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Question Number 78266 by msup trace by abdo last updated on 15/Jan/20
calculate f(a)=∫_0 ^1 ln(1−ax^3 )dx with 0<a<1
calculatef(a)=∫01ln(1−ax3)dxwith0<a<1
Commented by mathmax by abdo last updated on 18/Jan/20
f(a) =∫_0 ^1 ln(1−(^3 (√a)x)^3 )dx =_((^3 (√a)x)=t) ∫_0 ^((^3 (√a))) ln(1−t^3 )(dt/((^3 (√a)))) =(1/((^3 (√a)))) ∫_0 ^((^3 (√a))) ln(1−t^3 )dt we have ln(1−u))^((1)) =−(1/(1−u)) =−Σ_(n=0) ^∞ u^n ⇒ln(1−u) =−Σ_(n=0) ^∞ (u^(n+1) /(n+1)) =−Σ_(n=1) ^∞ (u^n /n) we have 0<(^3 (√a))<1 ⇒ln(1−t^3 )=−Σ_(n=1) ^∞ (t^(3n) /n) ⇒ ∫_0 ^((^3 (√a))) ln(1−t^3 )dt =−Σ_(n=1) ^∞ (1/n) ∫_0 ^((^3 (√a))) t^(3n) dt =−Σ_(n=1) ^∞ (1/(n(3n+1)))[ t^(3n+1) ]_0 ^((^3 (√a))) =−Σ_(n=1) ^∞ (((^3 (√a))a^n )/(n(3n+1))) ⇒ f(a) =−Σ_(n=1) ^∞ (a^n /(n(3n+1))) ⇒−(1/3)f(a) =Σ_(n=1) ^∞ (a^n /(3n(3n+1))) =Σ_(n=1) ^∞ ((1/(3n))−(1/(3n+1)))a^n =(1/3)Σ_(n=1) ^∞ (a^n /n)−Σ_(n=1) ^∞ (a^n /(3n+1)) =−(1/3)ln(1−a)−Σ_(n=1) ^∞ (a^n /(3n+1)) Σ_(n=1) ^∞ (a^n /(3n+1)) =Σ_(n=1) ^∞ ((((^3 (√a))^(3n+1) )/(3n+1)))×(1/((^3 (√a)))) =(1/((^3 (√a))))W(^3 (√a)) with w(x)=Σ_(n=1) ^∞ (x^(3n+1) /(3n+1)) ⇒w^′ (x)=Σ_(n=1) ^∞ x^(3n) =(1/(1−x^3 ))−1 ⇒ w(x)=∫_0 ^x ((1/(1−t^3 ))−1)dt +c =−x +∫_0 ^x (dt/((1−t^3 ))) =−x−∫_0 ^x (dt/(t^3 −1)) let decompose F(t) =(1/(t^3 −1)) =(1/((t−1)(t^2 +t+1))) F(t)=(a/(t−1)) +((bt+c)/(t^2 +t +1))
f(a)=∫01ln(1−(3ax)3)dx=(3ax)=t∫0(3a)ln(1−t3)dt(3a)=1(3a)∫0(3a)ln(1−t3)dtwehaveln(1−u))(1)=−11−u=−∑n=0∞un⇒ln(1−u)=−∑n=0∞un+1n+1=−∑n=1∞unnwehave0<(3a)<1⇒ln(1−t3)=−∑n=1∞t3nn⇒∫0(3a)ln(1−t3)dt=−∑n=1∞1n∫0(3a)t3ndt=−∑n=1∞1n(3n+1)[t3n+1]0(3a)=−∑n=1∞(3a)ann(3n+1)⇒f(a)=−∑n=1∞ann(3n+1)⇒−13f(a)=∑n=1∞an3n(3n+1)=∑n=1∞(13n−13n+1)an=13∑n=1∞ann−∑n=1∞an3n+1=−13ln(1−a)−∑n=1∞an3n+1∑n=1∞an3n+1=∑n=1∞((3a)3n+13n+1)×1(3a)=1(3a)W(3a)withw(x)=∑n=1∞x3n+13n+1⇒w′(x)=∑n=1∞x3n=11−x3−1⇒w(x)=∫0x(11−t3−1)dt+c=−x+∫0xdt(1−t3)=−x−∫0xdtt3−1letdecomposeF(t)=1t3−1=1(t−1)(t2+t+1)F(t)=at−1+bt+ct2+t+1
Commented by mathmax by abdo last updated on 18/Jan/20
a =(t−1)F(t)∣_(t=1) =(1/3) lim_(t→+∞) tF(t) =0 =a+b ⇒b =−(1/3) F(0)=−a +c =−1 ⇒c =a−1 =(1/3)−1 =−(2/3) ⇒ F(t)=(1/(3(t−1))) −(1/3)((t+2)/(t^2 +t +1)) ⇒ ∫ F(t)dt =(1/3)ln∣t−1∣−(1/6)∫((2t+1+3)/(t^2 +t+1))dt =(1/3)ln∣t−1∣−(1/6)ln(t^2 +t +1)−(1/2) ∫ (dt/(t^2 +t +1)) ∫ (dt/(t^2 +t +1)) =∫ (dt/((t+(1/2))^2 +(3/4))) =_(t+(1/2)=((√3)/2)u) (4/3) ∫ (1/(u^2 +1))×((√3)/2)du =(2/( (√3))) arctan(((2t+1)/( (√3)))) ⇒∫_0 ^x F(t)dt =[(1/3)ln∣t−1∣−(1/6)ln(t^2 +t +1)]_0 ^x −(1/( (√3)))[ arctan(((2t+1)/( (√3))))]_0 ^x =(1/3)ln∣x−1∣−(1/6)ln(x^2 +x+1) −(1/( (√3))){ arctan(((2x+1)/( (√3))))−arctan((1/( (√3)))) ⇒ w(x)=x−(1/3)ln∣x−1∣+(1/6)ln(x^2 +x+1)+(1/( (√3))){ arctan(((2x+1)/( (√3))))−(π/6)} ⇒f(a)=−(1/3)ln(1−a)−(1/((^3 (√a))))w(^3 (√a)) f(a) =−(1/3)ln(1−a)−(1/((^3 (√a)))){^3 (√a)−(1/3)ln∣^3 (√a)−1∣+(1/6)ln((^3 (√a))^2 +^3 (√a)+1) +(1/( (√3))) arctan(((2(^3 (√a))+1)/( (√3))))−(π/(6(√3)))}
