calculate-1-ln-lnx-1-x-2-dx-

Question Number 60727 by Mr X pcx last updated on 25/May/19

c a l c u l a t e \int_{1}^{+ \infty} \frac{l n (l n x)}{1 + x^{2}} d x

Commented by maxmathsup by imad last updated on 26/May/19

l e t A = \int_{1}^{+ \infty} \frac{l n (l n x)}{1 + x^{2}} d x c h a n g e m e n t l n (x) = t g i v e

A = \int_{0}^{+ \infty} \frac{l n (t)}{1 + e^{2 t}} e^{t} d t = \int_{0}^{\infty} \frac{e^{- t} l n (t)}{1 + e^{- 2 t}} d t

= \int_{0}^{\infty} e^{- t} l n (t) (\sum_{n = 0}^{\infty} {(- 1)}^{n} e^{- 2 n t}) d t

= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{\infty} e^{- (2 n + 1) t} l n (t) d t = \sum_{n = 0}^{\infty} {(- 1)}^{n} A_{n}

A_{n} = \int_{0}^{\infty} e^{- (2 n + 1) t} l n (t) d t =_{(2 n + 1) t = u} \int_{0}^{\infty} e^{- u} l n (\frac{u}{2 n + 1}) \frac{d u}{2 n + 1}

= \frac{1}{2 n + 1} \int_{0}^{\infty} e^{- u} {l n (u) - l n (2 n + 1)) d u

= \frac{1}{2 n + 1} {\int_{0}^{\infty} e^{- u} l n (u) d u - l n (2 n + 1) \int_{0}^{\infty} e^{- u} d u}

\int_{0}^{\infty} e^{- u} l n (u) d u = - γ (r e s u l t p r o v e d)

\int_{0}^{\infty} e^{- u} d u = {[- e^{- u}]}_{0}^{\infty} = 1 \Rightarrow A_{n} = \frac{- γ}{2 n + 1} - \frac{l n (2 n + 1)}{2 n + 1} \Rightarrow

A = \sum_{n = 0}^{\infty} {(- 1)}^{n} {- \frac{γ}{2 n + 1} - \frac{l n (2 n + 1)}{2 n + 1}}

= - γ \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} l n (2 n + 1)

= - \frac{γ π}{4} - \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{l n (2 n + 1)}{2 n + 1} w i t h γ c o n s t a n t o f E u l e r \dots b e c o n t i n u e d \dots

Commented by maxmathsup by imad last updated on 26/May/19

l e t t r y a n o t h e r w a y b y u s e o f c h a n g . x = \frac{1}{t} \Rightarrow

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

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

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

= \frac{1}{2} \int_{0}^{1} \frac{l n {l n (\frac{1}{t})}}{1 + i t} d t + \frac{1}{2} \int_{0}^{1} \frac{l n {l n (\frac{1}{t})}}{1 - i t} d t l e t W = \int_{0}^{1} \frac{l n {l n (\frac{1}{t})}}{1 + i t} d t

c h a n g . l n (\frac{1}{t}) = u g i v e \frac{1}{t} = e^{u} \Rightarrow t = e^{- u} \Rightarrow

W = - \int_{0}^{\infty} \frac{l n u}{1 + {i e}^{- u}} (- e^{- u}) d u = \int_{0}^{\infty} \frac{e^{- u} l n (u)}{1 + {i e}^{- u}} d u

= \int_{0}^{\infty} e^{- u} l n (u) (\sum_{n = 0}^{\infty} {(- {i e}^{- u})}^{n}) d u

= \sum_{n = 0}^{\infty} {(- i)}^{n} \int_{0}^{\infty} e^{- (n + 1) u} l n (u) d u

\int_{0}^{\infty} e^{- (n + 1) u} l n (u) d u =_{(n + 1) u = t} \int_{0}^{\infty} e^{- t} l n (\frac{t}{n + 1}) \frac{d t}{n + 1}

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

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

= \frac{- γ}{n + 1} - \frac{l n (n + 1)}{n + 1} \Rightarrow W = \sum_{n = 0}^{\infty} \frac{- γ {(- i)}^{n}}{n + 1} - \sum_{n = 0}^{\infty} \frac{{(- i)}^{n} l n (n + 1)}{n + 1}

H = \int_{0}^{1} \frac{l n {l n (\frac{1}{t})}}{1 - i t} d t =_{l n (\frac{1}{t}) = u} - \int_{0}^{\infty} \frac{l n (u)}{1 - i e^{- u}} (- e^{- u}) d u

= \int_{0}^{\infty} \frac{e^{- u} l n (u)}{1 - {i e}^{- u}} d u = \int_{0}^{\infty} e^{- u} l n u {\sum_{n = 0}^{\infty} i^{n} e^{- n u}} d u

= \sum_{n = 0}^{\infty} i^{n} \int_{0}^{\infty} e^{- (n + 1) u} l n (u) d u = \dots . b e c o n t i n u e d \dots .