Question Number 40156 by maxmathsup by imad last updated on 16/Jul/18
$${find}\:\:\:\int_{{e}^{\mathrm{2}} } ^{+\infty} \:\:\:\:\frac{{dt}}{{tln}\left({t}\right){ln}\left({ln}\left({t}\right)\right.} \\ $$
Commented by maxmathsup by imad last updated on 21/Jul/18
![let I = ∫_e^2 ^(+∞) (dt/(tln(t)ln(ln(t)))) changement ln(ln(t))=x give dx=(1/(tln(t))) dt ⇒(1/(tln(t))) =(dx/dt) ⇒ I = ∫_(ln(2)) ^(+∞) (dx/x) =lim_(ξ→+∞) ∫_(ln(2)) ^ξ (dx/x) =lim_(ξ→+∞) [ln∣x∣]_(ln(2)) ^ξ =lim_(ξ→+∞) ln((ξ/(ln(2)))) =+∞ the ntegral is divergent .? =](https://www.tinkutara.com/question/Q40452.png)
$${let}\:{I}\:=\:\int_{{e}^{\mathrm{2}} } ^{+\infty} \:\:\:\:\frac{{dt}}{{tln}\left({t}\right){ln}\left({ln}\left({t}\right)\right)}\:\:{changement}\:{ln}\left({ln}\left({t}\right)\right)={x}\:{give} \\ $$$${dx}=\frac{\mathrm{1}}{{tln}\left({t}\right)}\:{dt}\:\Rightarrow\frac{\mathrm{1}}{{tln}\left({t}\right)}\:=\frac{{dx}}{{dt}}\:\Rightarrow \\ $$$${I}\:=\:\int_{{ln}\left(\mathrm{2}\right)} ^{+\infty} \:\:\:\:\frac{{dx}}{{x}}\:={lim}_{\xi\rightarrow+\infty} \:\:\:\:\int_{{ln}\left(\mathrm{2}\right)} ^{\xi} \:\frac{{dx}}{{x}}\:={lim}_{\xi\rightarrow+\infty} \left[{ln}\mid{x}\mid\right]_{{ln}\left(\mathrm{2}\right)} ^{\xi} ={lim}_{\xi\rightarrow+\infty} {ln}\left(\frac{\xi}{{ln}\left(\mathrm{2}\right)}\right) \\ $$$$=+\infty\:\:{the}\:{ntegral}\:{is}\:{divergent}\:.? \\ $$$$= \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Jul/18
$${y}={ln}\left\{{ln}\left({t}\right)\right\} \\ $$$${dy}=\frac{{dt}}{{tlnt}} \\ $$$$\int_{{ln}\mathrm{2}} ^{\infty} \frac{{dy}}{{y}} \\ $$$$\mid{lny}\mid_{{ln}\mathrm{2}} ^{\infty} \:={ln}\infty−{ln}\left({ln}\mathrm{2}\right)\:\:{pls}\:{chrck}… \\ $$