How to evaluate $$\int_0^1 \frac{\ln \left(1+x+x^2+\cdots +x^{n} \right)}{x}dx$$ My attempt: \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*} but what's next?I have been stuck here for a while.
Evaluate $\int_0^1 \frac{\ln \left(1+x+x^2+\cdots +x^{n} \right)}{x}dx$
9
$\begingroup$
calculus
integration
-
2How to prove that $\ln{(1-x^{n+1})}=\frac{1}{n+1}\ln{(1-x)}$. I think you used this in your derivation but I am unable how to prove it. Please provide detail of the step required to prove it or provide some reference. Thanks in advance – 2017-01-12
-
2if $u=x^{n+1}$ then $\frac{du}{u}=(n+1)\frac{dx}{x}$. @FrankMoses But I agree, that step wasn't obvious. – 2017-01-12
-
2@Frank Moses The equality you mentioned is wrong. The idea of OP is correct. After doing the substitution suggested by Thomas Andrews, we eventually get $$\int_0^1\frac{\ln(1-x^{n+1})}{x}dx=\frac{1}{n+1}\int_0^1\frac{\ln(1-x)}{x}dx$$ – 2017-01-12
2 Answers
9
Notice that $$-\mathrm{Li}_{2}\left ( 1 \right )=\int_{0}^{1}\frac{\ln\left ( 1-x \right )}{x}\, \mathrm{d}x=-\sum_{n=1}^{\infty }\frac{1}{n}\int_{0}^{1}x^{n-1}\, \mathrm{d}x=-\sum_{n=1}^{\infty }\frac{1}{n^{2}}=-\frac{\pi ^{2}}{6}$$ and the answer will follow.
6
$$\int_0^1 \dfrac{\ln(1-x)}xdx = \int_0^1 \left(-\sum_{k=1}^{\infty} \dfrac{x^{k-1}}k\right)dx = -\sum_{k=1}^{\infty} \dfrac1k\int_0^1 x^{k-1}dx = - \sum_{k=1}^{\infty} \dfrac1{k^2} = -\zeta(2)$$