nice-calculus-prove-that-0-1-ln-1-x-ln-1-x-2-x-11-3-8-

Question Number 123462 by mnjuly1970 last updated on 25/Nov/20

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

p r o v e t h a t ::

Ω = \int_{0}^{1} \frac{l n (1 - x) l n (1 - x^{2})}{x} \overset{? ?}{=} \frac{11 ζ (3)}{8}

\dots \dots \dots \dots \dots . .

Answered by Lordose last updated on 25/Nov/20

Ω = \int_{0}^{1} \frac{\ln (1 - x) (\ln (1 - x) + \ln (1 + x))}{x} dx

Ω = \int_{0}^{1} \frac{\ln^{2} (1 - x)}{x} dx + \int_{0}^{1} \frac{\ln (1 - x) \ln (1 + x)}{x} dx

Ω = 2 ζ (3) - \frac{5 ζ (3)}{8}

Ω = \frac{11 ζ (3)}{8}

Commented by Dwaipayan Shikari last updated on 25/Nov/20

\int_{0}^{1} \frac{{l o g}^{2} (1 - x)}{x} d x

= \int_{0}^{1} \frac{{l o g}^{2} x}{1 - x} d x = \int_{0}^{1} {l o g}^{2} x \underset{n = 0}{\sum^{\infty}} x^{n} = \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} x^{n} {l o g}^{2} (x) l o g x = t

= \underset{n = 0}{\sum^{\infty}} \int_{- \infty}^{0} t^{2} e^{(n + 1) x} d x (n + 1) x = - Φ

= - \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + 1)}^{3}} \int_{\infty}^{0} Φ^{2} e^{- Φ} d Φ = \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + 1)}^{3}} . Γ (3) = 2 ζ (3)

Commented by mnjuly1970 last updated on 26/Nov/20

t h a n k y o u

Commented by mnjuly1970 last updated on 26/Nov/20

t h a n k y o u . .