Obtain-a-reduction-formulae-for-I-n-0-1-ln-x-n-dx-find-I-2-0-1-ln-x-2-dx-

Question Number 122186 by physicstutes last updated on 14/Nov/20

Obtain a reduction formulae for

I_{n} = \underset{0}{\int^{1}} {(\ln x)}^{n} d x

find I_{2} = \underset{0}{\int^{1}} {(\ln x)}^{2} d x

Answered by Olaf last updated on 14/Nov/20

I_{n} = \int_{0}^{1} \ln^{n} x d x

I_{0} = 1 and for n ⩾ 1 :

I_{n} = {[x \ln^{n} x]}_{0}^{1} - \int_{0}^{1} x (n \frac{1}{x} \ln^{n - 1} x) d x

I_{n} = - n \int_{0}^{1} \ln^{n - 1} x d x = - n I_{n - 1}

\Rightarrow I_{n} = {(- 1)}^{n} n! I_{0} = {(- 1)}^{n} n!

If n = 2, I_{2} = {(- 1)}^{2} 2! = 2

Commented by physicstutes last updated on 14/Nov/20

This is incredible . Thank you sir Olaf .

Answered by Dwaipayan Shikari last updated on 14/Nov/20

\int_{\infty}^{0} t^{n} x d t l o g x = t \Rightarrow \frac{1}{x} = \frac{d t}{d x}

= \int_{\infty}^{0} t^{n} e^{t} d t

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

= {(- 1)}^{n} Γ (n + 1) = {(- 1)}^{n} n!

a s p e r q u e s t i o n

I t i s {(- 1)}^{2} 2! = 2

Commented by physicstutes last updated on 14/Nov/20

thats right!