calculate-0-1-ln-x-ln-1-x-ln-1-x-2-dx-

Question Number 60658 by Mr X pcx last updated on 23/May/19

c a l c u l a t e \int_{0}^{1} l n (x) l n (1 - x) l n (1 - x^{2}) d x

Commented by maxmathsup by imad last updated on 29/May/19

w e h a v e l n (1 - x) = - \sum_{n = 1}^{\infty} \frac{x^{n}}{n} a n d l n (1 - x^{2}) = - \sum_{n = 1}^{\infty} \frac{x^{2 n}}{n} i f ∣ x ∣< 1

l n (1 - x) l n (1 - x^{2}) = (\sum_{n = 1}^{\infty} \frac{x^{n}}{n}) (\sum_{n = 1}^{\infty} \frac{x^{2 n}}{n}) l e t a_{n} = \frac{x^{n}}{n} a n d b_{n} = \frac{x^{2 n}}{n} \Rightarrow

l n (1 - x) l n (1 - x^{2}) = (Σ a_{n}) (Σ b_{n}) = \sum_{n = 1}^{\infty} c_{n} w i t h

c_{n} = \sum_{i + j = n} a_{i} b_{j} = \sum_{i = 1}^{n - 1} a i b_{n - i} = \sum_{i = 1}^{n - 1} \frac{x^{i}}{i} \frac{x^{2 n - 2 i}}{n - i} = \sum_{i = 1}^{n} \frac{x^{2 n - i}}{i (n - i)}

\Rightarrow \int_{0}^{1} l n (x) l n (1 - x) (1 - x^{2}) d x = \int_{0}^{1} (\sum_{n = 1}^{\infty} \sum_{i = 1}^{n - 1} \frac{x^{2 n - i}}{i (n - i)}) l n (x) d x

= \sum_{n = 1}^{\infty} (\sum_{i = 1}^{n - 1} \frac{1}{i (n - i)} \int_{0}^{1} x^{2 n - i} l n (x) d x)

\int_{0}^{1} x^{2 n - i} l n (x) d x =_{b y p a r t s} {[\frac{1}{2 n - i + 1} x^{2 n - i + 1} l n (x)]}_{0}^{1} - \int_{0}^{1} \frac{1}{(2 n - i + 1)} x^{2 n - i} d x

= - \frac{1}{{(2 n - i + 1)}^{2}} \Rightarrow

I = - \sum_{n = 1}^{\infty} (\sum_{i = 1}^{n - 1} \frac{1}{i (n - i) {(2 n - i + 1)}^{2}}) \dots b e c o n t i n u e d \dots .