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Question Number 36435 by prof Abdo imad last updated on 02/Jun/18
find the value of h(t)=∫_0 ^1 ln(1+tx^2 ) with ∣t∣≤1 2) calculate ∫_0 ^1 ln(1+x^2 )dx 3) calculate ∫_0 ^1 ln(1−x^2 )dx
findthevalueofh(t)=∫01ln(1+tx2)with∣t∣⩽12)calculate∫01ln(1+x2)dx3)calculate∫01ln(1−x2)dx
Commented by abdo.msup.com last updated on 03/Jun/18
another method but easy ∫_0 ^1 ln(1−x^2 )dx = ∫_0 ^1 ln(1−x)dx + ∫_0 ^1 ln(1+x)dx but ∫_0 ^1 ln(1−x)dx =_(1−x=t) −∫_1 ^0 lnt dt = ∫_0 ^1 ln(t)dt = [tln(t)]_0 ^1 =0 ∫_0 ^1 ln(1+x)dx =_(1+x=t) ∫_1 ^2 ln(t)dt =[ tln(t)]_1 ^2 =2ln(2) so ∫_0 ^1 ln(1−x^2 )dx =2ln(2) .
anothermethodbuteasy∫01ln(1−x2)dx=∫01ln(1−x)dx+∫01ln(1+x)dxbut∫01ln(1−x)dx=1−x=t−∫10lntdt=∫01ln(t)dt=[tln(t)]01=0∫01ln(1+x)dx=1+x=t∫12ln(t)dt=[tln(t)]12=2ln(2)so∫01ln(1−x2)dx=2ln(2).
Commented by abdo.msup.com last updated on 03/Jun/18
1)we have ∣tx^2 ∣≤1 ⇒ln(1+tx^2 ) =ln(1−(−tx^2 )) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)(tx^2 )^n =Σ_(n=1) ^∞ (((−1)^n )/n) t^(n−1) x^(2n) ⇒ h(t) = Σ_(n=1) ^∞ (((−1)^(n−1) t^n )/n) ∫_0 ^1 x^(2n) dx =Σ_(n=1) ^∞ (((−1)^(n−1) t^n )/(n(2n+1))) (1/2)h(t) =Σ_(n=1) ^∞ { (1/(2n))−(1/(2n+1))}(−1)^(n−1) t^n =(1/2) Σ_(n=1) ^∞ (((−1)^(n−1) )/n) t^n − Σ_(n=1) ^∞ (((−1)^(n−1) )/(2n+1)) t^n but Σ_(n=1) ^∞ (((−1)^(n−1) )/n) t^n =ln(1+t) Σ_(n=1) ^∞ (((−1)^(n−1) )/(2n+1)) t^n =(1/( (√t))) Σ_(n=1) ^∞ (((−1)^(n−1) )/(2n+1))((√t))^(2n+1) =(1/( (√t))) w((√t)) with w(x) =Σ_(n=1) ^∞ (((−1)^(n−1) )/(2n+1))x^(2n+1) w^′ (x) = Σ_(n=1) ^∞ (−1)^(n−1) x^(2n) = Σ_(n=0) ^∞ (−1)^n x^(2n +2) = x^2 (1/(1+x^2 )) ⇒ w(x) = ∫ (x^2 /(1+x^2 )) dx +λ =x −arctanx +λ but λ =w(0)=0 ⇒ w(x)=x −arctan(x) and Σ_(n=1) ^∞ (((−1)^(n−1) )/(2n+1)) t^n = (1/( (√t)))w((√t)) =(1/( (√t))){ (√t) −arctan((√t)) so ((h(t))/2) = (1/2)ln(1+t) −1 +((arctan((√t)))/( (√t))) ⇒ h(t) =ln(1+t) +((2arctan((√t)))/( (√t))) −2
1)wehave∣tx2∣⩽1⇒ln(1+tx2)=ln(1−(−tx2))=∑n=1∞(−1)n−1n(tx2)n=∑n=1∞(−1)nntn−1x2n⇒h(t)=∑n=1∞(−1)n−1tnn∫01x2ndx=∑n=1∞(−1)n−1tnn(2n+1)12h(t)=∑n=1∞{12n−12n+1}(−1)n−1tn=12∑n=1∞(−1)n−1ntn−∑n=1∞(−1)n−12n+1tnbut∑n=1∞(−1)n−1ntn=ln(1+t)∑n=1∞(−1)n−12n+1tn=1t∑n=1∞(−1)n−12n+1(t)2n+1=1tw(t)withw(x)=∑n=1∞(−1)n−12n+1x2n+1w′(x)=∑n=1∞(−1)n−1x2n=∑n=0∞(−1)nx2n+2=x211+x2⇒w(x)=∫x21+x2dx+λ=x−arctanx+λbutλ=w(0)=0⇒w(x)=x−arctan(x)and∑n=1∞(−1)n−12n+1tn=1tw(t)=1t{t−arctan(t)soh(t)2=12ln(1+t)−1+arctan(t)t⇒h(t)=ln(1+t)+2arctan(t)t−2
Commented by abdo.msup.com last updated on 03/Jun/18
we have proved that ∫_0 ^1 ln(1+tx^2 )dx = ln(1+t) +((2arctan((√t)))/( (√t))) −2 for t =1 we get ∫_0 ^1 ln(1+x^2 )dx =ln(2) + 2 (π/4) −2 =ln(2) +(π/2) −2 .
