1-find-f-x-0-x-ln-t-ln-1-t-dt-with-0-x-1-2-find-the-value-of-0-1-ln-t-ln-1-t-dt-

Question Number 43918 by maxmathsup by imad last updated on 17/Sep/18

1) f i n d f (x) = \int_{0}^{x} l n (t) l n (1 - t) d t w i t h 0 ⩽ x ⩽ 1

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

Commented by maxmathsup by imad last updated on 19/Sep/18

1) w e h a v e f o r 0 < t ⩽ x < 1 {l n}^{^{'}} (1 - t) = - \frac{1}{1 - t} = - \sum_{n = 0}^{\infty} t^{n} \Rightarrow

l n (1 - t) = - \sum_{n = 0}^{\infty} \frac{1}{n + 1} t^{n + 1} = - \sum_{n = 1}^{\infty} \frac{t^{n}}{n} \Rightarrow

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

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

= \frac{1}{n + 1} x^{n + 1} l n (x) - \frac{1}{n + 1} \int_{0}^{x} t^{n} d t = \frac{1}{n + 1} x^{n + 1} l n (x) - \frac{1}{{(n + 1)}^{2}} x^{n + 1} \Rightarrow

f (x) = - \sum_{n = 1}^{\infty} \frac{1}{n} {\frac{1}{n + 1} x^{n + 1} l n (x) - \frac{1}{{(n + 1)}^{2}} x^{n + 1}}

= - \sum_{n = 1}^{\infty} \frac{1}{n (n + 1)} x^{n + 1} l n (x) + \sum_{n = 1}^{\infty} \frac{1}{n {(n + 1)}^{2}} x^{n + 1} = H (x) - K (x)

K (x) = l n (x) \sum_{n = 1}^{\infty} {\frac{1}{n} - \frac{1}{n + 1}} x^{n + 1}

= l n (x) \sum_{n = 1}^{\infty} \frac{x^{n + 1}}{n} - l n (x) \sum_{n = 1}^{\infty} \frac{x^{n + 1}}{n + 1}

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

= - x l n (x) l n (1 - x) - l n (x) (- l n ∣ 1 - x ∣ - x)

= - x l n (x) l n ∣ 1 - x ∣ + l n (x) (l n ∣ 1 - x ∣ + x)

= (1 - x) l n (x) l n ∣ 1 - x ∣ + x l n (x) . a l s o w e h a v e

\frac{d}{d x} K (x) = \sum_{n = 1}^{\infty} \frac{1}{n (n + 1)} x^{n} = \sum_{n = 1}^{\infty} {\frac{1}{n} - \frac{1}{n + 1}} x^{n}

= \sum_{n = 1}^{\infty} \frac{x^{n}}{n} - \sum_{n = 1}^{\infty} \frac{x^{n}}{n + 1} = - l n ∣ 1 - x ∣ - \sum_{n = 2}^{\infty} \frac{x^{n - 1}}{n}

= - l n ∣ 1 - x ∣ - \frac{1}{x} {\sum_{n = 1}^{\infty} \frac{x^{n}}{n} - x} = - l n ∣ 1 - x ∣ + \frac{1}{x} l n ∣ 1 - x) + 1

= (\frac{1}{x} - 1) l n ∣ 1 - x ∣ + 1 = \frac{(1 - x) l n ∣ 1 - x ∣}{x} + 1 \Rightarrow

K (x) = x + \int \frac{(1 - x) l n ∣ 1 - x ∣}{x} d x + k \dots . b e c o n t i n u e d \dots

Commented by maxmathsup by imad last updated on 19/Sep/18

2) \int_{0}^{1} l n (t) l n (1 - t) d t = {l i m}_{x \to 0} \int_{0}^{x} l n (t) l n (1 - t) d t

= {l i m}_{x \to 1} {\sum_{n = 1}^{\infty} \frac{x^{n + 1}}{{(n + 1)}^{2}} - \sum_{n = 1}^{\infty} \frac{x^{n + 1} l n x}{n (n + 1)}} = \sum_{n = 1}^{\infty} \frac{1}{{(n + 1)}^{2}}

= \sum_{n = 2}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{6} - 1 .

Commented by maxmathsup by imad last updated on 19/Sep/18

\int_{0}^{1} l n (t) l n (1 - t) d t = {l i m}_{x \to 1} \int_{0}^{x} l n (t) l n (1 - t) d t = \dots