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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

letAn=∫01x2n+1ln(x)x2−1dx 1)justifytheexistenceofAn 2)calculateAn+1−An 3)provethatx∈]0,1[⇒0<xln(x)x2−1<12 4)findlimn→+∞An

Commented bymath 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)))

1)limx→0+x2n+1ln(x)x2−1=−limx→0+x2n+1ln(x)=0 alsolimx→1=limx→1x2n+1x+1lnxx−1=12 becauselimx→1ln(x)x−1=1⇒Anexist 2)An+1−An=∫01x2n+3ln(x)x2−1dx−∫01x2n+1ln(x)x2−1dx =∫01x2n+1(x2−1)ln(x)x2−1dx =∫01x2n+1ln(x)dx =[12n+2x2n+2lnx]01−∫0112n+2x2n+2dxx =−12n+2∫01x2ndx=−1(2n+1)(2n+2)

Commented bymath khazana by abdo last updated on 19/Jul/18

3) we have 0<x<1 ⇒ xln(x)<0 and x^2 −1<0 ⇒ 0< ((xln(x))/(x^2 −1)) let prove that ((xln(x))/(x^2 −1))<(1/2) let w(x)=(1/2) −((xln(x))/(x^2 −1)) with 0<x<1 w^′ (x)=−(((lnx+1)(x^2 −1)−xln(x)(2x))/((x^2 −1)^2 )) =−(((x^2 −1)ln(x) +x^2 −1 −2x^2 ln(x))/((x^2 −1)^2 )) =−((−(x^2 +1)ln(x) +x^2 −1)/((x^2 −1)^2 )) =((−x^2 +1 +(x^2 +1)ln(x))/((x^(2 ) −1)^2 )) = ((x^2 (ln(x)−1) +ln(x)+1)/((x^2 −1)^2 )) any way we have lim_(x→0^+ ) w(x)=(1/2)>0 lim_(x→1) w(x)= (1/2)>0 still to prove that w dont change the sign...

3)wehave0<x<1⇒xln(x)<0andx2−1<0⇒ 0<xln(x)x2−1letprovethatxln(x)x2−1<12let w(x)=12−xln(x)x2−1with0<x<1 w′(x)=−(lnx+1)(x2−1)−xln(x)(2x)(x2−1)2 =−(x2−1)ln(x)+x2−1−2x2ln(x)(x2−1)2 =−−(x2+1)ln(x)+x2−1(x2−1)2=−x2+1+(x2+1)ln(x)(x2−1)2 =x2(ln(x)−1)+ln(x)+1(x2−1)2anywaywehave limx→0+w(x)=12>0 limx→1w(x)=12>0stilltoprovethatwdont changethesign...

Commented bymath khazana by abdo last updated on 19/Jul/18

4) 0< x^(2n) ((xln(x))/(x^2 −1))< (x^(2n) /2) ⇒ 0< ∫_0 ^1 ((x^(2n+1) ln(x))/(x^2 −1))dx< ∫_0 ^1 (x^(2n) /2)dx ⇒ 0< A_n < (1/(2(2n+1))) →0(n→+∞) so lim_(n→+∞) A_n =0

4)0<x2nxln(x)x2−1<x2n2⇒ 0<∫01x2n+1ln(x)x2−1dx<∫01x2n2dx⇒ 0<An<12(2n+1)→0(n→+∞)so limn→+∞An=0

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