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

Question Number 184098 by paul2222 last updated on 02/Jan/23

\int_{0}^{1} \frac{\ln^{2} (1 - x)}{1 - x} dx

Commented by paul2222 last updated on 02/Jan/23

let 1 - x = u

\int_{0}^{1} \frac{\ln^{2} (u)}{u} du

let \ln (u) = - k du = - e^{- k} dk

\int_{0}^{\infty} \frac{k^{2}}{e^{- k}} e^{- k} dk = \int_{0}^{\infty} k^{2} dk

(the integral diverges)

Commented by mokys last updated on 03/Jan/23

e^{y} = 1 - x \to e^{y} d y = - d x

x = 1 \to y = - \infty, x = 0 \to y = 0

\int_{- \infty}^{0} y^{2} d y s o i s d i v e r g e

Answered by Ml last updated on 03/Jan/23

\int_{0}^{1} \frac{{(\ln (1 - x))}^{2}}{(1 - x)} dx

{\begin{cases} u = \ln (1 - x) \\ du = - \frac{1}{(1 - x)} dx \end{cases}, dx = - (1 - x) du

\int_{0}^{1} \frac{u^{2} (- (1 - x)) du}{(1 - x)} = - \int_{0}^{1} u^{2} du = - \frac{1}{3} u^{3}

\Rightarrow - \frac{1}{3} {(\ln (1 - x))}^{3} ∣_{0}^{1} = \infty

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