Question Number 62077 by naka3546 last updated on 15/Jun/19
$$\underset{\:\:\mathrm{1}} {\overset{\:\:\:\:\:\:\:\:\:\:\:\:\infty} {\int}}\:\left(\frac{\mathrm{ln}\:{x}}{{x}}\right)^{\mathrm{2018}} \:{dx}\:\:=\:\:? \\ $$$${Any}\:\:{trick}\left({s}\right)\:\:{to}\:\:{solve}\:\:{it}\:? \\ $$
Answered by Smail last updated on 15/Jun/19
![I_n =∫_1 ^∞ (((lnx)/x))^n dx By parts u=ln^n x⇒u′=n((ln^(n−1) x)/x) v′=x^(−n) ⇒v=(x^(1−n) /(1−n)) I_n =(1/(1−n))[((ln^n x)/x^(n−1) )]_1 ^∞ +(n/(n−1))∫_1 ^∞ ((ln^(n−1) x)/x^n )dx =(n/(n−1))∫_1 ^∞ ((ln^(n−1) x)/x^n )dx=(n/(n−1))(((n−1)/(n−1))∫_1 ^∞ ((ln^(n−2) x)/x^n )dx) =((n(n−1)(n−2)...(n−(n−2)))/((n−1)^(n−1) ))∫_1 ^∞ ((ln^(n−(n−1)) x)/x^n )dx =((n(n−1)...2)/((n−1)^(n−1) ))∫_1 ^∞ ((lnx)/x^n )dx ((n!)/((n−1)^n ))∫_1 ^∞ (dx/x^n )=((n!)/((n−1)^n ))[((−1)/((n−1)x^(n−1) ))]_1 ^∞ I_n =((n!)/((n−1)^(n+1) )) For n=2018 I_(2018) =∫_1 ^∞ (((lnx)/x))^(2018) =((2018!)/(2017^(2019) ))](Q62082.png)
$${I}_{{n}} =\int_{\mathrm{1}} ^{\infty} \left(\frac{{lnx}}{{x}}\right)^{{n}} {dx} \\ $$$${By}\:{parts} \\ $$$${u}={ln}^{{n}} {x}\Rightarrow{u}'={n}\frac{{ln}^{{n}−\mathrm{1}} {x}}{{x}} \\ $$$${v}'={x}^{−{n}} \Rightarrow{v}=\frac{{x}^{\mathrm{1}−{n}} }{\mathrm{1}−{n}} \\ $$$${I}_{{n}} =\frac{\mathrm{1}}{\mathrm{1}−{n}}\left[\frac{{ln}^{{n}} {x}}{{x}^{{n}−\mathrm{1}} }\right]_{\mathrm{1}} ^{\infty} +\frac{{n}}{{n}−\mathrm{1}}\int_{\mathrm{1}} ^{\infty} \frac{{ln}^{{n}−\mathrm{1}} {x}}{{x}^{{n}} }{dx} \\ $$$$=\frac{{n}}{{n}−\mathrm{1}}\int_{\mathrm{1}} ^{\infty} \frac{{ln}^{{n}−\mathrm{1}} {x}}{{x}^{{n}} }{dx}=\frac{{n}}{{n}−\mathrm{1}}\left(\frac{{n}−\mathrm{1}}{{n}−\mathrm{1}}\int_{\mathrm{1}} ^{\infty} \frac{{ln}^{{n}−\mathrm{2}} {x}}{{x}^{{n}} }{dx}\right) \\ $$$$=\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)...\left({n}−\left({n}−\mathrm{2}\right)\right)}{\left({n}−\mathrm{1}\right)^{{n}−\mathrm{1}} }\int_{\mathrm{1}} ^{\infty} \frac{{ln}^{{n}−\left({n}−\mathrm{1}\right)} {x}}{{x}^{{n}} }{dx} \\ $$$$=\frac{{n}\left({n}−\mathrm{1}\right)...\mathrm{2}}{\left({n}−\mathrm{1}\right)^{{n}−\mathrm{1}} }\int_{\mathrm{1}} ^{\infty} \frac{{lnx}}{{x}^{{n}} }{dx} \\ $$$$\frac{{n}!}{\left({n}−\mathrm{1}\right)^{{n}} }\int_{\mathrm{1}} ^{\infty} \frac{{dx}}{{x}^{{n}} }=\frac{{n}!}{\left({n}−\mathrm{1}\right)^{{n}} }\left[\frac{−\mathrm{1}}{\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{1}} }\right]_{\mathrm{1}} ^{\infty} \\ $$$${I}_{{n}} =\frac{{n}!}{\left({n}−\mathrm{1}\right)^{{n}+\mathrm{1}} } \\ $$$${For}\:{n}=\mathrm{2018} \\ $$$${I}_{\mathrm{2018}} =\int_{\mathrm{1}} ^{\infty} \left(\frac{{lnx}}{{x}}\right)^{\mathrm{2018}} =\frac{\mathrm{2018}!}{\mathrm{2017}^{\mathrm{2019}} } \\ $$$$ \\ $$