Menu Close

calculate-0-lnx-x-1-4-dx-

Tinku Tara 0 Comments
Pin




Question Number 98884 by mathmax by abdo last updated on 16/Jun/20
calculate ∫_0 ^∞ ((lnx)/((x+1)^4 ))dx
calculate∫0∞lnx(x+1)4dx
Answered by mathmax by abdo last updated on 17/Jun/20
I =∫_0 ^∞ ((lnx)/((x+1)^4 ))dx ⇒ I =∫_0 ^1 ((lnx)/((1+x)^4 ))dx +∫_1 ^∞ ((lnx)/((1+x)^4 ))dx(→x=(1/t)) =∫_0 ^1 ((lnx)/((1+x)^4 )) +−∫_0 ^1 ((−lnt)/((1+(1/t))^4 ))(((−dt)/t^2 ))=∫_0 ^1 ((lnx)/((1+x)^4 )) −∫_0 ^1 ((t^2 lnt)/((1+t)^4 ))dt =∫_0 ^1 (((1−x^2 )lnx)/((1+x)^4 )) dx we have (1/(1+x)) =Σ_(n=0) ^∞ (−1)^n x^n ⇒ −(1/((1+x)^2 )) =Σ_(n=1) ^∞ n(−1)^n x^(n−1) =Σ_(n=0) ^∞ (n+1)(−1)^(n+1) x^n ⇒ (1/((1+x)^2 )) =Σ_(n=0) ^∞ (n+1)(−1)^n x^n ⇒−((2(1+x))/((1+x)^4 )) =Σ_(n=1) ^∞ n(n+1)(−1)^n x^(n−1) ⇒ −(2/((1+x)^3 )) =Σ_(n=0) ^∞ (n+1)(n+2)(−1)^(n+1) x^n ⇒ (2/((1+x)^3 )) =Σ_(n=0) ^∞ (n+1)(n+2)(−1)^n x^n ⇒ −((2.3(1+x)^2 )/((1+x)^6 )) =Σ_(n=1) ^∞ n(n+1)(n+2)(−1)^n x^(n−1) ⇒ −(6/((1+x)^4 )) =Σ_(n=0) ^∞ (n+1)(n+2)(n+3)(−1)^(n+1) x^n ⇒ (6/((1+x)^4 )) =Σ_(n=0) ^∞ (n+1)(n+2)(n+3)(−1)^n x^n =Σ_(n=0) ^∞ (n^2 +3n +2)(n+3)(−1)^n x^n =Σ_(n=0) ^∞ (n^3 +3n^2 +3n^2 +9n +2n+3)(−1)^n x^n =Σ_(n=0) ^∞ (n^3 +6n^2 +11n +3)(−1)^n x^n ⇒ 6 ∫_0 ^1 (((1−x^2 )lnx)/((1+x)^4 ))dx =6 ∫_0 ^1 Σ_(n=0) ^∞ (n^3 +6n^2 +11n +3)(−1)^n x^n (1−x^2 )lnx dx =6Σ_(n=0) ^∞ (n^3 +6n^2 +11n +3)(−1)^n ∫_0 ^1 (x^n −x^(n+2) )lnx dx by parts ∫_0 ^1 (x^n −x^(n+2) )lnx dx =[((x^(n+1) /(n+1))−(x^(n+3) /(n+3)))lnx]_0 ^1 −∫_0 ^1 ((x^n /(n+1))−(x^(n+2) /(n+3)))dx =−(1/((n+1)^2 )) +(1/((n+3)^2 )) ⇒ 6 ∫_0 ^1 (((1−x^2 )lnx)/((1+x)^4 )) dx =6 Σ_(n=0) ^∞ (n^3 +6n^2 +11n +3)(−1)^n ((1/((n+3)^2 ))−(1/((n+1)^2 ))) ...be continued...
I=∫0∞lnx(x+1)4dx⇒I=∫01lnx(1+x)4dx+∫1∞lnx(1+x)4dx(→x=1t)=∫01lnx(1+x)4+−∫01−lnt(1+1t)4(−dtt2)=∫01lnx(1+x)4−∫01t2lnt(1+t)4dt=∫01(1−x2)lnx(1+x)4dxwehave11+x=∑n=0∞(−1)nxn⇒−1(1+x)2=∑n=1∞n(−1)nxn−1=∑n=0∞(n+1)(−1)n+1xn⇒1(1+x)2=∑n=0∞(n+1)(−1)nxn⇒−2(1+x)(1+x)4=∑n=1∞n(n+1)(−1)nxn−1⇒−2(1+x)3=∑n=0∞(n+1)(n+2)(−1)n+1xn⇒2(1+x)3=∑n=0∞(n+1)(n+2)(−1)nxn⇒−2.3(1+x)2(1+x)6=∑n=1∞n(n+1)(n+2)(−1)nxn−1⇒−6(1+x)4=∑n=0∞(n+1)(n+2)(n+3)(−1)n+1xn⇒6(1+x)4=∑n=0∞(n+1)(n+2)(n+3)(−1)nxn=∑n=0∞(n2+3n+2)(n+3)(−1)nxn=∑n=0∞(n3+3n2+3n2+9n+2n+3)(−1)nxn=∑n=0∞(n3+6n2+11n+3)(−1)nxn⇒6∫01(1−x2)lnx(1+x)4dx=6∫01∑n=0∞(n3+6n2+11n+3)(−1)nxn(1−x2)lnxdx=6∑n=0∞(n3+6n2+11n+3)(−1)n∫01(xn−xn+2)lnxdxbyparts∫01(xn−xn+2)lnxdx=[(xn+1n+1−xn+3n+3)lnx]01−∫01(xnn+1−xn+2n+3)dx=−1(n+1)2+1(n+3)2⇒6∫01(1−x2)lnx(1+x)4dx=6∑n=0∞(n3+6n2+11n+3)(−1)n(1(n+3)2−1(n+1)2)…becontinued…
