nice-calculus-find-0-1-ln-x-ln-1-x-x-1-x-dx-

Question Number 134852 by mnjuly1970 last updated on 07/Mar/21

\dots n i c e c a l c u l u s \dots

f i n d :::

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

Answered by mnjuly1970 last updated on 07/Mar/21

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

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

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

= ζ (3) + \frac{1}{2} \int_{0}^{1} \frac{{l n}^{2} (1 - t)}{t} d t = 2 ζ (3) \dots ✓ ✓

Answered by Ñï= last updated on 08/Mar/21

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

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

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

= 2 {L i}_{3} (x) ∣_{0}^{1}

= 2 {L i}_{3} (1) = 2 ζ (3)

Commented by mnjuly1970 last updated on 08/Mar/21

t h a n k s a l o t \dots