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Question Number 141623 by qaz last updated on 21/May/21
∑∞n=0(∫01xn1+xdx)2=ln2
Answered by mindispower last updated on 21/May/21
(∫01xn1+xdx)2=∫01∫01(xy)n(1+x)(1+y)dxdy⇔∑n⩾0∫∫01(xy)n(1+x)(1+y)dxdy=∫∫011(1+x)(1+y)(1−xy)dxdyS=∫011(1+y)∫01dx(1+x)(1−xy)dy∫01dx(1+x)(1−xy)=∫011(1+y).11+x+y1+y.11−xydx=ln(2)1+y−11+yln(1−y)S=∫01ln(2)(1+y)2−1(1+y)2ln(1−y)dy=[−ln(2)1+y]01+limx→1ln(1−y)1+y]0x+∫0x1(1+y)(1−y)dy=ln(2)2+limx→1ln(1−x)1+x+12ln(1+x1−x)=ln(2)2+limx→1ln(1+x)2+ln(1−x){11+x−12}=ln(2)+limx→1(1−x)ln(1−x)2(1+x)=ln(2)⇔∑n⩾0(∫01xn1+xdx)2=ln(2)
Commented by qaz last updated on 22/May/21
thankssirpower
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