wehaveprovedthat∫01ln(1+tx2)dx=ln(1+t)+2arctan(t)t−2fort=1weget∫01ln(1+x2)dx=ln(2)+2π4−2=ln(2)+π2−2.
Commented by abdo.msup.com last updated on 03/Jun/18
3) we have ln^′ (1−u) = ((−1)/(1−u)) =−Σ_(n=0) ^∞ u^n ⇒ ln(1−u) =−Σ_(n=0) ^∞ (u^(n+1) /(n+1)) +c =− Σ_(n=1) ^∞ (u^n /n) =Σ_(n=1) ^∞ (u^(n−1) /n) (c=0)⇒ ln(1−x^2 ) = Σ_(n=1) ^∞ (x^(2n−2) /n) ⇒ ∫_0 ^1 ln(1−x^2 )dx = Σ_(n=1) ^∞ (1/(n(2n−1))) =A (A/2) = Σ_(n=1) ^∞ (1/(2n(2n−1))) =Σ_(n=1) ^∞ ( (1/(2n−1)) −(1/(2n)))= lim_(n→+∞) Σ_(k=1) ^n ((1/(2k−1)) −(1/(2k))) but Σ_(k=1) ^n (1/(2k−1)) = 1 +(1/3) +(1/5) +... +(1/(2n−1)) =1 +(1/2) +(1/3) +(1/4) +......+(1/(2n−1)) +(1/(2n)) −(1/2) −(1/4) −....−(1/(2n)) = H_(2n) −(1/2) H_n Σ_(k=1) ^n (1/(2k−1)) −(1/2) Σ_(k=1) ^∞ (1/k) = H_(2n) −(1/2) H_n −(1/2) H_n = H_(2n) −H_n =ln(2n)+γ +o((1/n)) −ln(n)−δ −o((1/n)) =ln(((2n)/n)) +o((1/n)) →ln(2) (n→+∞) so so ∫_0 ^1 ln(1−x^2 )dx =2ln2 .
3)wehaveln′(1−u)=−11−u=−∑n=0∞un⇒ln(1−u)=−∑n=0∞un+1n+1+c=−∑n=1∞unn=∑n=1∞un−1n(c=0)⇒ln(1−x2)=∑n=1∞x2n−2n⇒∫01ln(1−x2)dx=∑n=1∞1n(2n−1)=AA2=∑n=1∞12n(2n−1)=∑n=1∞(12n−1−12n)=limn→+∞∑k=1n(12k−1−12k)but∑k=1n12k−1=1+13+15+…+12n−1=1+12+13+14+……+12n−1+12n−12−14−….−12n=H2n−12Hn∑k=1n12k−1−12∑k=1∞1k=H2n−12Hn−12Hn=H2n−Hn=ln(2n)+γ+o(1n)−ln(n)−δ−o(1n)=ln(2nn)+o(1n)→ln(2)(n→+∞)soso∫01ln(1−x2)dx=2ln2.

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