0-1-ln-2-1-x-ln-x-x-dx-

Question Number 160252 by mnjuly1970 last updated on 26/Nov/21

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

Answered by TheSupreme last updated on 26/Nov/21

I =

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

t = 1 - x \to x = 1 - t \to d x = - d t

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

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

{2 I = \frac{1}{2} {l n}^{2} (1 - x) {l n}^{2} (x)]}_{0}^{1}

{\int_{0}^{1} {l n}^{2} (1 - x) \frac{l n (x)}{x} = \frac{1}{4} {l n}^{2} (1 - x) {l n}^{2} (x)]}_{0}^{1}

l i \underset{x \to 0}{m} \frac{1}{4} {l n}^{2} (1 - x) {l n}^{2} (x) = \frac{1}{4} x^{2} {l n}^{2} (x) = 0

I = 0

Commented by mr W last updated on 27/Nov/21

p l e a s e r e c h e c k s i r!

i {d o n}^{'} t k n o w h o w t o s o l v e t h i s d e f i n i t e

i n t e g r a l, b u t i^{'} m s u r e i t m u s t b e

n e g a t i v e a n d t h e r e f o r e c a n n o t b e

z e r o .

0 ⩽ x ⩽ 1

\ln^{2} (1 - x) > 0

\ln (x) < 0

x > 0

\Rightarrow \frac{\ln^{2} (1 - x) \ln (x)}{x} < 0

\Rightarrow \int_{0}^{1} \frac{\ln^{2} (1 - x) \ln (x)}{x} d x < 0

\Rightarrow \int_{0}^{1} \frac{\ln^{2} (1 - x) \ln (x)}{x} d x \neq 0

Commented by MJS_new last updated on 27/Nov/21

\underset{0}{\int^{1}} \frac{\ln^{2} (1 - x) \ln x}{x} d x =

[by parts]

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

[t = 1 - x \to d x = - d t]

= {[\frac{1}{2} \ln^{2} x \ln^{2} (1 - x)]}_{0}^{1} - \underset{1}{\int^{0}} \frac{\ln^{2} (1 - t) \ln t}{t} d t =

= {[\frac{1}{2} \ln^{2} x \ln^{2} (1 - x)]}_{0}^{1} + \underset{0}{\int^{1}} \frac{\ln^{2} (1 - t) \ln t}{t} d t

so we have

I = {[\frac{1}{2} \ln^{2} x \ln^{2} (1 - x)]}_{0}^{1} + I

we cannot solve it this way (or am I too tired

to understand anything ? {it}^{'} s 2 : 30 a . m .)

Commented by MJS_new last updated on 27/Nov/21

approximately

\underset{0}{\int^{1}} \frac{\ln^{2} (1 - x) \ln x}{x} d x \approx - . 541161616604

Answered by mnjuly1970 last updated on 27/Nov/21