a=(t−1)F(t)∣t=1=13limt→+∞tF(t)=0=a+b⇒b=−13F(0)=−a+c=−1⇒c=a−1=13−1=−23⇒F(t)=13(t−1)−13t+2t2+t+1⇒∫F(t)dt=13ln∣t−1∣−16∫2t+1+3t2+t+1dt=13ln∣t−1∣−16ln(t2+t+1)−12∫dtt2+t+1∫dtt2+t+1=∫dt(t+12)2+34=t+12=32u43∫1u2+1×32du=23arctan(2t+13)⇒∫0xF(t)dt=[13ln∣t−1∣−16ln(t2+t+1)]0x−13[arctan(2t+13)]0x=13ln∣x−1∣−16ln(x2+x+1)−13{arctan(2x+13)−arctan(13)⇒w(x)=x−13ln∣x−1∣+16ln(x2+x+1)+13{arctan(2x+13)−π6}⇒f(a)=−13ln(1−a)−1(3a)w(3a)f(a)=−13ln(1−a)−1(3a){3a−13ln∣3a−1∣+16ln((3a)2+3a+1)+13arctan(2(3a)+13)−π63}
Answered by mind is power last updated on 15/Jan/20
ln(1−ax^3 )= ln(1−ax^3 )=−Σ_(k≥1) ((a^k x^(3k) )/k) f(a)=−∫_0 ^1 Σ_(k≥1) (a^k /k).x^(3k) dx =−Σ_(k≥1) (a^k /(k(3k+1))) =−(1/3){Σ_(k≥1) (a^k /k)−Σ_(k≥1) (a^(3k) /(3k+1))} =−(1/3)Σ_(k≥1) (a^k /k)+(1/3)Σ_(k≥1) (a^(3k) /(3k+1)) =((ln(1−a))/3)+(1/(3a)).Σ_(k≥1) (a^(3k+1) /(3k+1)) g(a)=Σ_(k≥1) (a^(3k+1) /(3k+1))⇒g′(a)=Σ_(k≥1) a^(3k) =(a^3 /(1−a^3 )) g(a)=∫_0 ^a (x^3 /(1−x^3 ))dx=∫_0 ^a (−1+(1/(1−x^3 )))dx =−a+∫_0 ^a (dx/((1−x)(1+x+x^2 )))=−a+∫_0 ^a (1/3){(1/((1−x)))+((x+2)/(1+x+x^2 ))}dx =−a+(1/3)∫_0 ^a ((1/(1−x))+((x+(1/2))/(1+x+x^2 ))+(3/2).(1/((x+(1/2))^2 +(3/4))))dx =−a+((ln(1−a))/3)+(1/6)ln(1+a+a^2 )+(2/3)∫_0 ^a (dx/((((2x+1)/( (√3))))^2 +1)) g(a)=−a+((ln(1−a))/3)+((ln(1+a+a^2 ))/6)+(1/( (√3))){arctan(((2a+1)/( (√3))))−(π/6)} f(a)=((ln(1−a))/3)+(1/(3a))g(a) =−(1/3)+(1/(3a))(aln(1−a)+ln(1−a))+((ln(1+a+a^2 ))/(18a))+(1/(3a(√3))){arctan(((2a+1)/( (√3))))−(π/6)}
ln(1−ax3)=ln(1−ax3)=−∑k⩾1akx3kkf(a)=−∫01∑k⩾1akk.x3kdx=−∑k⩾1akk(3k+1)=−13{∑k⩾1akk−∑k⩾1a3k3k+1}=−13∑k⩾1akk+13∑k⩾1a3k3k+1=ln(1−a)3+13a.∑k⩾1a3k+13k+1g(a)=∑k⩾1a3k+13k+1⇒g′(a)=∑k⩾1a3k=a31−a3g(a)=∫0ax31−x3dx=∫0a(−1+11−x3)dx=−a+∫0adx(1−x)(1+x+x2)=−a+∫0a13{1(1−x)+x+21+x+x2}dx=−a+13∫0a(11−x+x+121+x+x2+32.1(x+12)2+34)dx=−a+ln(1−a)3+16ln(1+a+a2)+23∫0adx(2x+13)2+1g(a)=−a+ln(1−a)3+ln(1+a+a2)6+13{arctan(2a+13)−π6}f(a)=ln(1−a)3+13ag(a)=−13+13a(aln(1−a)+ln(1−a))+ln(1+a+a2)18a+13a3{arctan(2a+13)−π6}
Commented by msup trace by abdo last updated on 15/Jan/20
thank you sir.
thankyousir.
Commented by mind is power last updated on 15/Jan/20
y′re welcom
y′rewelcom

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