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Question Number 33259 by prof Abdo imad last updated on 14/Apr/18
find the value of ∫_0 ^∞ ((arctan(2x))/(a^2 +x^2 )) dx with a≠0
findthevalueof∫0∞arctan(2x)a2+x2dxwitha≠0
Commented by prof Abdo imad last updated on 27/Apr/18
let put I= ∫_0 ^∞ ((arctan(2x))/(a^2 +x^2 ))dx if a>0 ch.x=at give I = ∫_0 ^∞ ((arctan(2at))/(a^2 (1+t^2 ))) adt = (1/a)∫_0 ^∞ ((arctan(2at))/(1+t^2 ))dt ⇒aI = ∫_0 ^∞ ((arctan(2at))/(1+t^2 ))dt =f(a) we have f^′ (a) = ∫_0 ^∞ ((2t)/((1+4a^2 t^2 )(1+t^2 )))dt let decompose F(t) = ((2t)/((1+4a^2 t^2 )(1+t^2 ))) = ((αt +b)/(t^2 +1)) + ((ct +d)/(4a^2 t^2 +1)) F(−t)=−F(t) ⇒((−αt+b)/(t^2 +1)) +((−ct +d)/(4a^2 t^2 +1)) =((−αt−b)/(t^2 +1)) +((−ct −d)/(4a^2 t^2 +1)) ⇒ b=d=0 ⇒ F(t)= ((αt)/(t^2 +1)) +((ct)/(4a^2 t^2 +1)) lim_(t→+∞) t F(t) =0 =α +(c/(4a^2 )) ⇒4a^2 α +c =0 ⇒ c=−4a^2 α ⇒F(t)= ((αt)/(t^2 +1)) −4a^2 ((αt)/(4a^2 t^2 +1)) F(1) = (2/((1+4a^2 )2)) =(1/(4a^2 +1)) = (α/2) −((4a^2 α)/(4a^2 +1)) ⇒ 1 =(1/2)(4a^2 +1)α −4a^2 α =(2a^2 +(1/2)−4a^2 )α =((1/2) −2a^2 )α =((1−4a^2 )/2) α ⇒α= (2/(1−4a^2 )) F(t) =(2/(1−4a^2 )) (t/(t^2 +1)) −4a^2 (2/(1−4a^2 )) (t/(4a^2 t^2 +1)) F(t) = (2/(1−4a^2 )) (t/(t^2 +1)) −((8a^2 )/(1−4a^2 )) (t/(4a^2 t^2 +1)) f^′ (a) = (1/(1−4a^2 )) ∫_0 ^∞ ((2tdt)/(t^2 +1)) −((8a^2 )/(1−4a^2 ))∫_0 ^∞ ((tdt)/(4a^2 t^2 +1)) but ∫_0 ^∞ ((tdt)/(4a^2 t^2 +1)) =(1/(8a^2 ))∫_0 ^∞ ((8a^2 t)/(4a^2 t^2 +1)) f^′ (a)= (1/(1−4a^2 ))[ln(((1+t^2 )/(4a^2 t^2 +1)))]_0 ^(+∞) =(1/(1−4a^2 ))ln((1/(4a^2 ))) = ((−ln(4a^2 ))/(1−4a^2 )) ⇒f(a) = ∫_0 ^a ((−ln(4x^2 ))/(1−4x^2 ))dx +λ but λ =f(0)=0 ⇒f(a) =−∫_0 ^(a ) ((2ln(2x))/(1−4x^2 ))dx −f(a)=_(2x=t) 2 ∫_0 ^(2a) ((ln(t))/(1−t^2 )) (dt/2) = ∫_0 ^(2a) ((ln(t))/(1−t^2 ))dt if0 <2a<1 ⇔ 0<a<(1/2) ∫_0 ^(2a) ((ln(t))/(1−t^2 ))dt = ∫_0 ^(2a) (Σ_(n=0) ^∞ t^(2n) )ln(t)dt = Σ_(n=0) ^∞ ∫_0 ^(2a) t^(2n) ln(t)dt =Σ_(n=0) ^∞ A_n A_n =∫_0 ^(2a) t^(2n) ln(t)dt be calculated by recurrence ....be continued....
letputI=∫0∞arctan(2x)a2+x2dxifa>0ch.x=atgiveI=∫0∞arctan(2at)a2(1+t2)adt=1a∫0∞arctan(2at)1+t2dt⇒aI=∫0∞arctan(2at)1+t2dt=f(a)wehavef′(a)=∫0∞2t(1+4a2t2)(1+t2)dtletdecomposeF(t)=2t(1+4a2t2)(1+t2)=αt+bt2+1+ct+d4a2t2+1F(−t)=−F(t)⇒−αt+bt2+1+−ct+d4a2t2+1=−αt−bt2+1+−ct−d4a2t2+1⇒b=d=0⇒F(t)=αtt2+1+ct4a2t2+1limt→+∞tF(t)=0=α+c4a2⇒4a2α+c=0⇒c=−4a2α⇒F(t)=αtt2+1−4a2αt4a2t2+1F(1)=2(1+4a2)2=14a2+1=α2−4a2α4a2+1⇒1=12(4a2+1)α−4a2α=(2a2+12−4a2)α=(12−2a2)α=1−4a22α⇒α=21−4a2F(t)=21−4a2tt2+1−4a221−4a2t4a2t2+1F(t)=21−4a2tt2+1−8a21−4a2t4a2t2+1f′(a)=11−4a2∫0∞2tdtt2+1−8a21−4a2∫0∞tdt4a2t2+1but∫0∞tdt4a2t2+1=18a2∫0∞8a2t4a2t2+1f′(a)=11−4a2[ln(1+t24a2t2+1)]0+∞=11−4a2ln(14a2)=−ln(4a2)1−4a2⇒f(a)=∫0a−ln(4x2)1−4x2dx+λbutλ=f(0)=0⇒f(a)=−∫0a2ln(2x)1−4x2dx−f(a)=2x=t2∫02aln(t)1−t2dt2=∫02aln(t)1−t2dtif0<2a<1⇔0<a<12∫02aln(t)1−t2dt=∫02a(∑n=0∞t2n)ln(t)dt=∑n=0∞∫02at2nln(t)dt=∑n=0∞AnAn=∫02at2nln(t)dtbecalculatedbyrecurrence….becontinued….

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