nice-calculus-evaluate-0-1-ln-1-x-x-3-

Question Number 137874 by mnjuly1970 last updated on 07/Apr/21

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

e v a l u a t e ::

Ω = \int_{0}^{1} {(\frac{l n (1 - x)}{x})}^{3} = ? \dots .

Answered by EnterUsername last updated on 07/Apr/21

Ω = \int_{0}^{1} {(\frac{l n (1 - x)}{x})}^{3} d x

= \int_{0}^{1} \frac{{l n}^{3} x}{{(1 - x)}^{3}} d x = {[\frac{{l n}^{3} x}{2 {(1 - x)}^{2}} - \frac{3}{2} \int \frac{{l n}^{2} x}{x {(1 - x)}^{2}} d x]}_{0}^{1}

= {[\frac{{l n}^{3} x}{2 {(1 - x)}^{2}} - \frac{3}{2} \int (\frac{1}{x} + \frac{1}{1 - x} + \frac{1}{{(1 - x)}^{2}}) {l n}^{2} x d x]}_{0}^{1}

= {[\frac{{l n}^{3} x}{2 {(1 - x)}^{2}} - \frac{{l n}^{3} x}{2} - \frac{3}{2} \int \frac{{l n}^{2} x}{1 - x} d x - \frac{3}{2} \int \frac{{l n}^{2} x}{{(1 - x)}^{2}} d x]}_{0}^{1}

= - \frac{3}{2} (\int_{0}^{1} \frac{{l n}^{2} x}{1 - x} d x + \int_{0}^{1} \frac{{l n}^{2} x}{{(1 - x)}^{2}} d x) = - \frac{3}{2} (- ψ^{″} (1) + 2 ζ (2))

= - \frac{3}{2} (2 ζ (3) + 2 ζ (2)) = - 3 (ζ (3) + ζ (2))

Commented by EnterUsername last updated on 07/Apr/21

S e e Q 137829 f o r \int_{0}^{1} \frac{{l n}^{2} x}{{(1 - x)}^{2}} d x = 2 ζ (2)

Commented by mnjuly1970 last updated on 07/Apr/21

v e r y n i c e s o l u t i o n t h a n k s a l o t

m a s t e r \dots .