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Question Number 128276 by john_santu last updated on 06/Jan/21

 ∫_e^2  ^( ∞)  (dx/(x^3  ln x)) ?

$$\:\int_{{e}^{\mathrm{2}} } ^{\:\infty} \:\frac{{dx}}{{x}^{\mathrm{3}} \:\mathrm{ln}\:{x}}\:? \\ $$

Answered by liberty last updated on 06/Jan/21

 ϝ = lim_(Y→+∞) ∫_e^2  ^Y (dx/(x^3 .ln x)) = lim_(Y→+∞) (−(1/(2 ln^2 x)))_e^2  ^Y    ϝ = lim_(Y→+∞) ((1/8)−(1/(2 ln^2 Y)))=(1/8)

$$\:\digamma\:=\:\underset{\mathrm{Y}\rightarrow+\infty} {\mathrm{lim}}\int_{\mathrm{e}^{\mathrm{2}} } ^{\mathrm{Y}} \frac{{dx}}{{x}^{\mathrm{3}} .\mathrm{ln}\:{x}}\:=\:\underset{\mathrm{Y}\rightarrow+\infty} {\mathrm{lim}}\left(−\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{ln}\:^{\mathrm{2}} {x}}\right)_{\mathrm{e}^{\mathrm{2}} } ^{\mathrm{Y}} \\ $$$$\:\digamma\:=\:\underset{\mathrm{Y}\rightarrow+\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{ln}\:^{\mathrm{2}} \mathrm{Y}}\right)=\frac{\mathrm{1}}{\mathrm{8}} \\ $$

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