Question Number 147344 by mathdanisur last updated on 20/Jul/21
$$\underset{{n}\rightarrow\infty} {{lim}}\:\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:{log}\:\left(\frac{\mathrm{1}+{sin}^{\boldsymbol{{n}}} \boldsymbol{{x}}}{\mathrm{1}+{x}+{x}^{\boldsymbol{{n}}} }\right){dx}\:=\:? \\ $$
Answered by mathmax by abdo last updated on 20/Jul/21
![f_n (x)=log(((1+sin^n x)/(1+x+x^n )))→cs to f(x)=log((1/(1+x)))=−log(1+x) ⇒ lim_(n→+∞) ∫_0 ^1 f_n (x)dx=∫_0 ^1 lim_(n→+∞) f_n (x)dx=−∫_0 ^1 log(1+x)dx =_(1+x=t) −∫_1 ^2 log(t)dt =−[tlot−t]_1 ^2 =−(2log2−2+1) =1−2log2](Q147345.png)
$$\mathrm{f}_{\mathrm{n}} \left(\mathrm{x}\right)=\mathrm{log}\left(\frac{\mathrm{1}+\mathrm{sin}^{\mathrm{n}} \mathrm{x}}{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{n}} }\right)\rightarrow\mathrm{cs}\:\mathrm{to}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{log}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}}\right)=−\mathrm{log}\left(\mathrm{1}+\mathrm{x}\right)\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{f}_{\mathrm{n}} \left(\mathrm{x}\right)\mathrm{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{f}_{\mathrm{n}} \left(\mathrm{x}\right)\mathrm{dx}=−\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{log}\left(\mathrm{1}+\mathrm{x}\right)\mathrm{dx} \\ $$$$=_{\mathrm{1}+\mathrm{x}=\mathrm{t}} \:\:\:−\int_{\mathrm{1}} ^{\mathrm{2}} \:\mathrm{log}\left(\mathrm{t}\right)\mathrm{dt}\:=−\left[\mathrm{tlot}−\mathrm{t}\right]_{\mathrm{1}} ^{\mathrm{2}} \:=−\left(\mathrm{2log2}−\mathrm{2}+\mathrm{1}\right) \\ $$$$=\mathrm{1}−\mathrm{2log2} \\ $$
Commented by mathdanisur last updated on 20/Jul/21
$${thank}\:{you}\:{Ser} \\ $$
Commented by mathmax by abdo last updated on 20/Jul/21
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir} \\ $$
Commented by mathdanisur last updated on 20/Jul/21
$${Thanks}\:{Ser},\:{but}\:{how}\:{did}\:{you}\:{change} \\ $$$$\left({interchange}\:{of}\right)\:{the}\:{limit}\:{to}\:{the} \\ $$$${integral},\:{please}\:{note}\:{Ser}... \\ $$
Commented by mathmax by abdo last updated on 21/Jul/21
$$\mathrm{theoreme}\:\mathrm{de}\:\mathrm{convergence}\:\mathrm{dominee}... \\ $$
Commented by mathdanisur last updated on 21/Jul/21
$${Ser},\:{can}\:{you}\:{state}\:{that}\:{theorem},\:{please} \\ $$