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Question Number 77995 by mathmax by abdo last updated on 12/Jan/20
calculate ∫_(−∞) ^(+∞) ((arctan(2x+1))/((x^2 +3)^2 ))dx
$${calculate}\:\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left(\mathrm{2}{x}+\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }{dx} \\ $$
Commented by msup trace by abdo last updated on 13/Jan/20
I =∫_(−∞) ^(+∞) ((arctan(2x+1))/((x^2 +3)^2 ))dx ⇒ I=_(x=(√3)t) ∫_(−∞) ^(+∞) ((arctan(2(√3)t+1))/(9(t^2 +1)^2 ))(√3)dt =((√3)/9) ∫_(−∞) ^(+∞) ((arctan(2(√3)t+1))/((t^2 +1)^2 ))dt let ϕ(z)=((arctan(2(√3)z+1))/((z^2 +1)^2 )) ⇒ ϕ(z)=((arctan(2(√3)z +1))/((z−i)^2 (z+i)^2 )) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i)=lim_(z→i) (1/((2−1)!)){(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) {((arctan(2(√3)z +1))/((z+i)^2 ))}^((1)) =lim_(z→i) ((((2(√3))/(1+(2(√3)z+1)^2 ))(z+i)^2 −2(z+i)arctan(2(√3)z +1))/((z+i)^4 )) =lim_(z→i) ((((2(√3)(z+i))/(1+(2(√3)z +1)^2 ))−2arctan(2(√3)z+1))/((z+i)^3 )) =((((4i(√3))/(1+(2(√3)i +1)^2 ))−2arctan(2(√3)i+1))/(−8i)) =−((√3)/2)×(1/(1−12+4(√3)i +1))+(1/(4i))arctan(2(√3)i+1) =((−(√3))/(2(−10+4(√3)i))) +(1/(4i))arctan(2(√3)i+1) =((√3)/(4(5−2(√3)i)))−(i/4) arctan(2(√3)i+1)
$${I}\:=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left(\mathrm{2}{x}+\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }{dx}\:\Rightarrow \\ $$$${I}=_{{x}=\sqrt{\mathrm{3}}{t}} \:\:\:\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{t}+\mathrm{1}\right)}{\mathrm{9}\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\sqrt{\mathrm{3}}{dt} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{9}}\:\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{t}+\mathrm{1}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$${let}\:\varphi\left({z}\right)=\frac{{arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{z}+\mathrm{1}\right)}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\varphi\left({z}\right)=\frac{{arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{z}\:+\mathrm{1}\right)}{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} } \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)={lim}_{{z}\rightarrow{i}} \frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\left\{\frac{{arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{z}\:+\mathrm{1}\right)}{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\frac{\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{1}+\left(\mathrm{2}\sqrt{\mathrm{3}}{z}+\mathrm{1}\right)^{\mathrm{2}} }\left({z}+{i}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{i}\right){arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{z}\:+\mathrm{1}\right)}{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\frac{\frac{\mathrm{2}\sqrt{\mathrm{3}}\left({z}+{i}\right)}{\mathrm{1}+\left(\mathrm{2}\sqrt{\mathrm{3}}{z}\:+\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{2}{arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{z}+\mathrm{1}\right)}{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\frac{\mathrm{4}{i}\sqrt{\mathrm{3}}}{\mathrm{1}+\left(\mathrm{2}\sqrt{\mathrm{3}}{i}\:+\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{2}{arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{i}+\mathrm{1}\right)}{−\mathrm{8}{i}} \\ $$$$=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{1}−\mathrm{12}+\mathrm{4}\sqrt{\mathrm{3}}{i}\:+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{4}{i}}{arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{i}+\mathrm{1}\right) \\ $$$$=\frac{−\sqrt{\mathrm{3}}}{\mathrm{2}\left(−\mathrm{10}+\mathrm{4}\sqrt{\mathrm{3}}{i}\right)}\:+\frac{\mathrm{1}}{\mathrm{4}{i}}{arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{i}+\mathrm{1}\right) \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}\left(\mathrm{5}−\mathrm{2}\sqrt{\mathrm{3}}{i}\right)}−\frac{{i}}{\mathrm{4}}\:{arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{i}+\mathrm{1}\right) \\ $$
Commented by msup trace by abdo last updated on 13/Jan/20
we have arctan(z)=(1/(2i))ln(((1+iz)/(1−iz))) ⇒ arctan(2(√3)i +1)=(1/(2i))ln(((1+i(2(√3)i+1))/(1−i(2(√3)i +1)))) =(1/(2i))ln(((1−2(√3)+i)/(1+2(√3)−i))) also 1−2(√3)+i =(√((1−2(√3))^2 +1))e^(iarctan((1/(1−2(√3))))) 1+2(√3)−i =(√((1+2(√3))^2 +1))e^(−iarctan((1/(1+2(√3))))) ⇒ln(((...)/(...)))=(1/2)ln((1−2(√3))^2 +1)+iarctan((1/(1−2(√3)))) −(1/2)ln((1+2(√3))^2 +1) +i arctan((1/(1+2(√3))))
$${we}\:{have}\:{arctan}\left({z}\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{1}+{iz}}{\mathrm{1}−{iz}}\right)\:\Rightarrow \\ $$$${arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{i}\:+\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{1}+{i}\left(\mathrm{2}\sqrt{\mathrm{3}}{i}+\mathrm{1}\right)}{\mathrm{1}−{i}\left(\mathrm{2}\sqrt{\mathrm{3}}{i}\:+\mathrm{1}\right)}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}+{i}}{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}−{i}}\right)\:{also} \\ $$$$\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}+{i}\:=\sqrt{\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{1}}{e}^{{iarctan}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}}\right)} \\ $$$$\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}−{i}\:=\sqrt{\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:+\mathrm{1}}{e}^{−{iarctan}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}}\right)} \\ $$$$\Rightarrow{ln}\left(\frac{…}{…}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{1}\right)+{iarctan}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}}\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{1}\right)\:+{i}\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}}\right) \\ $$
Commented by mathmax by abdo last updated on 13/Jan/20
⇒arctan(2(√3)i+1)=(1/(4i))ln(((1−4(√3)+12+1)/(1+4(√3)+12+1))) +(1/2){ arctan((1/(1−2(√3))))+arctan((1/(1+2(√3))))} =(1/(4i))ln(((7−2(√3))/(7+2(√3))))+(1/2){(π/2)−arctan(2(√3)+1)−((π/2)−arctan(2(√3)−1)} =(1/(4i))ln(((7−2(√3))/(7+2(√3))))+(1/2)(arctan(2(√3)−1)−arctan(2(√3)+1))
$$\Rightarrow{arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{i}+\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{4}{i}}{ln}\left(\frac{\mathrm{1}−\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{12}+\mathrm{1}}{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{12}+\mathrm{1}}\right) \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}}\right)+{arctan}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{i}}{ln}\left(\frac{\mathrm{7}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{7}+\mathrm{2}\sqrt{\mathrm{3}}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\pi}{\mathrm{2}}−{arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}\right)−\left(\frac{\pi}{\mathrm{2}}−{arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}\right)\right\}\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{i}}{ln}\left(\frac{\mathrm{7}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{7}+\mathrm{2}\sqrt{\mathrm{3}}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left({arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}\right)−{arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}\right)\right) \\ $$

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