Question Number 33259 by prof Abdo imad last updated on 14/Apr/18
$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left(\mathrm{2}{x}\right)}{{a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }\:{dx}\:{with}\:{a}\neq\mathrm{0} \\ $$
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....](https://www.tinkutara.com/question/Q33906.png)
$${let}\:{put}\:\:{I}=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{2}{x}\right)}{{a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }{dx} \\ $$$${if}\:{a}>\mathrm{0}\:\:{ch}.{x}={at}\:{give}\: \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{2}{at}\right)}{{a}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:{adt}\:=\:\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left(\mathrm{2}{at}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\Rightarrow{aI}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{2}{at}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:={f}\left({a}\right)\:{we}\:{have} \\ $$$${f}^{'} \left({a}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\:\frac{\mathrm{2}{t}}{\left(\mathrm{1}+\mathrm{4}{a}^{\mathrm{2}} {t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}\:\:{let}\:{decompose} \\ $$$${F}\left({t}\right)\:=\:\:\frac{\mathrm{2}{t}}{\left(\mathrm{1}+\mathrm{4}{a}^{\mathrm{2}} {t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:=\:\frac{\alpha{t}\:+{b}}{{t}^{\mathrm{2}} +\mathrm{1}}\:+\:\frac{{ct}\:+{d}}{\mathrm{4}{a}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left(−{t}\right)=−{F}\left({t}\right)\:\Rightarrow\frac{−\alpha{t}+{b}}{{t}^{\mathrm{2}} +\mathrm{1}}\:+\frac{−{ct}\:+{d}}{\mathrm{4}{a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\frac{−\alpha{t}−{b}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{−{ct}\:−{d}}{\mathrm{4}{a}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\:{b}={d}=\mathrm{0}\:\Rightarrow \\ $$$${F}\left({t}\right)=\:\frac{\alpha{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\:\:+\frac{{ct}}{\mathrm{4}{a}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${lim}_{{t}\rightarrow+\infty} {t}\:{F}\left({t}\right)\:=\mathrm{0}\:=\alpha\:+\frac{{c}}{\mathrm{4}{a}^{\mathrm{2}} }\:\Rightarrow\mathrm{4}{a}^{\mathrm{2}} \alpha\:+{c}\:=\mathrm{0}\:\Rightarrow \\ $$$${c}=−\mathrm{4}{a}^{\mathrm{2}} \alpha\:\Rightarrow{F}\left({t}\right)=\:\frac{\alpha{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\:−\mathrm{4}{a}^{\mathrm{2}} \:\:\frac{\alpha{t}}{\mathrm{4}{a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}} \\ $$$${F}\left(\mathrm{1}\right)\:=\:\frac{\mathrm{2}}{\left(\mathrm{1}+\mathrm{4}{a}^{\mathrm{2}} \right)\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{4}{a}^{\mathrm{2}} +\mathrm{1}}\:=\:\frac{\alpha}{\mathrm{2}}\:−\frac{\mathrm{4}{a}^{\mathrm{2}} \alpha}{\mathrm{4}{a}^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}{a}^{\mathrm{2}} +\mathrm{1}\right)\alpha\:−\mathrm{4}{a}^{\mathrm{2}} \alpha\:=\left(\mathrm{2}{a}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{4}{a}^{\mathrm{2}} \right)\alpha \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{2}}\:−\mathrm{2}{a}^{\mathrm{2}} \right)\alpha\:=\frac{\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{2}}\:\alpha\:\Rightarrow\alpha=\:\frac{\mathrm{2}}{\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} } \\ $$$${F}\left({t}\right)\:=\frac{\mathrm{2}}{\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} }\:\frac{{t}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:−\mathrm{4}{a}^{\mathrm{2}} \:\frac{\mathrm{2}}{\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} }\:\:\frac{{t}}{\mathrm{4}{a}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left({t}\right)\:=\:\frac{\mathrm{2}}{\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} }\:\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\:−\frac{\mathrm{8}{a}^{\mathrm{2}} }{\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} }\:\:\frac{{t}}{\mathrm{4}{a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}} \\ $$$${f}^{'} \left({a}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}{tdt}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\:−\frac{\mathrm{8}{a}^{\mathrm{2}} }{\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \:\frac{{tdt}}{\mathrm{4}{a}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${but}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{tdt}}{\mathrm{4}{a}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{8}{a}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{8}{a}^{\mathrm{2}} {t}}{\mathrm{4}{a}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${f}^{'} \left({a}\right)=\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} }\left[{ln}\left(\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}}\right)\right]_{\mathrm{0}} ^{+\infty} \:=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} }{ln}\left(\frac{\mathrm{1}}{\mathrm{4}{a}^{\mathrm{2}} }\right) \\ $$$$=\:\frac{−{ln}\left(\mathrm{4}{a}^{\mathrm{2}} \right)}{\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} }\:\Rightarrow{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{{a}} \:\:\frac{−{ln}\left(\mathrm{4}{x}^{\mathrm{2}} \right)}{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }{dx}\:+\lambda \\ $$$${but}\:\lambda\:={f}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow{f}\left({a}\right)\:=−\int_{\mathrm{0}} ^{{a}\:} \frac{\mathrm{2}{ln}\left(\mathrm{2}{x}\right)}{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }{dx} \\ $$$$−{f}\left({a}\right)=_{\mathrm{2}{x}={t}} \:\:\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{2}{a}} \:\:\:\frac{{ln}\left({t}\right)}{\mathrm{1}−{t}^{\mathrm{2}} }\:\frac{{dt}}{\mathrm{2}}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}{a}} \:\:\frac{{ln}\left({t}\right)}{\mathrm{1}−{t}^{\mathrm{2}} }{dt} \\ $$$${if}\mathrm{0}\:<\mathrm{2}{a}<\mathrm{1}\:\Leftrightarrow\:\mathrm{0}<{a}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}{a}} \:\:\frac{{ln}\left({t}\right)}{\mathrm{1}−{t}^{\mathrm{2}} }{dt}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}{a}} \left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{t}^{\mathrm{2}{n}} \right){ln}\left({t}\right){dt} \\ $$$$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\int_{\mathrm{0}} ^{\mathrm{2}{a}} \:{t}^{\mathrm{2}{n}} \:{ln}\left({t}\right){dt}\:\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{A}_{{n}} \\ $$$${A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{2}{a}} \:{t}^{\mathrm{2}{n}} {ln}\left({t}\right){dt}\:\:{be}\:{calculated}\:{by}\:{recurrence} \\ $$$$….{be}\:{continued}…. \\ $$$$ \\ $$