Question-61809

Question Number 61809 by aliesam last updated on 09/Jun/19

Commented by maxmathsup by imad last updated on 10/Jun/19

l e t A = \int_{0}^{1} \frac{x^{n}}{\sqrt{l n (x)}} d x \Rightarrow A = \int_{0}^{1} \frac{x^{n}}{\sqrt{- (- l n x)}} d x = \frac{1}{\sqrt{- 1}} \int_{0}^{1} \frac{x^{n}}{\sqrt{- l n (x)}} d x

c h a n g e m e n t \sqrt{- l n (x)} = t g i v e - l n (x) = t^{2} \Rightarrow l n (x) = - t^{2} \Rightarrow x = e^{- t^{2}} \Rightarrow

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

=_{\sqrt{n + 1} t = u} - 2 i \int_{0}^{\infty} e^{- u^{2}} \frac{d u}{\sqrt{n + 1}} = \frac{- 2 i}{\sqrt{n + 1}} \int_{0}^{\infty} e^{- u^{2}} d u = \frac{- 2 i}{\sqrt{n + 1}} \frac{\sqrt{π}}{2} \Rightarrow

A = \frac{- i \sqrt{π}}{\sqrt{n + 1}} .

Answered by perlman last updated on 09/Jun/19

\sqrt{l n (x)} = i \sqrt{- l n (x)} \forall x \in] 0; 1]

w i t h e \sqrt{- l n (x)} = y

x = e^{- y^{2}}

= \int_{+ \infty}^{0} \frac{e^{- {n y}^{2}}}{i y} (- 2 y) e^{- y^{2}} d y

= i \int_{0}^{+ \infty} e^{- (n + 1) y^{2}} d y

l e t z = \sqrt{(n + 1)} y

d z = \sqrt{n + 1} d y

= i \int_{0}^{+ \infty} \sqrt{n + 1} e^{- z^{2}} d z = i \sqrt{n + 1} \int_{0}^{+ \infty} e^{- z^{2}} d z = i \sqrt{n + 1} \frac{\sqrt{π}}{2}

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