Question Number 78622 by mathmax by abdo last updated on 19/Jan/20
$${calculate}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\:\:\int_{\mathrm{0}} ^{{n}} \left(\mathrm{1}−\frac{{t}}{{n}}\right)^{{n}} {ln}\left(\mathrm{1}+{nt}\right){dt} \\ $$
Commented by mathmax by abdo last updated on 22/Jan/20
![let A_n =∫_0 ^n (1−(t/n))^n ln(1+nt)dt changement nt =x give A_n =∫_0 ^n^2 (1/n)(1−(x/n^2 ))^n ln(1+x)dx =∫_0 ^n^2 f_n (x)dx f_n (x)=(1/n)(1−(x/n^2 ))^n ln(1+x) ⇒f_n (x)=(1/n)e^(nln(1−(x/n^2 ))) ln(1+x) ln(1−(x/n^2 )) ∼−(x/n^2 ) ⇒nln(1−(x/n^2 ))∼−(x/n) ⇒e^(nln(1−(x/n))) ∼e^(−(x/n)) ∼1−(x/n) ⇒f_n (x)∼(1/n)(1−(x/n))ln(1+x) (n→+∞) ⇒ f_n →^(cs) 0 on]0,+∞[ ⇒lim_(n→+∞) A_n =lim_(n→+∞) ∫_R f_n (x)χ_([0,n[) (x)dx =∫_R lim_(n→+∞) f_n (x)χ_([0,n[) (x)dx =0](https://www.tinkutara.com/question/Q79050.png)
$${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{{n}} \left(\mathrm{1}−\frac{{t}}{{n}}\right)^{{n}} {ln}\left(\mathrm{1}+{nt}\right){dt}\:\:{changement}\:{nt}\:={x}\:{give} \\ $$$${A}_{{n}} =\int_{\mathrm{0}} ^{{n}^{\mathrm{2}} } \frac{\mathrm{1}}{{n}}\left(\mathrm{1}−\frac{{x}}{{n}^{\mathrm{2}} }\right)^{{n}} {ln}\left(\mathrm{1}+{x}\right){dx}\:\:\:=\int_{\mathrm{0}} ^{{n}^{\mathrm{2}} } {f}_{{n}} \left({x}\right){dx} \\ $$$${f}_{{n}} \left({x}\right)=\frac{\mathrm{1}}{{n}}\left(\mathrm{1}−\frac{{x}}{{n}^{\mathrm{2}} }\right)^{{n}} {ln}\left(\mathrm{1}+{x}\right)\:\Rightarrow{f}_{{n}} \left({x}\right)=\frac{\mathrm{1}}{{n}}{e}^{{nln}\left(\mathrm{1}−\frac{{x}}{{n}^{\mathrm{2}} }\right)} {ln}\left(\mathrm{1}+{x}\right) \\ $$$${ln}\left(\mathrm{1}−\frac{{x}}{{n}^{\mathrm{2}} }\right)\:\sim−\frac{{x}}{{n}^{\mathrm{2}} }\:\Rightarrow{nln}\left(\mathrm{1}−\frac{{x}}{{n}^{\mathrm{2}} }\right)\sim−\frac{{x}}{{n}}\:\Rightarrow{e}^{{nln}\left(\mathrm{1}−\frac{{x}}{{n}}\right)} \:\sim{e}^{−\frac{{x}}{{n}}} \\ $$$$\sim\mathrm{1}−\frac{{x}}{{n}}\:\Rightarrow{f}_{{n}} \left({x}\right)\sim\frac{\mathrm{1}}{{n}}\left(\mathrm{1}−\frac{{x}}{{n}}\right){ln}\left(\mathrm{1}+{x}\right)\:\:\:\left({n}\rightarrow+\infty\right)\:\Rightarrow \\ $$$$\left.{f}_{{n}} \:\rightarrow^{{cs}} \:\:\mathrm{0}\:\:{on}\right]\mathrm{0},+\infty\left[\:\:\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} ={lim}_{{n}\rightarrow+\infty} \int_{{R}} {f}_{{n}} \left({x}\right)\chi_{\left[\mathrm{0},{n}\left[\right.\right.} \:\left({x}\right){dx}\right. \\ $$$$=\int_{{R}} \:{lim}_{{n}\rightarrow+\infty} \:{f}_{{n}} \left({x}\right)\chi_{\left[\mathrm{0},{n}\left[\right.\right.} \left({x}\right){dx}\:=\mathrm{0} \\ $$