study-the-integral-0-1-x-ln-1-x-dx-

Question Number 60680 by maxmathsup by imad last updated on 24/May/19

s t u d y t h e i n t e g r a l \int_{0}^{1} \frac{x}{l n (1 - x)} d x

Commented by maxmathsup by imad last updated on 29/May/19

l e t I = \int_{0}^{1} \frac{x}{l n (1 - x)} d x c h a n g e m e n t l n (1 - x) = - t g i v e 1 - x = e^{- t}

I = \int_{0}^{\infty} \frac{1 - e^{- t}}{- t} (e^{- t} d t) = - \int_{0}^{\infty} \frac{e^{- t} - e^{- 2 t}}{t} d t = \int_{0}^{\infty} \frac{e^{- 2 t} - e^{- t}}{t} d t

a t v (0) e^{- 2 t} \sim 1 - 2 t, e^{- t} \sim 1 - t \Rightarrow e^{- 2 t} - e^{- t} \sim - t \Rightarrow

\frac{e^{- 2 t} - e^{- t}}{t} \sim - 1 a l s o {l i m}_{\to + \infty} t^{2} \frac{e^{- 2 t} - e^{- t}}{t} = 0 \Rightarrow I c o n v e r g e s

Commented by maxmathsup by imad last updated on 29/May/19

w e h a v e I = {l i m}_{ξ \to 0} I (ξ) w i t h I (ξ) = \int_{ξ}^{\infty} \frac{e^{- 2 t} - e^{- t}}{t} d t

I (ξ) = \int_{ξ}^{+ \infty} \frac{e^{- 2 t}}{t} d t =_{2 t = u} \int_{2 ξ}^{+ \infty} \frac{e^{- u}}{\frac{u}{2}} \frac{d u}{2} = \int_{2 ξ}^{+ \infty} \frac{e^{- u}}{u} d u \Rightarrow

I (ξ) = \int_{2 ξ}^{+ \infty} \frac{e^{- t}}{t} d t - \int_{ξ}^{+ \infty} \frac{e^{- t}}{t} d t = \int_{2 ξ}^{ξ} \frac{e^{- t}}{t} d t = - \int_{ξ}^{2 ξ} \frac{e^{- t}}{t} d t

\exists c \in] ξ, 2 ξ [/ I (ξ) = - e^{- ξ} \int_{ξ}^{2 ξ} \frac{d t}{t} = - e^{- ξ} l n (\frac{2 ξ}{ξ}) \Rightarrow {l i m}_{ξ \to 0} I (ξ) = - l n (2) \Rightarrow

I = - l n (2) .