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Question Number 36689 by prof Abdo imad last updated on 04/Jun/18

1) f i n d t h e v a l u e o f \int_{0}^{1} l n (1 - x^{3}) d x t h e n

f i n d t h e v a l u e o f \int_{0}^{1} l n (1 + x + x^{2}) d x

2) f i n d t h e v a l u e o f \int_{0}^{1} l n (1 + x^{3}) d x t h e n

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

Commented by maxmathsup by imad last updated on 04/Aug/18

1) l e t A = \int_{0}^{1} l n (1 - x^{3}) d x w e h a v e {l n}^{^{'}} {(1 - u)}^{} = - \frac{1}{1 - u} = - \sum_{n = 0}^{\infty} u^{n}

w i t h ∣ u ∣< 1 \Rightarrow l n (1 - u) = - \sum_{n = 0}^{\infty} \frac{u^{n + 1}}{n + 1} = - \sum_{n = 1}^{\infty} \frac{u^{n}}{n} \Rightarrow

A = - \int_{0}^{1} (\sum_{n = 1}^{\infty} \frac{x^{3 n}}{n}) d x = - \sum_{n = 1}^{\infty} \frac{1}{n} \int_{0}^{1} x^{3 n} d x = - \sum_{n = 1}^{\infty} \frac{1}{n (3 n + 1)} \Rightarrow

- \frac{A}{3} = \sum_{n = 1}^{\infty} \frac{1}{3 n (3 n + 1)} = \sum_{n = 1}^{\infty} (\frac{1}{3 n} - \frac{1}{3 n + 1}) l e t

S_{n} = \sum_{k = 1}^{n} (\frac{1}{3 k} - \frac{1}{3 k + 1}) \Rightarrow S_{n} = \frac{1}{3} H_{n} - \sum_{k = 1}^{n} \frac{1}{3 k + 1} b u t

\sum_{k = 1}^{n} \frac{1}{3 k + 1} ? w e h a v e \sum_{k = 1}^{n} \frac{1}{k} = \sum_{k = 3 p} \frac{1}{k} + \sum_{k = 3 p + 1} \frac{1}{k} + \sum_{k = 3 p + 2} \frac{1}{k}

= \sum_{p = 1}^{[\frac{n}{3}]} \frac{1}{3 p} + \sum_{p = 0}^{[\frac{n - 1}{3}]} \frac{1}{3 p + 1} + \sum_{p = 0}^{[\frac{n - 2}{3}]} \frac{1}{3 p + 2} b u t

\sum_{p = 0}^{[\frac{n - 2}{3}]} \frac{1}{3 p + 2} =_{p = j - 1} \sum_{j = 1}^{[\frac{n - 2}{3}] - 1} \frac{1}{3 j - 1} \Rightarrow

\sum_{p = 0}^{[\frac{n - 1}{3}]} \frac{1}{3 p + 1} + \sum_{p = 1}^{[\frac{n - 5}{3}]} \frac{1}{3 p - 1} = H_{n} - \frac{1}{3} H_{[\frac{n}{3}]} \dots b e c o n t i n u e d \dots