advanced-calculus-i-prove-that-0-1-ln-1-ln-1-x-ln-1-x-dx-n-1-n-1-n-2-ii-prove-that-0-1-ln-1-x-x-1

Question Number 114045 by mnjuly1970 last updated on 17/Sep/20

\dots a d v a n c e d c a l c u l u s \dots

i : p r o v e t h a t ::

\int_{0}^{1} \frac{l n (1 + l n (1 - x))}{l n (1 - x)} d x \overset{?}{=} \underset{n = 1}{\sum^{\infty}} \frac{Γ (n + 1)}{n^{2}}

i i :

p r o v e t h a t ::

Ω = \int_{0}^{1} \frac{l n (1 + x)}{x (1 + x^{2})} d x \overset{?}{=} \frac{5 π^{2}}{48}

You can't use 'macro parameter character #' in math mode

Answered by Dwaipayan Shikari last updated on 16/Sep/20

\int_{0}^{1} \frac{l o g (1 + x)}{x (1 + x)} = \int_{0}^{1} \frac{l o g (1 + x)}{x} - \frac{l o g (1 + x)}{(1 + x)} d x

= \int_{0}^{1} {(- 1)}^{n} \underset{n = 1}{\sum^{\infty}} \frac{x^{n - 1}}{n} - \int_{1}^{2} \frac{l o g u}{u} d u

= \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} \frac{1}{n^{2}} - \frac{1}{2} {[{(l o g u)}^{2}]}_{1}^{2} = \frac{π^{2}}{12} - \frac{1}{2} {(l o g 2)}^{2}

Answered by mathmax by abdo last updated on 16/Sep/20

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

= \int_{0}^{1} \frac{\ln (1 + x)}{x} dx - \int_{0}^{1} \frac{\ln (1 + x)}{1 + x} dx but

\int_{0}^{1} \frac{\ln (1 + x)}{1 + x} dx =_{1 + x = t} \int_{1}^{2} \frac{\ln (t)}{x > t} dt = {[\frac{1}{2} \ln^{2} t]}_{1}^{2} = \frac{\ln^{2} (2)}{2}

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

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

\int_{0}^{1} \frac{\ln (1 + x)}{x} dx = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}} = - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}}

= - {2^{1 - 2} - 1} ξ (2) = \frac{π^{2}}{12} \Rightarrow I = \frac{1}{2} \ln^{2} (2) - \frac{π^{2}}{12}

Commented by mathmax by abdo last updated on 16/Sep/20

sorry I = \frac{π^{2}}{12} - \frac{1}{2} \ln^{2} (2)

Commented by mnjuly1970 last updated on 17/Sep/20

t h a n k y o u s i r

t h a t w a s m y m i s t a k e

d e n o m i n a t o r i s x (1 + x^{2})

i e d i t e d i t .

Answered by mathmax by abdo last updated on 16/Sep/20

A = \int_{0}^{1} \frac{\ln (1 + \ln (1 - x))}{\ln (1 - x)} dx we do tbe changement \ln (1 - x) = - t \Rightarrow

1 - x = e^{- t} \Rightarrow x = 1 - e^{- t} \Rightarrow A = \int_{0}^{+ \infty} \frac{\ln (1 + t)}{- t} (e^{- t}) dt

= - \int_{0}^{\infty} \frac{e^{- t} \ln (1 + t)}{t} dt = - \int_{0}^{\infty} e^{- t} (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} t^{n - 1}}{n}) dt

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} \int_{0}^{\infty} t^{n - 1} e^{- t} dt = \sum_{n = 2}^{\infty} {(- 1)}^{n} \frac{Γ (n - 1)}{n} - 1

Commented by mnjuly1970 last updated on 17/Sep/20

A = - \int_{0}^{\infty} \frac{l n (1 - t) e^{- t}}{t} d t i s

c o r r e c t . p l e a s e c h e c k i t

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

y o u r e f f o r t .

Answered by mindispower last updated on 17/Sep/20

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

= \int_{0}^{\infty} \frac{l n (1 - x)}{- x} e^{- x} d x s o m t h i n g w r o n g

l n (1 - x) \in C x > 1