Question Number 33343 by prof Abdo imad last updated on 14/Apr/18
![prove that ∀ α ∈]1,+∞[ lim_(n→∞) ∫_0 ^n (1+(x/n))^n e^(−αx) dx = (1/(α−1)) .](https://www.tinkutara.com/question/Q33343.png)
$$\left.{prove}\:{that}\:\forall\:\alpha\:\in\right]\mathrm{1},+\infty\left[\right. \\ $$$${lim}_{{n}\rightarrow\infty} \:\:\int_{\mathrm{0}} ^{{n}} \:\left(\mathrm{1}+\frac{{x}}{{n}}\right)^{{n}} \:{e}^{−\alpha{x}} {dx}\:=\:\frac{\mathrm{1}}{\alpha−\mathrm{1}}\:. \\ $$
Commented by abdo imad last updated on 17/Apr/18
![let put I_n = ∫_0 ^n (1+(x/n))^n e^(−αx) dx = ∫_R f_n (x)dx with f_n (x) = (1+(x/n))^n e^(−αx) χ_([0,n[) (x)dx we have f_n (x)→^(c.s) f(x) = e^((1−α)x) if x≥0 and f(x)=0 if x<0 but we have (1+(x/n))^n =e^(nln(1+(x/n))) but ln(1+(x/n))≤(x/n) ⇒ nln(1+(x/n)) ≤x ⇒ (1+(x/n))^n e^(−αx) ≤ g(x)=_ e^((1−α)x) theorem of convergence dominee give ∫_R f_n (x)dx _(n→∞) → ∫_0 ^∞ e^((1−α)x) dx =(1/(1−α))[ e^((1−α)x) ]_0 ^(+∞) =((−1)/(1−α)) = (1/(α−1)) ⇒ lim_(n→∞) I_n = (1/(1−α)) .](https://www.tinkutara.com/question/Q33491.png)
$${let}\:{put}\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{{n}} \:\left(\mathrm{1}+\frac{{x}}{{n}}\right)^{{n}} \:{e}^{−\alpha{x}} {dx}\:=\:\int_{{R}} {f}_{{n}} \left({x}\right){dx}\:\:{with} \\ $$$${f}_{{n}} \left({x}\right)\:=\:\left(\mathrm{1}+\frac{{x}}{{n}}\right)^{{n}} \:{e}^{−\alpha{x}} \:\chi_{\left[\mathrm{0},{n}\left[\right.\right.} \left({x}\right){dx}\:\:{we}\:{have} \\ $$$${f}_{{n}} \left({x}\right)\rightarrow^{{c}.{s}} \:\:{f}\left({x}\right)\:=\:{e}^{\left(\mathrm{1}−\alpha\right){x}} \:{if}\:{x}\geqslant\mathrm{0}\:{and}\:{f}\left({x}\right)=\mathrm{0}\:{if}\:{x}<\mathrm{0}\: \\ $$$${but}\:{we}\:{have}\:\left(\mathrm{1}+\frac{{x}}{{n}}\right)^{{n}} ={e}^{{nln}\left(\mathrm{1}+\frac{{x}}{{n}}\right)} \:\:\:{but}\:{ln}\left(\mathrm{1}+\frac{{x}}{{n}}\right)\leqslant\frac{{x}}{{n}}\:\Rightarrow \\ $$$${nln}\left(\mathrm{1}+\frac{{x}}{{n}}\right)\:\leqslant{x}\:\:\Rightarrow\:\left(\mathrm{1}+\frac{{x}}{{n}}\right)^{{n}} \:{e}^{−\alpha{x}} \:\leqslant\:{g}\left({x}\right)=_{} {e}^{\left(\mathrm{1}−\alpha\right){x}} \: \\ $$$${theorem}\:{of}\:{convergence}\:{dominee}\:{give} \\ $$$$\int_{{R}} {f}_{{n}} \left({x}\right){dx}\:_{{n}\rightarrow\infty} \rightarrow\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{\left(\mathrm{1}−\alpha\right){x}} {dx}\:=\frac{\mathrm{1}}{\mathrm{1}−\alpha}\left[\:{e}^{\left(\mathrm{1}−\alpha\right){x}} \right]_{\mathrm{0}} ^{+\infty} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{1}−\alpha}\:=\:\frac{\mathrm{1}}{\alpha−\mathrm{1}}\:\Rightarrow\:{lim}_{{n}\rightarrow\infty} \:{I}_{{n}} \:\:=\:\frac{\mathrm{1}}{\mathrm{1}−\alpha}\:\:. \\ $$
Commented by prof Abdo imad last updated on 17/Apr/18
$${lim}\:{I}_{{n}} =\:\frac{\mathrm{1}}{\alpha−\mathrm{1}}\:. \\ $$