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2-1-x-2-lnx-dx-converges-or-diverges-




Question Number 138716 by floor(10²Eta[1]) last updated on 17/Apr/21
∫_2 ^∞ (1/(x^2 lnx))dx   converges or diverges?
$$\int_{\mathrm{2}} ^{\infty} \frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \mathrm{lnx}}\mathrm{dx}\: \\ $$$$\mathrm{converges}\:\mathrm{or}\:\mathrm{diverges}? \\ $$
Answered by mathmax by abdo last updated on 17/Apr/21
I=∫_2 ^∞  (dx/(x^2 lnx))? ⇒I=_(lnx=t)   ∫_(ln2) ^∞    ((e^t dt)/(e^(2t) t)) =∫_(ln2) ^(+∞)  (e^(−t) /t)dt  lim_(t→+∞) t^2  .(e^(−t) /t)=lim_(t→+∞) te^(−t)  =0 ⇒I is convergent
$$\mathrm{I}=\int_{\mathrm{2}} ^{\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \mathrm{lnx}}?\:\Rightarrow\mathrm{I}=_{\mathrm{lnx}=\mathrm{t}} \:\:\int_{\mathrm{ln2}} ^{\infty} \:\:\:\frac{\mathrm{e}^{\mathrm{t}} \mathrm{dt}}{\mathrm{e}^{\mathrm{2t}} \mathrm{t}}\:=\int_{\mathrm{ln2}} ^{+\infty} \:\frac{\mathrm{e}^{−\mathrm{t}} }{\mathrm{t}}\mathrm{dt} \\ $$$$\mathrm{lim}_{\mathrm{t}\rightarrow+\infty} \mathrm{t}^{\mathrm{2}} \:.\frac{\mathrm{e}^{−\mathrm{t}} }{\mathrm{t}}=\mathrm{lim}_{\mathrm{t}\rightarrow+\infty} \mathrm{te}^{−\mathrm{t}} \:=\mathrm{0}\:\Rightarrow\mathrm{I}\:\mathrm{is}\:\mathrm{convergent} \\ $$

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