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Question Number 66796 by mathmax by abdo last updated on 19/Aug/19
find the value of ∫_0 ^1 ((ln(1+x^2 ))/x^2 )dx
findthevalueof∫01ln(1+x2)x2dx
Commented by mathmax by abdo last updated on 20/Aug/19
let I =∫_0 ^1 ((ln(1+x^2 ))/x^2 )dx by parts u^′ =(1/x^2 ) and v=ln(1+x^2 ) ⇒ I =[−(1/x)ln(1+x^2 )]_0 ^1 −∫_0 ^1 (−(1/x))((2x)/(1+x^2 ))dx =−ln(2) +∫_0 ^1 (2/(1+x^2 ))dx =−ln(2)+2[arctanx]_0 ^1 =−ln(2)+2((π/4)) =(π/2)−ln(2) rest to prove that lim_(x→0) ((ln(1+x^2 ))/x) =0 =lim_(x→0) x ((ln(1+x^2 ))/x^2 ) =0 because lim_(u→0) ((ln(1+u))/u) =1
letI=∫01ln(1+x2)x2dxbypartsu′=1x2andv=ln(1+x2)⇒I=[−1xln(1+x2)]01−∫01(−1x)2x1+x2dx=−ln(2)+∫0121+x2dx=−ln(2)+2[arctanx]01=−ln(2)+2(π4)=π2−ln(2)resttoprovethatlimx→0ln(1+x2)x=0=limx→0xln(1+x2)x2=0becauselimu→0ln(1+u)u=1
Commented by mathmax by abdo last updated on 20/Aug/19
remark we have ln^′ (1+u) =(1/(1+u)) =Σ_(n=0) ^∞ (−1)^n u^n if ∣u∣<1 ⇒ ln(1+u) =Σ_(n=0) ^∞ (((−1)^n )/(n+1))u^(n+1) =Σ_(n=1) ^∞ (((−1)^(n−1) u^n )/n) =−Σ_(n=1) ^∞ (((−1)^n )/n)u^n ⇒ln(1+x^2 ) =−Σ_(n=1) ^∞ (((−1)^n )/n) x^(2n) ⇒((ln(1+x^2 ))/x^2 ) =−Σ_(n=1) ^∞ (((−1)^n )/n)x^(2n−2) ⇒∫_0 ^1 ((ln(1+x^2 ))/x^2 )dx =−Σ_(n=1) ^∞ (((−1)^n )/n) [(1/(2n−1))x^(2n−1) ]_0 ^1 =−Σ_(n=1) ^∞ (((−1)^n )/(n(2n−1))) =(π/2)−ln(2) ⇒Σ_(n=1) ^∞ (((−1)^n )/(n(2n−1))) =ln(2)−(π/2)
remarkwehaveln′(1+u)=11+u=∑n=0∞(−1)nunif∣u∣<1⇒ln(1+u)=∑n=0∞(−1)nn+1un+1=∑n=1∞(−1)n−1unn=−∑n=1∞(−1)nnun⇒ln(1+x2)=−∑n=1∞(−1)nnx2n⇒ln(1+x2)x2=−∑n=1∞(−1)nnx2n−2⇒∫01ln(1+x2)x2dx=−∑n=1∞(−1)nn[12n−1x2n−1]01=−∑n=1∞(−1)nn(2n−1)=π2−ln(2)⇒∑n=1∞(−1)nn(2n−1)=ln(2)−π2

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