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Question Number 55229 by maxmathsup by imad last updated on 19/Feb/19
calculate lim_(n→+∞) ∫_0 ^n (e^(nx) /(1+nx^2 )) dx .
$${calculate}\:{lim}_{{n}\rightarrow+\infty} \:\int_{\mathrm{0}} ^{{n}} \:\:\frac{{e}^{{nx}} }{\mathrm{1}+{nx}^{\mathrm{2}} }\:{dx}\:\:. \\ $$
Commented by maxmathsup by imad last updated on 25/Feb/19
let I_n =∫_0 ^n (e^(nx) /(1+nx^2 )) dx we have 1+nx^2 ≤n(1+x^2 ) ⇒(1/(1+nx^2 )) ≥(1/(n(1+x^2 ))) ⇒ (e^(nx) /(1+nx^2 )) ≥ (e^(nx) /(n(1+x^2 ))) ⇒ I_n ≥(1/n) ∫_0 ^n (e^(nx) /(1+x^2 ))dx but e^(nx) =Σ_(p=0) ^∞ (((nx)^p )/(p!)) ⇒ (e^(nx) /(1+x^2 )) =Σ_(p=0) ^∞ ((n^p x^p )/(p!(1+x^2 ))) ⇒(1/n)∫_0 ^n (e^(nx) /(1+x^2 )) dx =(1/n)Σ_(p=0) ^∞ n^p ∫_0 ^n (x^p /(p!(1+x^2 )))dx =∫_0 ^n (dx/(1+x^2 )) +Σ_(p=1) ^∞ n^(p−1) ∫_0 ^n (x^p /(p!(1+x^2 )))dx ⇒ lim_(n→+∞) (1/n) ∫_0 ^n (e^(nx) /(1+x^2 ))dx =+∞ ⇒ lim_(n→+∞) I_n = +∞ .
$$\:{let}\:{I}_{{n}} =\int_{\mathrm{0}} ^{{n}} \:\:\frac{{e}^{{nx}} }{\mathrm{1}+{nx}^{\mathrm{2}} }\:{dx}\:\:{we}\:{have}\:\mathrm{1}+{nx}^{\mathrm{2}} \leqslant{n}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:\Rightarrow\frac{\mathrm{1}}{\mathrm{1}+{nx}^{\mathrm{2}} }\:\geqslant\frac{\mathrm{1}}{{n}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$$\frac{{e}^{{nx}} }{\mathrm{1}+{nx}^{\mathrm{2}} }\:\geqslant\:\frac{{e}^{{nx}} }{{n}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:\Rightarrow\:{I}_{{n}} \:\geqslant\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{{n}} \:\:\frac{{e}^{{nx}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:{but} \\ $$$${e}^{{nx}} \:=\sum_{{p}=\mathrm{0}} ^{\infty} \:\frac{\left({nx}\right)^{{p}} }{{p}!}\:\Rightarrow\:\frac{{e}^{{nx}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:=\sum_{{p}=\mathrm{0}} ^{\infty} \:\frac{{n}^{{p}} \:{x}^{{p}} }{{p}!\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:\Rightarrow\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{{n}} \:\frac{{e}^{{nx}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$$$=\frac{\mathrm{1}}{{n}}\sum_{{p}=\mathrm{0}} ^{\infty} \:{n}^{{p}} \:\int_{\mathrm{0}} ^{{n}} \:\frac{{x}^{{p}} }{{p}!\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}\:=\int_{\mathrm{0}} ^{{n}} \:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:+\sum_{{p}=\mathrm{1}} ^{\infty} \:{n}^{{p}−\mathrm{1}} \:\int_{\mathrm{0}} ^{{n}} \:\frac{{x}^{{p}} }{{p}!\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{{n}} \:\frac{{e}^{{nx}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=+\infty\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} \:{I}_{{n}} =\:+\infty\:. \\ $$

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