advanced-calculus-prove-that-0-1-ln-x-ln-1-x-1-x-dx-13-8-3-pi-2-4-ln-2-

Question Number 137501 by mnjuly1970 last updated on 03/Apr/21

\dots . a d v a n c e d \dots . c a l c u l u s \dots .

p r o v e t h a t ::

ϕ = \int_{0}^{1} \frac{l n (x) . l n (1 - x)}{1 + x} d x = \frac{13}{8} ζ (3) - \frac{π^{2}}{4} l n (2) \dots .

Answered by Ñï= last updated on 04/Apr/21

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

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

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

= \frac{\partial}{\partial a \partial b} ∣_{a = 0, b = - 1} \int_{0}^{1} x^{a} {(1 - x^{2})}^{b} d x - \frac{\partial}{\partial a \partial b} ∣_{a = 1, b = - 1} \int_{0}^{1} x^{a} {(1 - x^{2})}^{b} d x - {\int \frac{l n (1 + x) {l n (1 + x) + l n \frac{x}{1 + x}}}{1 + x}}_{0}^{1}

= \frac{\partial}{\partial a \partial b} ∣_{a = 0, b = - 1} \frac{1}{2} \int_{0}^{1} u^{(a - 1) / 2} {(1 - u)}^{b} d u - \frac{\partial}{\partial a \partial b} ∣_{a = 1, b = - 1} \frac{1}{2} \int_{0}^{1} u^{(a - 1) / 2} {(1 - u)}^{b} d u - {\int \frac{{l n}^{2} (1 + x)}{1 + x} d x + \int \frac{l n (1 - \frac{1}{1 + x})}{1 + x} l n (1 + x) d x}_{0}^{1}

= \frac{\partial}{\partial a \partial b} ∣_{a = 0, b = - 1} \frac{1}{2} B (\frac{a + 1}{2}, b + 1) - \frac{\partial}{\partial a \partial b} ∣_{a = 1, b = - 1} \frac{1}{2} B (\frac{a + 1}{2}, b + 1) - \frac{1}{3} {l n}^{3} 2 + \underset{n = 1}{\sum^{\infty}} \int_{0}^{1} \frac{{(1 + x)}^{- n}}{n} \cdot \frac{l n (1 + x)}{1 + x} d x

= \frac{3}{2} ζ (3) - \frac{π^{2}}{4} l n 2 - \frac{1}{3} {l n}^{3} 2 + \underset{n = 1}{\sum^{\infty}} \frac{1}{n} {\int_{0}^{1} {(1 + x)}^{- n - 1} l n (1 + x) d x}

= \frac{3}{2} ζ (3) - \frac{π^{2}}{4} l n 2 - \frac{1}{3} {l n}^{3} 2 + \underset{n = 1}{\sum^{\infty}} \frac{1}{n} {\frac{{(1 + x)}^{- n}}{- n} l n (1 + x) - \int \frac{{(1 + x)}^{- n - 1}}{- n} d x}_{0}^{1}

= \frac{3}{2} ζ (3) - \frac{π^{2}}{4} l n 2 - \frac{1}{3} {l n}^{3} 2 - \underset{n = 1}{\sum^{\infty}} \frac{{(1 + x)}^{- n} l n (1 + x)}{n^{2}} ∣_{0}^{1} + \underset{n = 1}{\sum^{\infty}} \frac{{(1 + x)}^{- n}}{n^{3}} ∣_{0}^{1}

= \frac{3}{2} ζ (3) - \frac{π^{2}}{4} l n 2 - \frac{1}{3} {l n}^{3} 2 - {L i}_{2} (\frac{1}{1 + x}) l n (1 + x) ∣_{0}^{1} - {L i}_{3} (\frac{1}{1 + x}) ∣_{0}^{1}

= \frac{3}{2} ζ (3) - \frac{π^{2}}{4} l n 2 - \frac{1}{3} {l n}^{3} 2 - l n 2 {L i}_{2} (\frac{1}{2}) - {L i}_{3} (\frac{1}{2}) + {L i}_{3} (1)

= a n s \dots

Commented by mnjuly1970 last updated on 04/Apr/21

g r e a t \dots t h a n k s a l o t \dots