nice-calculus-prove-that-0-1-log-1-x-x-2-dx-2-2-

Question Number 137829 by mnjuly1970 last updated on 07/Apr/21

\dots \dots . n i c e \dots \dots \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 o g (1 - x)}{x})}^{2} d x = 2 ζ (2) \dots .

Answered by EnterUsername last updated on 07/Apr/21

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

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

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

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

= 2 ψ^{'} (1) = 2 \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + 1)}^{2}} = 2 \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} = 2 ζ (2) = \frac{π^{2}}{3}

Commented by mnjuly1970 last updated on 07/Apr/21

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