1-x} \right)}{x}dx \\ &=\frac{1}{n+1} \int_0^1 \frac{\ln \left({1-x} \right)}{x}dx -\int_0^1 \frac{\ln \left({1-x} \right)}{x}dx \end{align*> pero ¿qué sigue? Llevo atascado aquí un rato.

Answer 1

Tenga en cuenta que $$-\mathrm{Li}_{2}\left ( 1 \right )=\int_{0}^{1}\frac{\ln\left ( 1-x \right )}{x}\, \mathrm{d}x=-\sum_{n=1}^{\infty }\frac{1}{n}\int_{0}^{1}x^{n-1}\, \mathrm{d}x=-\sum_{n=1}^{\infty }\frac{1}{n^{2}}=-\frac{\pi ^{2}}{6}$$ y la respuesta seguirá.

Answer 2

$$\int_0^1 \dfrac{\ln(1-x)}xdx = \int_0^1 \left(-\sum_{k=1}^{\infty} \dfrac{x^{k-1}}k\right)dx = -\sum_{k=1}^{\infty} \dfrac1k\int_0^1 x^{k-1}dx = - \sum_{k=1}^{\infty} \dfrac1{k^2} = -\zeta(2)$$