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Question Number 72012 by mathmax by abdo last updated on 23/Oct/19

find f(α) =∫_0 ^∞ ((arctan(αx))/((x^2 +3)^2 ))dx with α real.

$${find}\:{f}\left(\alpha\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\alpha{x}\right)}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }{dx}\:\:{with}\:\alpha\:{real}. \\ $$

Commented by mathmax by abdo last updated on 07/Nov/19

we have f(α)=∫_0 ^∞ ((arctan(αx))/((x^2 +3)^2 ))dx ⇒ f^′ (α) =∫_0 ^∞ (x/((1+α^2 x^2 )(x^2 +3)^2 ))dx =_(αx =t) ∫_0 ^∞ (t/(α(1+t^2 )((t^2 /α^2 ) +3)^2 ))(dt/α) =(α^4 /α^2 )∫_0 ^∞ ((tdt)/((t^2 +1)(t^2 +3α^2 )^2 )) =α^(2 ) ∫_0 ^∞ ((tdt)/((t^2 +1)(t^2 +3α^2 )^2 )) let decompose F(t) =(t/((t^2 +1)(t^2 +3α^2 )^2 )) F(t)=((at +b)/(t^(2 ) +1)) +((ct+d)/(t^2 +3α^2 )) +((et +f)/((t^2 +3α^2 )^2 )) F(−t)=F(t) ⇒((−at+b)/(t^2 +1))+((−ct+d)/(t^2 +3α^2 )) +((−et+f)/((t^2 +3α^2 )^2 )) =((−at−b)/(t^2 +1))+((−ct−d)/(t^2 +3α^2 )) +((−et−f)/((t^2 +3α^2 )^2 )) ⇒b=d=f=0 ⇒ F(t) =((at)/(t^2 +1)) +((ct)/(t^(2 ) +3α^2 )) +((et)/((t^2 +3α^2 )^2 )) lim_(t→+∞) tF(t) =0 =a+c+e ⇒e =−a−c ⇒ F(t)=((at)/(t^2 +1)) +((ct)/(t^2 +3α^2 ))−(((a+c)t)/((t^2 +3α^2 )^2 )) F(1) =(1/(2(1+3α^2 )^2 )) =(a/2) +(c/(1+3α^2 ))−((a+c)/((1+3α^2 )^2 )) ⇒ (1/2) =(1/2)(1+3α^2 )^2 a +(1+3α^2 )c−a−c ⇒ 1 =(1+3α^2 )^2 a+2(1+3α^2 )c−2a−2c ⇒ {(1+3α^2 )^2 −2}a +{2(1+3α^2 )−2}c =1 ....be continued...

$${we}\:{have}\:{f}\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}\left(\alpha{x}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }{dx}\:\Rightarrow \\ $$$${f}^{'} \left(\alpha\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{x}}{\left(\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }{dx}\:=_{\alpha{x}\:={t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{t}}{\alpha\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\frac{{t}^{\mathrm{2}} }{\alpha^{\mathrm{2}} }\:+\mathrm{3}\right)^{\mathrm{2}} }\frac{{dt}}{\alpha} \\ $$$$=\frac{\alpha^{\mathrm{4}} }{\alpha^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{tdt}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\alpha^{\mathrm{2}\:} \:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{tdt}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${let}\:{decompose}\:{F}\left({t}\right)\:=\frac{{t}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${F}\left({t}\right)=\frac{{at}\:+{b}}{{t}^{\mathrm{2}\:} +\mathrm{1}}\:+\frac{{ct}+{d}}{{t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} }\:+\frac{{et}\:+{f}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${F}\left(−{t}\right)={F}\left({t}\right)\:\Rightarrow\frac{−{at}+{b}}{{t}^{\mathrm{2}} \:+\mathrm{1}}+\frac{−{ct}+{d}}{{t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} }\:+\frac{−{et}+{f}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{−{at}−{b}}{{t}^{\mathrm{2}} \:+\mathrm{1}}+\frac{−{ct}−{d}}{{t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} } \\ $$$$+\frac{−{et}−{f}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} }\:\Rightarrow{b}={d}={f}=\mathrm{0}\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\frac{{at}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{ct}}{{t}^{\mathrm{2}\:} +\mathrm{3}\alpha^{\mathrm{2}} }\:+\frac{{et}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${lim}_{{t}\rightarrow+\infty} {tF}\left({t}\right)\:=\mathrm{0}\:={a}+{c}+{e}\:\Rightarrow{e}\:=−{a}−{c}\:\Rightarrow \\ $$$${F}\left({t}\right)=\frac{{at}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{ct}}{{t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} }−\frac{\left({a}+{c}\right){t}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{{a}}{\mathrm{2}}\:+\frac{{c}}{\mathrm{1}+\mathrm{3}\alpha^{\mathrm{2}} }−\frac{{a}+{c}}{\left(\mathrm{1}+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} {a}\:+\left(\mathrm{1}+\mathrm{3}\alpha^{\mathrm{2}} \right){c}−{a}−{c}\:\Rightarrow \\ $$$$\mathrm{1}\:=\left(\mathrm{1}+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} {a}+\mathrm{2}\left(\mathrm{1}+\mathrm{3}\alpha^{\mathrm{2}} \right){c}−\mathrm{2}{a}−\mathrm{2}{c}\:\Rightarrow \\ $$$$\left\{\left(\mathrm{1}+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\right\}{a}\:+\left\{\mathrm{2}\left(\mathrm{1}+\mathrm{3}\alpha^{\mathrm{2}} \right)−\mathrm{2}\right\}{c}\:=\mathrm{1} \\ $$$$....{be}\:{continued}... \\ $$

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