∫_( 1) ^( ∞) (((ln x)/x))^(2018) dx = ? Any trick(s) to solve it ?


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Question Number 62077 by naka3546 last updated on 15/Jun/19

$\underset{1}{\int^{\infty}} {(\frac{\ln x}{x})}^{2018} d x = ?$ $A n y t r i c k (s) t o s o l v e i t ?$
	Answered by Smail last updated on 15/Jun/19

	$I_{n} = \int_{1}^{\infty} {(\frac{l n x}{x})}^{n} d x$ $B y p a r t s$ $u = {l n}^{n} x \Rightarrow u^{'} = n \frac{{l n}^{n - 1} x}{x}$ $v^{'} = x^{- n} \Rightarrow v = \frac{x^{1 - n}}{1 - n}$ $I_{n} = \frac{1}{1 - n} {[\frac{{l n}^{n} x}{x^{n - 1}}]}_{1}^{\infty} + \frac{n}{n - 1} \int_{1}^{\infty} \frac{{l n}^{n - 1} x}{x^{n}} d x$ $= \frac{n}{n - 1} \int_{1}^{\infty} \frac{{l n}^{n - 1} x}{x^{n}} d x = \frac{n}{n - 1} (\frac{n - 1}{n - 1} \int_{1}^{\infty} \frac{{l n}^{n - 2} x}{x^{n}} d x)$ $= \frac{n (n - 1) (n - 2) . . . (n - (n - 2))}{{(n - 1)}^{n - 1}} \int_{1}^{\infty} \frac{{l n}^{n - (n - 1)} x}{x^{n}} d x$ $= \frac{n (n - 1) . . . 2}{{(n - 1)}^{n - 1}} \int_{1}^{\infty} \frac{l n x}{x^{n}} d x$ $\frac{n!}{{(n - 1)}^{n}} \int_{1}^{\infty} \frac{d x}{x^{n}} = \frac{n!}{{(n - 1)}^{n}} {[\frac{- 1}{(n - 1) x^{n - 1}}]}_{1}^{\infty}$ $I_{n} = \frac{n!}{{(n - 1)}^{n + 1}}$ $F o r n = 2018$ $I_{2018} = \int_{1}^{\infty} {(\frac{l n x}{x})}^{2018} = \frac{2018!}{2017^{2019}}$
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