Answered by maths mind last updated on 18/Jun/20
let f(n)=∫_0 ^(+∞) ((ln(x))/((x+1)^n ))dx....cv for n≥2 we want f(4) f(n+1)=∫_0 ^(+∞) ((ln(x)dx)/((1+x)^(n+1) )) =[((xln(x)−x)/((1+x)^(n+1) ))]_0 ^(+∞) −(n+1)∫_0 ^(+∞) ((xln(x)−x)/((1+x)^(n+2) ))dx ⇒∫_0 ^(+∞) ((ln(x)dx)/((1+x)^(n+1) ))=−(n+1)∫_0 ^(+∞) {((ln(x))/((1+x)^(n+1) )) −((ln(x))/((1+x)^(n+2) ))}dx +(n+1)∫_0 ^(+∞) ((1/((1+x)^(n+1) ))−(1/((1+x)^(n+2) )))dx ⇒f(n+1)=−(n+1)f(n+1)+(n+1)f(n+2)+(n+1)[((−1)/(n(1+x)^n ))+(1/((n+1)(1+x)^(n+1) ))]_0 ^(+∞) ⇒(n+2)f(n+1)=(n+1)f(n+2)+(n+1)((1/n)−(1/(n+1))) ⇒(n+1)f(n)=nf(n+1)+(1/(n−1)) ⇒((f(n))/n)=((f(n+1))/(n+1))+(1/(n(n−1)(n+1))) ⇒Σ_(k≥2) (((f(k+1))/(k+1))−((f(k))/k))=Σ_(k≥2) (1/(k(k−1)(k+1))) ⇒((f(n))/n)−((f(2))/2)=Σ_(k=2) ^(n−1) .(1/((k−1)k(k+1)))=Σ_(k=2) ^(n−1) ((1/(2(k−1)))−(1/k)+(1/(2(k+1)))) =−(1/(2(n−1)))+(1/2)−(1/4)+(1/(2n))=((−2n+n(n−1)+2(n−1))/(4n(n−1)))=((n^2 −n−2)/(4n(n−1))) f(2)=∫_0 ^(+∞) ((ln(x))/((1+x)^2 ))dx=∫_∞ ^0 ((ln((1/x))x^2 )/((1+x)^2 )).((−dx)/x^2 )⇒f(2)=0 ((f(n))/n)=((n^2 −n−2)/(4n(n−1)))⇒f(n)=((n^2 −n−2)/(4(n−1))) ∫_0 ^(+∞) ((ln(x)dx)/((1+x)^4 ))=f(4)=((4^2 −4−2)/(4(4−1)))=((10)/(4.3))=(5/6)
letf(n)=∫0+∞ln(x)(x+1)ndx….cvforn⩾2wewantf(4)f(n+1)=∫0+∞ln(x)dx(1+x)n+1=[xln(x)−x(1+x)n+1]0+∞−(n+1)∫0+∞xln(x)−x(1+x)n+2dx⇒∫0+∞ln(x)dx(1+x)n+1=−(n+1)∫0+∞{ln(x)(1+x)n+1−ln(x)(1+x)n+2}dx+(n+1)∫0+∞(1(1+x)n+1−1(1+x)n+2)dx⇒f(n+1)=−(n+1)f(n+1)+(n+1)f(n+2)+(n+1)[−1n(1+x)n+1(n+1)(1+x)n+1]0+∞⇒(n+2)f(n+1)=(n+1)f(n+2)+(n+1)(1n−1n+1)⇒(n+1)f(n)=nf(n+1)+1n−1⇒f(n)n=f(n+1)n+1+1n(n−1)(n+1)⇒∑k⩾2(f(k+1)k+1−f(k)k)=∑k⩾21k(k−1)(k+1)⇒f(n)n−f(2)2=∑n−1k=2.1(k−1)k(k+1)=∑n−1k=2(12(k−1)−1k+12(k+1))=−12(n−1)+12−14+12n=−2n+n(n−1)+2(n−1)4n(n−1)=n2−n−24n(n−1)f(2)=∫0+∞ln(x)(1+x)2dx=∫∞0ln(1x)x2(1+x)2.−dxx2⇒f(2)=0f(n)n=n2−n−24n(n−1)⇒f(n)=n2−n−24(n−1)∫0+∞ln(x)dx(1+x)4=f(4)=42−4−24(4−1)=104.3=56

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *