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Question Number 141623 by qaz last updated on 21/May/21

Σ_(n=0) ^∞ (∫_0 ^1 (x^n /(1+x))dx)^2 =ln 2

n=0(01xn1+xdx)2=ln2

Answered by mindispower last updated on 21/May/21

(∫_0 ^1 (x^n /(1+x))dx)^2 =∫_0 ^1 ∫_0 ^1 (((xy)^n )/((1+x)(1+y)))dxdy  ⇔Σ_(n≥0) ∫∫_0 ^1 (((xy)^n )/((1+x)(1+y)))dxdy  =∫∫_0 ^1 (1/((1+x)(1+y)(1−xy)))dxdy  S=∫_0 ^1 (1/((1+y)))∫_0 ^1 (dx/((1+x)(1−xy)))dy  ∫_0 ^1 (dx/((1+x)(1−xy)))=∫_0 ^1 (1/((1+y))).(1/(1+x))+(y/(1+y)).(1/(1−xy))dx  =((ln(2))/(1+y))−(1/(1+y))ln(1−y)  S=∫_0 ^1 ((ln(2))/((1+y)^2 ))−(1/((1+y)^2 ))ln(1−y)dy  =[−((ln(2))/(1+y))]_0 ^1 +lim_(x→1) ((ln(1−y))/(1+y))]_0 ^x +∫_0 ^x (1/((1+y)(1−y)))dy  =((ln(2))/2)+lim_(x→1) ((ln(1−x))/(1+x))+(1/2)ln(((1+x)/(1−x)))  =((ln(2))/2)+lim_(x→1) ((ln(1+x))/2)+ln(1−x){(1/(1+x))−(1/2)}  =ln(2)+lim_(x→1) (((1−x)ln(1−x))/(2(1+x)))=ln(2)  ⇔Σ_(n≥0) (∫_0 ^1 (x^n /(1+x))dx)^2 =ln(2)

(01xn1+xdx)2=0101(xy)n(1+x)(1+y)dxdyn001(xy)n(1+x)(1+y)dxdy=011(1+x)(1+y)(1xy)dxdyS=011(1+y)01dx(1+x)(1xy)dy01dx(1+x)(1xy)=011(1+y).11+x+y1+y.11xydx=ln(2)1+y11+yln(1y)S=01ln(2)(1+y)21(1+y)2ln(1y)dy=[ln(2)1+y]01+limx1ln(1y)1+y]0x+0x1(1+y)(1y)dy=ln(2)2+limx1ln(1x)1+x+12ln(1+x1x)=ln(2)2+limx1ln(1+x)2+ln(1x){11+x12}=ln(2)+limx1(1x)ln(1x)2(1+x)=ln(2)n0(01xn1+xdx)2=ln(2)

Commented by qaz last updated on 22/May/21

thanks sir power

thankssirpower

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