Question Number 66351 by mathmax by abdo last updated on 12/Aug/19
$${let}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{{nt}} }{\left(\mathrm{1}+{e}^{{t}} \right)^{{n}+\mathrm{1}} }{dt}\:\:\:\:\:\left({n}\:{from}\:{N}^{\bigstar} \right) \\ $$$$\left.\right){prove}\:{the}\:{existence}\:{of}\:{I}_{{n}} \\ $$$$\left.\mathrm{2}\right){find}\:{lim}_{{n}\rightarrow+\infty} \:\:\:{I}_{{n}} \\ $$
Commented by mathmax by abdo last updated on 15/Aug/19
![∀A>0 thefunction t→(e^(nt) /((1+e^t )^(n+1) )) is continue on [0,A] so integrable let see what happenat +∞ we have lim_(t→+∞) t^2 (e^(nt) /((1+e^t )^(n+1) )) =lim_(t→+∞) t^2 (e^(nt) /e^((n+1)t) ) =lim_(n→+∞) t^2 e^(−t) =0 so I_n is convergent 2)changement e^(nt) =x give nt=ln(x) ⇒ I_n =∫_1 ^(+∞) (x/((1+x^(1/n) )^(n+1) )) (dx/(nx)) =∫_1 ^(+∞) (dx/((1+x^(1/n) )^(n+1) )) =∫_1 ^(+∞) e^(−(n+1)ln(1+x^(1/n) )) dx =∫_R f_n (x)dx with f_n (x)=e^(−(n+1)ln(1+x^(1/n) )) χ_([1,+∞[) (x) we have f_n )continues and f_n →^(cs) 0 ⇒lim_(n→+∞) I_n =∫_R lim_(n→+∞) f_n (x)dx =0](https://www.tinkutara.com/question/Q66454.png)
$$\forall{A}>\mathrm{0}\:{thefunction}\:{t}\rightarrow\frac{{e}^{{nt}} }{\left(\mathrm{1}+{e}^{{t}} \right)^{{n}+\mathrm{1}} }\:{is}\:{continue}\:{on}\:\left[\mathrm{0},{A}\right]\:{so} \\ $$$${integrable}\:\:{let}\:{see}\:{what}\:{happenat}\:+\infty\:{we}\:{have}\: \\ $$$${lim}_{{t}\rightarrow+\infty} \:\:\:{t}^{\mathrm{2}} \:\frac{{e}^{{nt}} }{\left(\mathrm{1}+{e}^{{t}} \right)^{{n}+\mathrm{1}} }\:={lim}_{{t}\rightarrow+\infty} {t}^{\mathrm{2}} \:\frac{{e}^{{nt}} }{{e}^{\left({n}+\mathrm{1}\right){t}} }\:={lim}_{{n}\rightarrow+\infty} {t}^{\mathrm{2}} {e}^{−{t}} =\mathrm{0} \\ $$$${so}\:{I}_{{n}} {is}\:{convergent} \\ $$$$\left.\mathrm{2}\right){changement}\:{e}^{{nt}} ={x}\:{give}\:{nt}={ln}\left({x}\right)\:\Rightarrow \\ $$$${I}_{{n}} =\int_{\mathrm{1}} ^{+\infty} \:\:\:\:\frac{{x}}{\left(\mathrm{1}+{x}^{\frac{\mathrm{1}}{{n}}} \right)^{{n}+\mathrm{1}} }\:\frac{{dx}}{{nx}}\:=\int_{\mathrm{1}} ^{+\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\frac{\mathrm{1}}{{n}}} \right)^{{n}+\mathrm{1}} } \\ $$$$=\int_{\mathrm{1}} ^{+\infty} \:{e}^{−\left({n}+\mathrm{1}\right){ln}\left(\mathrm{1}+{x}^{\frac{\mathrm{1}}{{n}}} \right)} {dx}\:=\int_{{R}} \:{f}_{{n}} \left({x}\right){dx}\:{with} \\ $$$$\left.{f}_{{n}} \left({x}\right)={e}^{−\left({n}+\mathrm{1}\right){ln}\left(\mathrm{1}+{x}^{\frac{\mathrm{1}}{{n}}} \right)} \chi_{\left[\mathrm{1},+\infty\left[\right.\right.} \:\left({x}\right)\:\:{we}\:{have}\:{f}_{{n}} \right){continues} \\ $$$${and}\:{f}_{{n}} \rightarrow^{{cs}} \:\mathrm{0}\:\:\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{I}_{{n}} =\int_{{R}} {lim}_{{n}\rightarrow+\infty} {f}_{{n}} \left({x}\right){dx}\:=\mathrm{0} \\ $$