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Question Number 37888 by abdo mathsup 649 cc last updated on 19/Jun/18
find f(α) = ∫_0 ^1 arctan(e^(−αx) )dx with α≥0
findf(α)=∫01arctan(e−αx)dxwithα⩾0
Commented by math khazana by abdo last updated on 21/Jun/18
we have f^′ (α)=∫_0 ^1 ((−x e^(−αx) )/(1+e^(−2αx) ))dx = ∫_0 ^1 −x e^(−αx) { Σ_(n=0) ^∞ (−1)^n e^(−2nαx) } =Σ_(n=0) ^∞ (−1)^(n+1) ∫_0 ^1 x e^(−(2n+1)αx) dx but A_n =∫_0 ^1 x e^(−(2n+1)αx) dx=_((2n+1)αx=t) ∫_0 ^((2n+1)α) (t/((2n+1)α)) e^(−t) (dt/((2n+1)α)) = (1/((2n+1)^2 α^2 )) ∫_0 ^((2n+1)α) t .e^(−t) dt and by parts ∫_0 ^((2n+1)α) t.e^(−t) dt =[−t e^(−t) ]_0 ^((2n+1)α) +∫_0 ^((2n+1)α) e^(−t) dt =−(2n+1)α e^(−(2n+1)α) +[−e^(−t) ]_0 ^((2n+1)α) =−(2n+1)α e^(−(2n+1)α) +(1−e^(−(2n+1)α) ) ⇒ A_n = (1/((2n+1)^2 α^2 )){ −(2n+1)α e^(−(2n+1)α) +(1−e^(−(2n+1)α) ) =−(1/((2n+1)α)) e^(−(2n+1)α) +((1−e^(−(2n+1)α) )/((2n+1)^2 α^2 )) I =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)α)) e^(−(2n+1)α) +Σ_(n=0) ^∞ (−1)^n ((e^(−(2n+1)α) −1)/((2n+1)^2 α^2 )) ...be continued...
wehavef′(α)=∫01−xe−αx1+e−2αxdx=∫01−xe−αx{∑n=0∞(−1)ne−2nαx}=∑n=0∞(−1)n+1∫01xe−(2n+1)αxdxbutAn=∫01xe−(2n+1)αxdx=(2n+1)αx=t∫0(2n+1)αt(2n+1)αe−tdt(2n+1)α=1(2n+1)2α2∫0(2n+1)αt.e−tdtandbyparts∫0(2n+1)αt.e−tdt=[−te−t]0(2n+1)α+∫0(2n+1)αe−tdt=−(2n+1)αe−(2n+1)α+[−e−t]0(2n+1)α=−(2n+1)αe−(2n+1)α+(1−e−(2n+1)α)⇒An=1(2n+1)2α2{−(2n+1)αe−(2n+1)α+(1−e−(2n+1)α)=−1(2n+1)αe−(2n+1)α+1−e−(2n+1)α(2n+1)2α2I=∑n=0∞(−1)n(2n+1)αe−(2n+1)α+∑n=0∞(−1)ne−(2n+1)α−1(2n+1)2α2…becontinued…
Commented by math khazana by abdo last updated on 21/Jun/18
we have artan^′ (u)= (1/(1+u^2 )) =Σ_(n=0) ^∞ (−1)^n u^(2n) ⇒ arctan(u) = Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) u^(2n+1) +c (c=0)⇒ arctan(e^(−αx) ) =Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) e^(−(2n+1)αx) and f(α)= ∫_0 ^1 arctan(e^(−αx) )dx =Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) ∫_0 ^1 e^(−(2n+1)αx) dx =Σ_(n=0) ^∞ (((−1)^(n+1) )/((2n+1)^2 α))[ e^(−(2n+1)αx) ]_0 ^1 =(1/α) Σ_(n=0) ^∞ (((−1)^(n+1) )/((2n+1)^2 )){ e^(−(2n+1)α) −1} = (1/α) Σ_(n=0) ^∞ (((−1)^(n+1) )/((2n+1)^2 )) e^(−(2n+1)α) +(1/α) Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 )) ...be continued...
wehaveartan′(u)=11+u2=∑n=0∞(−1)nu2n⇒arctan(u)=∑n=0∞(−1)n2n+1u2n+1+c(c=0)⇒arctan(e−αx)=∑n=0∞(−1)n2n+1e−(2n+1)αxandf(α)=∫01arctan(e−αx)dx=∑n=0∞(−1)n2n+1∫01e−(2n+1)αxdx=∑n=0∞(−1)n+1(2n+1)2α[e−(2n+1)αx]01=1α∑n=0∞(−1)n+1(2n+1)2{e−(2n+1)α−1}=1α∑n=0∞(−1)n+1(2n+1)2e−(2n+1)α+1α∑n=0∞(−1)n(2n+1)2…becontinued…
Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jun/18
let t=e^(−αx ) dt=e^(−αx) ×−α dx dx=(dt/(−αt)) ∫_1 ^e^(−α) tan^− (t)×(dt/(−αt)) =(1/(−α))∫_1 ^e^(−α) (1/t)(t−(t^3 /3)+(t^5 /5)...)dt ((−1)/α)∫_1 ^e^(−α) (1−(t^2 /3)+(t^4 /5)..)dt =((−1)/α)×∣t−(t^3 /9)+(t^5 /(25))...)∣_1 ^e^(−α)
lett=e−αxdt=e−αx×−αdxdx=dt−αt∫1e−αtan−(t)×dt−αt=1−α∫1e−α1t(t−t33+t55…)dt−1α∫1e−α(1−t23+t45..)dt=−1α×∣t−t39+t525…)∣1e−α

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