Question Number 40158 by maxmathsup by imad last updated on 16/Jul/18
![let A_n = ∫_0 ^1 ((x^(2n+1) ln(x))/(x^2 −1))dx 1) justify the existence of A_n 2)calculate A_(n+1) −A_n 3) prove that x∈]0,1[ ⇒0<((xln(x))/(x^2 −1))<(1/2) 4) find lim_(n→+∞) A_n](https://www.tinkutara.com/question/Q40158.png)
$${let}\:\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} \:{ln}\left({x}\right)}{{x}^{\mathrm{2}} \:−\mathrm{1}}{dx} \\ $$$$\left.\mathrm{1}\right)\:{justify}\:{the}\:{existence}\:{of}\:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right){calculate}\:{A}_{{n}+\mathrm{1}} \:−{A}_{{n}} \\ $$$$\left.\mathrm{3}\left.\right)\:{prove}\:{that}\:\:{x}\in\right]\mathrm{0},\mathrm{1}\left[\:\Rightarrow\mathrm{0}<\frac{{xln}\left({x}\right)}{{x}^{\mathrm{2}} \:−\mathrm{1}}<\frac{\mathrm{1}}{\mathrm{2}}\:\:\right. \\ $$$$\left.\mathrm{4}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} {A}_{{n}} \\ $$
Commented by math khazana by abdo last updated on 19/Jul/18
![1) lim_(x→0^+ ) ((x^(2n+1) ln(x))/(x^2 −1)) =−lim_(x→0^+ ) x^(2n+1) ln(x)=0 also lim_(x→1) =lim_(x→1) (x^(2n+1) /(x+1)) ((lnx)/(x−1)) =(1/2) because lim_(x→1) ((ln(x))/(x−1)) =1 ⇒ A_n exist 2) A_(n+1) −A_n = ∫_0 ^1 ((x^(2n+3) ln(x))/(x^2 −1))dx−∫_0 ^1 ((x^(2n+1) ln(x))/(x^2 −1))dx = ∫_0 ^1 ((x^(2n+1) (x^2 −1)ln(x))/(x^2 −1))dx = ∫_0 ^1 x^(2n+1) ln(x)dx =[ (1/(2n+2)) x^(2n+2) lnx]_0 ^1 −∫_0 ^1 (1/(2n+2))x^(2n+2) (dx/x) =− (1/(2n+2))∫_0 ^1 x^(2n) dx =−(1/((2n+1)(2n+2)))](https://www.tinkutara.com/question/Q40328.png)
$$\left.\mathrm{1}\right)\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} {ln}\left({x}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}\:=−{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:{x}^{\mathrm{2}{n}+\mathrm{1}} {ln}\left({x}\right)=\mathrm{0} \\ $$$${also}\:{lim}_{{x}\rightarrow\mathrm{1}} ={lim}_{{x}\rightarrow\mathrm{1}} \:\:\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{{x}+\mathrm{1}}\:\frac{{lnx}}{{x}−\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${because}\:{lim}_{{x}\rightarrow\mathrm{1}} \:\:\frac{{ln}\left({x}\right)}{{x}−\mathrm{1}}\:=\mathrm{1}\:\:\Rightarrow\:{A}_{{n}} \:\:{exist} \\ $$$$\left.\mathrm{2}\right)\:{A}_{{n}+\mathrm{1}} \:−{A}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{\mathrm{2}{n}+\mathrm{3}} \:{ln}\left({x}\right)}{{x}^{\mathrm{2}} \:−\mathrm{1}}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} \:{ln}\left({x}\right)}{{x}^{\mathrm{2}} \:−\mathrm{1}}{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} \left({x}^{\mathrm{2}} −\mathrm{1}\right){ln}\left({x}\right)}{{x}^{\mathrm{2}} \:−\mathrm{1}}{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{x}^{\mathrm{2}{n}+\mathrm{1}} \:{ln}\left({x}\right){dx}\: \\ $$$$=\left[\:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}}\:{x}^{\mathrm{2}{n}+\mathrm{2}} {lnx}\right]_{\mathrm{0}} ^{\mathrm{1}} \:\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}}{x}^{\mathrm{2}{n}+\mathrm{2}} \:\frac{{dx}}{{x}} \\ $$$$=−\:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:{x}^{\mathrm{2}{n}} {dx}\:=−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)} \\ $$
Commented by math khazana by abdo last updated on 19/Jul/18
$$\left.\mathrm{3}\right)\:{we}\:{have}\:\mathrm{0}<{x}<\mathrm{1}\:\:\Rightarrow\:{xln}\left({x}\right)<\mathrm{0}\:{and}\:{x}^{\mathrm{2}} \:−\mathrm{1}<\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{0}<\:\frac{{xln}\left({x}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}\:{let}\:{prove}\:{that}\:\:\frac{{xln}\left({x}\right)}{{x}^{\mathrm{2}} \:−\mathrm{1}}<\frac{\mathrm{1}}{\mathrm{2}}\:{let} \\ $$$${w}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{{xln}\left({x}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}\:{with}\:\mathrm{0}<{x}<\mathrm{1} \\ $$$${w}^{'} \left({x}\right)=−\frac{\left({lnx}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{1}\right)−{xln}\left({x}\right)\left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right){ln}\left({x}\right)\:+{x}^{\mathrm{2}} \:−\mathrm{1}\:−\mathrm{2}{x}^{\mathrm{2}} {ln}\left({x}\right)}{\left({x}^{\mathrm{2}} \:−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−\frac{−\left({x}^{\mathrm{2}} \:+\mathrm{1}\right){ln}\left({x}\right)\:+{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{−{x}^{\mathrm{2}} \:+\mathrm{1}\:+\left({x}^{\mathrm{2}} \:+\mathrm{1}\right){ln}\left({x}\right)}{\left({x}^{\mathrm{2}\:} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\:\frac{{x}^{\mathrm{2}} \left({ln}\left({x}\right)−\mathrm{1}\right)\:+{ln}\left({x}\right)+\mathrm{1}}{\left({x}^{\mathrm{2}} \:−\mathrm{1}\right)^{\mathrm{2}} }\:\:{any}\:{way}\:{we}\:{have} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:{w}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}>\mathrm{0} \\ $$$${lim}_{{x}\rightarrow\mathrm{1}} \:\:{w}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}>\mathrm{0}\:{still}\:{to}\:{prove}\:{that}\:{w}\:{dont} \\ $$$${change}\:{the}\:{sign}… \\ $$
Commented by math khazana by abdo last updated on 19/Jul/18
$$\left.\mathrm{4}\right)\:\mathrm{0}<\:{x}^{\mathrm{2}{n}} \:\:\frac{{xln}\left({x}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}<\:\frac{{x}^{\mathrm{2}{n}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{0}<\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} {ln}\left({x}\right)}{{x}^{\mathrm{2}} \:−\mathrm{1}}{dx}<\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{\mathrm{2}{n}} }{\mathrm{2}}{dx}\:\Rightarrow \\ $$$$\mathrm{0}<\:{A}_{{n}} <\:\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)}\:\rightarrow\mathrm{0}\left({n}\rightarrow+\infty\right)\:{so} \\ $$$${lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =\mathrm{0} \\ $$