Calculus-Evaluate-0-1-log-1-x-1-x-3-dx-

Question Number 142469 by mnjuly1970 last updated on 01/Jun/21

\dots \dots C a l c u l u s \dots . .

E v a l u a t e : \int_{0}^{1} {(\frac{l o g (\frac{1}{x})}{1 - x})}^{3} d x = ? ?

Answered by mindispower last updated on 01/Jun/21

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

\frac{1}{{(1 - x)}^{3}} = \frac{d^{2}}{2 {d x}^{2}} . \frac{1}{1 - x} = \sum_{k ⩾ 2} k (k - 1) x^{k - 2}

= \sum_{k ⩾ 0} (k + 1) (k + 2) x^{k}

= \sum_{k ⩾ 0} (k + 1) (k + 2) \int_{0}^{1} - {(l n (x))}^{3} x^{k}

= \sum_{k ⩾ 0} (k + 1) (k + 2) \int_{0}^{\infty} t^{3} e^{- (k + 1) t} d t

= \frac{1}{2} \sum_{k ⩾ 0} \frac{k + 2}{{(k + 1)}^{3}} \int_{0}^{\infty} t^{3} e^{- t} d t

= 3 \sum_{k ⩾ 0} (\frac{1}{{(k + 1)}^{2}} + \frac{1}{{(k + 1)}^{3}}) = 3 (ζ (2) + ζ (3))

Commented by mnjuly1970 last updated on 02/Jun/21

t h a n k y o u s o m u c h m r p o w e r

e x c e l l e n t a s a l w a y s \dots

Commented by mindispower last updated on 02/Jun/21

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