nice-calculus-evaluation-0-1-log-1-x-log-1-x-m-n-july-197o-

Question Number 115055 by mnjuly1970 last updated on 23/Sep/20

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

e v a l u a t i o n :

χ = \int_{0}^{1} l o g (1 - x) . l o g (1 + x) = ? ? ?

\dots m . n . j u l y . 197 o \dots

Answered by mathmax by abdo last updated on 23/Sep/20

I = \int_{0}^{1} \ln (1 - x) . \ln (1 + x) dx we have \frac{d}{dx} \ln (1 - x) = \frac{- 1}{1 - x}

= - \sum_{n = 0}^{\infty} x^{n} \Rightarrow \ln (1 - x) = - \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n + 1} = - \sum_{n = 1}^{\infty} \frac{x^{n}}{n}

also \ln (1 + x) = - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} x^{n}}{n} \Rightarrow

\ln (1 - x) . \ln (1 + x) = (\sum_{n = 1}^{\infty} \frac{x^{n}}{n}) . (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} x^{n})

= \sum_{n = 1}^{\infty} c_{n} x^{n} with c_{n} = \sum_{i + j = n} a_{i} b_{j} = \sum_{i + j = n} \frac{1}{i} \times \frac{{(- 1)}^{j}}{j}

= \sum_{i = 1}^{n - 1} \frac{1}{i} \times \frac{{(- 1)}^{n - i}}{n - i} \Rightarrow \ln (1 - x) . \ln (1 + x) = \sum_{n = 1}^{\infty} (\sum_{i = 1}^{n - 1} \frac{{(- 1)}^{n - i}}{i (n - i)}) x^{n}

\Rightarrow \int_{0}^{1} \ln (1 - x) \ln (1 + x) dx = \sum_{n = 1}^{\infty} (\sum_{i = 1}^{n} \frac{{(- 1)}^{n - i}}{i (n - i)}) \times \frac{1}{n + 1}

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n + 1} (\sum_{i = 1}^{n - 1} \frac{{(- 1)}^{i}}{i (n - i)}) but

\sum_{i = 1}^{n - 1} \frac{{(- 1)}^{i}}{i (n - i)} = \frac{1}{n} \sum_{i = 1}^{n - 1} {(- 1)}^{i} (\frac{1}{i} - \frac{1}{n - i}) = \frac{1}{n} \sum_{i = 1}^{n - 1} \frac{{(- 1)}^{i}}{i}

- \frac{1}{n} \sum_{i = 1}^{n - 1} \frac{{(- 1)}^{i}}{n - i} (n - i = p) = \frac{1}{n} \sum_{i = 1}^{n - 1} \frac{{(- 1)}^{i}}{i} - \frac{1}{n} \sum_{1}^{n - 1} \frac{{(- 1)}^{n - p}}{p}

= \frac{1}{n} \sum_{i = 1}^{n - 1} \frac{{(- 1)}^{i}}{i} - \frac{{(- 1)}^{n}}{n} \sum_{i = 1}^{n - 1} \frac{{(- 1)}^{i}}{i}

= \frac{1}{n} (1 - {(- 1)}^{n}) \sum_{i = 1}^{n - 1} \frac{{(- 1)}^{i}}{i} = \frac{1}{n} \sum_{i = 1}^{2 n} \frac{{(- 1)}^{i}}{i} \dots . be continued \dots .

Commented by mnjuly1970 last updated on 23/Sep/20

t h a n k y o u s o m u c h s i r

m a x \dots

Commented by mathmax by abdo last updated on 23/Sep/20

you are welcome sir

Answered by mathdave last updated on 23/Sep/20

s o l u t i o n

i f l e t I_{1} = \int_{0}^{1} \ln (1 - x) \ln (1 + x) d x \dots . . (1)

p u t x = - x

I_{2} = \int_{- 1}^{0} \ln (1 + x) \ln (1 - x) d x \dots \dots . . (2) s i n c e t h e i n t e g r a n d a r e u n c h a n g e d t h e n w e a d d t h e m u p

w h i c h m e a n t h a t I_{1} = I_{2}

I = I_{1} + I_{2} = 2 I_{1}, I_{1} = \frac{1}{2} I

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

p u t x = 2 y - 1, d x = 2 d y

I_{1} = \frac{1}{2} \int_{0}^{1} \ln (2 - 2 y) \ln (2 y) 2 d y = \int_{0}^{1} [(\ln 2 + \ln (1 - y)) (\ln 2 + \ln y)] d y

I = \int_{0}^{1} (\ln^{2} 2 + \ln 2 (\ln (1 - y) + \ln y) + \ln y \ln (1 - y)) d x

I = \ln^{2} (2) \int_{0}^{1} d y + \ln 2 \int_{0}^{1} (\ln y + \ln (1 - y)) d y + \int_{0}^{1} \ln y \ln (1 - y) d y

l e t A = \ln^{2} (2) \int_{0}^{1} d y = \ln^{2} (2) \dots \dots (1)

l e t

B = \ln 2 \int_{0}^{1} (\ln y + \ln (1 - y)) d y = 2 \int_{0}^{1} \ln y d y

B = 2 \ln 2 {(y \ln y - y)}_{0}^{1} = - 2 \ln 2 \dots . . (2)

l e t

C = \int_{0}^{1} \ln y \ln (1 - y) d y = - \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \int_{0}^{1} y^{n} \ln y d y

C = - \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \frac{\partial}{\partial a} ∣_{a = 1} \int_{0}^{1} y^{n} . y^{a - 1} d y = - \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \frac{\partial}{\partial a} ∣_{a = 1} (\frac{1}{n + a})

C = \underset{n = 1}{\sum^{\infty}} \frac{1}{n {(n + 1)}^{2}} = \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + 1} - \frac{1}{{(n + 1)}^{2}})

C = \underset{n = 1}{\sum^{\infty}} \frac{1}{n} - \underset{n = 2}{\sum^{\infty}} \frac{1}{n} - \underset{n = 2}{\sum^{\infty}} \frac{1}{n^{2}} = \underset{n = 1}{\sum^{\infty}} (\frac{1}{n}) - \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - 1) - \underset{n = 1}{\sum^{\infty}} (\frac{π^{2}}{6} - 1)

C = 1 - \frac{π^{2}}{6} + 1 = (2 - \frac{π^{2}}{6}) \dots \dots \dots (3)

b u t

I = A + B + C = \ln^{2} (2) - 2 \ln 2 + 2 - \frac{π^{2}}{6} = 2 - \frac{π^{2}}{6} + \ln^{2} (2) - \ln 4

∵ \int_{0}^{1} \ln (1 - x) \ln (1 + x) d x = 2 - \frac{π^{2}}{6} + \ln^{2} (2) - \ln 4

b y m a t h d a v e (23 / 09 / 2020)

Commented by mnjuly1970 last updated on 23/Sep/20

t h a n k y o u m r d a v e

n i c e s o l u t i o n t h a t w a s i

w a n t e d \dots

Commented by mnjuly1970 last updated on 23/Sep/20

f i n a l a n s w e r :

{l n}^{2} (2) - l n (4) + 2 - \frac{π^{2}}{6}

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

d a v e .

Commented by mnjuly1970 last updated on 23/Sep/20

y o u a r e v e r y g e n t l m a n

m r d a v e . t h a n k y o u f o r

y o u r e f f o r t .

Commented by Tawa11 last updated on 06/Sep/21

great

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