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

Question Number 45232 by maxmathsup by imad last updated on 10/Oct/18

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

Commented by maxmathsup by imad last updated on 11/Oct/18

l e t I = \int_{0}^{1} l n (x) l n (1 + x) d x w e h a v e f o r ∣ x ∣< 1

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

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

I = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{1} x^{n} l n (x) d x b y p a r t s

A_{n} = \int_{0}^{1} x^{n} l n (x) d x = {[\frac{1}{n + 1} x^{n + 1} l n (x)]}_{x \to 0}^{1} - \int_{0}^{1} \frac{1}{(n + 1)} x^{n} d x

= - \frac{1}{{(n + 1)}^{2}} \Rightarrow I = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n {(n + 1)}^{2}} l e t d e c o m p o s e F (x) = \frac{1}{x {(x + 1)}^{2}}

F (x) = \frac{a}{x} + \frac{b}{x + 1} + \frac{c}{{(x + 1)}^{2}}

a = {l i m}_{x \to 0} x F (x) = 1

c = {l i m}_{x \to - 1} {(x + 1)}^{2} F (x) = - 1 \Rightarrow F (x) = \frac{1}{x} + \frac{b}{x + 1} - \frac{1}{{(x + 1)}^{2}}

F (1) = \frac{1}{4} = 1 + \frac{b}{2} - \frac{1}{4} \Rightarrow 1 = 4 + 2 b - 1 = 3 + 2 b \Rightarrow 2 b = - 2 \Rightarrow b = - 1 \Rightarrow

F (x) = \frac{1}{x} - \frac{1}{x + 1} - \frac{1}{{(x + 1)}^{2}} \Rightarrow

I = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n + 1} - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{2}} b u t

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

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

\sum_{n = 1}^{\infty} \frac{1}{{(n + 1)}^{2}} = \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} + 1 l e t f i n d \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}}

w e h a v e \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = \sum_{n = 1}^{\infty} \frac{1}{{(2 n)}^{2}} - \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}}

= \frac{1}{4} \frac{π^{2}}{6} - \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} b u t \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{1}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} \Rightarrow

\sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{3}{4} Σ \frac{1}{n^{2}} = \frac{3}{4} \frac{π^{2}}{6} = \frac{π^{2}}{8} \Rightarrow \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = \frac{π^{2}}{24} - \frac{π^{2}}{8} = - \frac{π^{2}}{12} \Rightarrow

\int_{0}^{1} l n (x) l n (1 + x) d x = - l n (2) - l n (2) + 1 - (- \frac{π^{2}}{12} + 1)

= \frac{π^{2}}{12} - 2 l n (2) .

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

\int_{0}^{1} l n (1 + x) \times l n (x) d x \approx - 0.21 (a p p r o x)