Question Number 33694 by math khazana by abdo last updated on 22/Apr/18
$${calculate}\:{lim}_{{n}\rightarrow+\infty} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{{x}^{{n}} \:\:+{e}^{{x}} }\:\:. \\ $$
Commented by math khazana by abdo last updated on 29/Apr/18
![let put A_n = ∫_0 ^n (dx/(x^n +e^x )) A_n = ∫_0 ^1 (dx/(x^n +e^x )) + ∫_1 ^n (dx/(x^n +e^x )) but ∫_0 ^1 (dx/(x^n +e^x ))→ ∫_0 ^1 e^(−x) dx=[−e^(−x) ]_0 ^1 = 1−(1/e) ch. x^n =t give ∫_1 ^n (dx/(x^n +e^x )) = ∫_1 ^n^n (1/(t + e^t^(1/n) )) (1/n) t^((1/n)−1) dt = (1/n) ∫_1 ^n^n (t^(1/n) /(t^2 +t e^t^(1/n) ))dt ≤ (1/n) ∫_1 ^(+∞) (t^(1/n) /(t^2 +t e^t^(1/n) ))dt ≤ (1/n) ∫_1 ^(+∞) (t^(1/n) /t^2 ) dt →0(n→+∞) so lim_(n→+∞) A_n =1−(1/e) .](https://www.tinkutara.com/question/Q34027.png)
$${let}\:{put}\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{{n}} \:\:\:\:\:\:\:\frac{{dx}}{{x}^{{n}} \:\:+{e}^{{x}} }\:\: \\ $$$${A}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{{x}^{{n}} \:+{e}^{{x}} }\:\:+\:\int_{\mathrm{1}} ^{{n}} \:\:\:\:\:\frac{{dx}}{{x}^{{n}} \:\:+{e}^{{x}} }\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{{x}^{{n}} \:+{e}^{{x}} }\rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}} {dx}=\left[−{e}^{−{x}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:=\:\mathrm{1}−\frac{\mathrm{1}}{{e}} \\ $$$${ch}.\:{x}^{{n}} ={t}\:{give} \\ $$$$\int_{\mathrm{1}} ^{{n}} \:\:\:\frac{{dx}}{{x}^{{n}} \:+{e}^{{x}} }\:=\:\int_{\mathrm{1}} ^{{n}^{{n}} } \:\:\:\:\:\frac{\mathrm{1}}{{t}\:\:+\:{e}^{{t}^{\frac{\mathrm{1}}{{n}}} } }\:\frac{\mathrm{1}}{{n}}\:{t}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} {dt} \\ $$$$=\:\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{1}} ^{{n}^{{n}} } \:\:\:\:\:\:\frac{{t}^{\frac{\mathrm{1}}{{n}}} }{{t}^{\mathrm{2}} \:+{t}\:{e}^{{t}^{\frac{\mathrm{1}}{{n}}} } }{dt}\:\leqslant\:\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{{t}^{\frac{\mathrm{1}}{{n}}} }{{t}^{\mathrm{2}} \:+{t}\:{e}^{{t}^{\frac{\mathrm{1}}{{n}}} } }{dt} \\ $$$$\leqslant\:\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{{t}^{\frac{\mathrm{1}}{{n}}} }{{t}^{\mathrm{2}} }\:{dt}\:\rightarrow\mathrm{0}\left({n}\rightarrow+\infty\right)\:{so} \\ $$$${lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =\mathrm{1}−\frac{\mathrm{1}}{{e}}\:. \\ $$$$ \\ $$