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Question Number 192933 by mnjuly1970 last updated on 31/May/23

Answered by MM42 last updated on 01/Jun/23

ln(1−x)=u⇒((−1)/(1−x))dx=du &i x^(n−1) dx=dv⇒(x^n /n)=v ⇒I_n =∫x^(n−1) ln(1−x)dx=((x^n ln(1−x))/n) +(1/n)∫ (x^n /(1−x))dx ∫ (x^n /(1−x))dx=−∫((1−x^n −1)/(1−x))dx =−∫(1+x^2 +x^3 +...+x^(n−1) )dx+∫(dx/(1−x)) =x+(x^2 /2)+(x^3 /3)+...+(x^n /n)−ln(1−x) ⇒φ_n =lim_(x→1) ( ((x^n ln(1−x))/n)−((ln(1−x))/n))−(1/n)(1+(1/2)+(1/3)+...+(1/n)) =lim_(x→1) (((1+x+x^2 +...+x^(n−1) )/n))((ln(1−x))/(1/(x−1)))=0 ⇒φ_n =−(1/n)((1/1)+(1/2)+(1/3)+...+(1/n))

ln(1−x)=u⇒−11−xdx=du&ixn−1dx=dv⇒xnn=v⇒In=∫xn−1ln(1−x)dx=xnln(1−x)n+1n∫xn1−xdx∫xn1−xdx=−∫1−xn−11−xdx=−∫(1+x2+x3+...+xn−1)dx+∫dx1−x=x+x22+x33+...+xnn−ln(1−x)⇒ϕn=limx→1(xnln(1−x)n−ln(1−x)n)−1n(1+12+13+...+1n)=limx→1(1+x+x2+...+xn−1n)ln(1−x)1x−1=0⇒ϕn=−1n(11+12+13+...+1n)

Answered by witcher3 last updated on 01/Jun/23

Σ_(n≥1) (−1)^(n−1) x^(n−1) =(1/(1+x)) Ω=∫_0 ^1 ((ln(1−x))/(1+x))dx x→((1−t)/(1+t))⇒∫_0 ^1 ((ln(((2t)/(1+t))))/(2/(1+t))).(2/((1+t)^2 ))dt =∫_0 ^1 ((ln(2))/(1+t))+((ln(t))/(1+t))−((ln(1+t))/(1+t))dt =(1/2)ln^2 (2)+[_0 ^1 ln(t)ln(1+t)]−∫_0 ^1 ((ln(1+t))/t)dt =((ln^2 (2))/2)−∫_0 ^1 ((ln(1−(−t)))/((−t)))d(−t) =((ln^2 (2))/2)+Li_2 (−1)=((ln^2 (2))/2)−(𝛑^2 /(12))

∑n⩾1(−1)n−1xn−1=11+xΩ=∫01ln(1−x)1+xdxx→1−t1+t⇒∫01ln(2t1+t)21+t.2(1+t)2dt=∫01ln(2)1+t+ln(t)1+t−ln(1+t)1+tdt=12ln2(2)+[01ln(t)ln(1+t)]−∫01ln(1+t)tdt=ln2(2)2−∫01ln(1−(−t))(−t)d(−t)=ln2(2)2+Li2(−1)=ln2(2)2−π212

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