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Question Number 87527 by mathmax by abdo last updated on 04/Apr/20
find ∫_0 ^∞ ((arctan(3x))/(x^2 +x+1))dx
$${find}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{3}{x}\right)}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx} \\ $$
Commented by mathmax by abdo last updated on 05/Apr/20
let f(t) =∫_0 ^∞ ((arctan(tx))/(x^2 +x+1))dx with t>0 we have f^′ (t) =∫_0 ^∞ (x/((1+t^2 x^2 )(x^2 +x+1)))dx =_(tx=u) ∫_0 ^∞ (u/(t(1+u^2 )((u^2 /t^2 )+(u/t)+1)))(du/t) =∫_0 ^∞ ((udu)/((1+u^2 )(u^2 +tu +t^2 ))) u^2 +tu +t^2 →Δ=t^2 −4t^2 =−3t^2 <0 ⇒ F(u) =(u/((u^2 +1)(u^2 +tu +t^2 ))) =((au+b)/(u^2 +1)) +((cu +d)/(u^2 +tu +t^2 )) lim_(u→+∞) uF(u) =0 =a+c ⇒c=−a ⇒ F(u) =((au+b)/(u^2 +1)) +((−au +d)/(u^2 +tu +t^2 )) F(0) =0 =b+d ⇒d=−b ⇒F(u) =((au+b)/(u^2 +1))−((au+b)/(u^2 +tu +t^2 )) F(1) =(1/(2(t^2 +t+1))) =((a+b)/2)−((a+b)/(t^2 +t+1)) ⇒ (1/2) =((a+b)/2)(t^2 +t+1)−(a+b) ⇒1 =(a+b)(t^2 +t+1)−2a−2b ⇒ (t^2 +t−1)a +(t^2 +t−1)b =1 F(−1)=((−1)/(2(t^2 −t+1))) =((−a+b)/2) +((a+b)/(t^2 −t +1)) ⇒ −1 =(t^2 −t+1)(−a+b) +2(a+b) ⇒ 1 =(t^2 −t+1)(a−b) −2a−2b ⇒ 1 =(t^2 −t−1)a −(t^2 −t−1)b we get the system { (((t^2 +t−1)a +(t^2 +t−1)b =1)),(((t^2 −t−1)a−(t^2 −t−1)b =1)) :} Δ =−(t^2 +t−1)(t^2 −t−1)−(t^2 −t−1)(t^2 +t−1) =−2((t^2 −1)^2 −t^2 ) =−2(t^4 −2t^2 +1−t^2 )=−2(t^4 −3t^2 +1) Δ_a = determinant (((1 t^2 +t−1)),((1 −t^2 +t+1)))=−t^2 +t+1−t^2 −t +1 =−2t^2 +2 Δ_b = determinant (((t^2 +t−1 1)),((t^2 −t−1 1)))=t^2 +t−1−t^2 +t+1 =2t ⇒ a =((−2t^2 +2)/(−2(t^4 −3t^2 +1))) =((t^2 −1)/(t^4 −3t^2 +1)) b =((2t)/(−2(t^4 −3t^2 +1))) =((−t)/(t^4 −3t^2 +1)) ....be continued...
$${let}\:{f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({tx}\right)}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}{dx}\:\:\:{with}\:{t}>\mathrm{0} \\ $$$${we}\:{have}\:{f}^{'} \left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}}{\left(\mathrm{1}+{t}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)}{dx} \\ $$$$=_{{tx}={u}} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{u}}{{t}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\frac{{u}^{\mathrm{2}} }{{t}^{\mathrm{2}} }+\frac{{u}}{{t}}+\mathrm{1}\right)}\frac{{du}}{{t}}\:\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{udu}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left({u}^{\mathrm{2}} +{tu}\:+{t}^{\mathrm{2}} \right)} \\ $$$${u}^{\mathrm{2}} \:+{tu}\:+{t}^{\mathrm{2}} \:\rightarrow\Delta={t}^{\mathrm{2}} −\mathrm{4}{t}^{\mathrm{2}} =−\mathrm{3}{t}^{\mathrm{2}} <\mathrm{0}\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\frac{{u}}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\left({u}^{\mathrm{2}} \:+{tu}\:+{t}^{\mathrm{2}} \right)}\:=\frac{{au}+{b}}{{u}^{\mathrm{2}} +\mathrm{1}}\:+\frac{{cu}\:+{d}}{{u}^{\mathrm{2}} +{tu}\:+{t}^{\mathrm{2}} } \\ $$$${lim}_{{u}\rightarrow+\infty} {uF}\left({u}\right)\:=\mathrm{0}\:={a}+{c}\:\Rightarrow{c}=−{a}\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\frac{{au}+{b}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{−{au}\:+{d}}{{u}^{\mathrm{2}} \:+{tu}\:+{t}^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{0}\:={b}+{d}\:\Rightarrow{d}=−{b}\:\Rightarrow{F}\left({u}\right)\:=\frac{{au}+{b}}{{u}^{\mathrm{2}} \:+\mathrm{1}}−\frac{{au}+{b}}{{u}^{\mathrm{2}} \:+{tu}\:+{t}^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)}\:=\frac{{a}+{b}}{\mathrm{2}}−\frac{{a}+{b}}{{t}^{\mathrm{2}} \:+{t}+\mathrm{1}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{{a}+{b}}{\mathrm{2}}\left({t}^{\mathrm{2}} \:+{t}+\mathrm{1}\right)−\left({a}+{b}\right)\:\Rightarrow\mathrm{1}\:=\left({a}+{b}\right)\left({t}^{\mathrm{2}} \:+{t}+\mathrm{1}\right)−\mathrm{2}{a}−\mathrm{2}{b}\:\Rightarrow \\ $$$$\left({t}^{\mathrm{2}} \:+{t}−\mathrm{1}\right){a}\:+\left({t}^{\mathrm{2}} +{t}−\mathrm{1}\right){b}\:=\mathrm{1} \\ $$$${F}\left(−\mathrm{1}\right)=\frac{−\mathrm{1}}{\mathrm{2}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)}\:=\frac{−{a}+{b}}{\mathrm{2}}\:+\frac{{a}+{b}}{{t}^{\mathrm{2}} −{t}\:+\mathrm{1}}\:\Rightarrow \\ $$$$−\mathrm{1}\:=\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)\left(−{a}+{b}\right)\:+\mathrm{2}\left({a}+{b}\right)\:\Rightarrow \\ $$$$\mathrm{1}\:=\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)\left({a}−{b}\right)\:−\mathrm{2}{a}−\mathrm{2}{b}\:\Rightarrow \\ $$$$\mathrm{1}\:=\left({t}^{\mathrm{2}} −{t}−\mathrm{1}\right){a}\:−\left({t}^{\mathrm{2}} −{t}−\mathrm{1}\right){b}\:\:{we}\:{get}\:{the}\:{system} \\ $$$$\begin{cases}{\left({t}^{\mathrm{2}} \:+{t}−\mathrm{1}\right){a}\:+\left({t}^{\mathrm{2}} \:+{t}−\mathrm{1}\right){b}\:=\mathrm{1}}\\{\left({t}^{\mathrm{2}} −{t}−\mathrm{1}\right){a}−\left({t}^{\mathrm{2}} −{t}−\mathrm{1}\right){b}\:=\mathrm{1}}\end{cases} \\ $$$$\Delta\:=−\left({t}^{\mathrm{2}} +{t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} −{t}−\mathrm{1}\right)−\left({t}^{\mathrm{2}} \:−{t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +{t}−\mathrm{1}\right) \\ $$$$=−\mathrm{2}\left(\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} −{t}^{\mathrm{2}} \right)\:=−\mathrm{2}\left({t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}−{t}^{\mathrm{2}} \right)=−\mathrm{2}\left({t}^{\mathrm{4}} −\mathrm{3}{t}^{\mathrm{2}} \:+\mathrm{1}\right) \\ $$$$\Delta_{{a}} =\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:{t}^{\mathrm{2}} \:+{t}−\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{t}^{\mathrm{2}} \:+{t}+\mathrm{1}}\end{vmatrix}=−{t}^{\mathrm{2}} +{t}+\mathrm{1}−{t}^{\mathrm{2}} −{t}\:+\mathrm{1}\:=−\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2} \\ $$$$\Delta_{{b}} =\begin{vmatrix}{{t}^{\mathrm{2}} +{t}−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{{t}^{\mathrm{2}} −{t}−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{vmatrix}={t}^{\mathrm{2}} \:+{t}−\mathrm{1}−{t}^{\mathrm{2}} +{t}+\mathrm{1}\:=\mathrm{2}{t}\:\Rightarrow \\ $$$${a}\:=\frac{−\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{2}}{−\mathrm{2}\left({t}^{\mathrm{4}} −\mathrm{3}{t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{4}} −\mathrm{3}{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${b}\:=\frac{\mathrm{2}{t}}{−\mathrm{2}\left({t}^{\mathrm{4}} −\mathrm{3}{t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{−{t}}{{t}^{\mathrm{4}} −\mathrm{3}{t}^{\mathrm{2}} \:+\mathrm{1}}\:….{be}\:{continued}… \\ $$

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