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Question Number 73337 by mathmax by abdo last updated on 10/Nov/19
calculate ∫_0 ^∞ ((cos(artan(2x)))/((3+x^2 )^2 ))dx
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left({artan}\left(\mathrm{2}{x}\right)\right)}{\left(\mathrm{3}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 11/Nov/19
let I =∫_0 ^∞ ((cos(arctan(2x)))/((x^2 +3)^2 ))dx ⇒I =_(x=(√3)t) ∫_0 ^∞ ((cos(arctan(2(√3)t)))/(9(t^2 +1)^2 ))(√3)dt ⇒2 I =((√3)/9)∫_(−∞) ^(+∞) ((cos(arctan(2(√3)t))/((t^2 +1)^2 ))dt =((√3)/9) Re(∫_(−∞) ^(+∞) (e^(iarctan(2(√3)t)) /((t^2 +1)^2 ))dt) let W(z)=(e^(iarctan(2(√3)z)) /((z^2 +1)^2 )) ⇒W(z)=(e^(iarctan(2(√3)z)) /((z−i)^2 (z+i)^2 )) and ∫_(−∞) ^(+∞) W(z)dz =2iπ Res(W,i) Res(W,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 W(z)}^((1)) =lim_(z→i) {(e^(iarctan(2(√3)z)) /((z+i)^2 ))}^((1)) =lim_(z→i) ((i((2(√3))/(1+12z^2 ))(z+i)^2 e^(iarctan(2(√3)z)) −2(z+i)e^(iarctan(2(√3)z)) )/((z+i)^4 )) =lim_(z→i) (({((2i(√3)(z+i))/(1+12z^2 ))−2}e^(iarctan(2(√3)z)) )/((z+i)^3 )) =(((((−4(√3))/(−11))−2}e^(i arctan(2(√3)i)) )/(−8i)) =(((4(√3)+22)e^(i arctan(2(√3)i)) )/(88i)) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ×(((4(√3)+22)e^(i arctan(2(√3)i)) )/(88i)) =(π/(44)){ cos(arctan(2(√3)i)+isin(arctan(2(√3)i))} ⇒ I =((√3)/(18))×(π/(44)) cos(arctan(2(√3)i)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({arctan}\left(\mathrm{2}{x}\right)\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }{dx}\:\Rightarrow{I}\:=_{{x}=\sqrt{\mathrm{3}}{t}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left({arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{t}\right)\right)}{\mathrm{9}\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\sqrt{\mathrm{3}}{dt} \\ $$$$\Rightarrow\mathrm{2}\:{I}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{9}}\int_{−\infty} ^{+\infty} \:\frac{{cos}\left({arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{t}\right)\right.}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{9}}\:{Re}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{iarctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{t}\right)} }{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\right) \\ $$$${let}\:{W}\left({z}\right)=\frac{{e}^{{iarctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{z}\right)} }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{W}\left({z}\right)=\frac{{e}^{{iarctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{z}\right)} }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} }\:{and} \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},{i}\right) \\ $$$${Res}\left({W},{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{2}} {W}\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\:\:\left\{\frac{{e}^{{iarctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{z}\right)} }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\:\frac{{i}\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{1}+\mathrm{12}{z}^{\mathrm{2}} }\left({z}+{i}\right)^{\mathrm{2}} {e}^{{iarctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{z}\right)} −\mathrm{2}\left({z}+{i}\right){e}^{{iarctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{z}\right)} }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\:\:\:\:\:\frac{\left\{\frac{\mathrm{2}{i}\sqrt{\mathrm{3}}\left({z}+{i}\right)}{\mathrm{1}+\mathrm{12}{z}^{\mathrm{2}} }−\mathrm{2}\right\}{e}^{{iarctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{z}\right)} }{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left(\frac{−\mathrm{4}\sqrt{\mathrm{3}}}{−\mathrm{11}}−\mathrm{2}\right\}{e}^{{i}\:{arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{i}\right)} }{−\mathrm{8}{i}}\:=\frac{\left(\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{22}\right){e}^{{i}\:{arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{i}\right)} }{\mathrm{88}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi×\frac{\left(\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{22}\right){e}^{{i}\:{arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{i}\right)} }{\mathrm{88}{i}} \\ $$$$=\frac{\pi}{\mathrm{44}}\left\{\:{cos}\left({arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{i}\right)+{isin}\left({arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{i}\right)\right)\right\}\:\Rightarrow\right. \\ $$$${I}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{18}}×\frac{\pi}{\mathrm{44}}\:{cos}\left({arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{i}\right)\right. \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 11/Nov/19
forgive we have ∫_(−∞) ^(+∞) W(z)dz =(π/(44))(4(√3)+22) e^(iarctan(2(√3)i)) ⇒ I =((√3)/(18))×(π/(22))(2(√3)+11)cos(arctan(2(√3)i)
$${forgive}\:\:{we}\:{have}\:\int_{−\infty} ^{+\infty} {W}\left({z}\right){dz}\:=\frac{\pi}{\mathrm{44}}\left(\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{22}\right)\:{e}^{{iarctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{i}\right)} \:\Rightarrow \\ $$$${I}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{18}}×\frac{\pi}{\mathrm{22}}\left(\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{11}\right){cos}\left({arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{i}\right)\right. \\ $$
Commented by mathmax by abdo last updated on 11/Nov/19
we have arctan(z)=(1/(2i))ln(((1+iz)/(1−iz))) ⇒arctan(2(√3)i) =(1/(2i))ln(((1−2(√3))/(1+2(√3)))) =(1/(2i))ln(−((2(√3)−1)/(2(√3)+1))) =(1/(2i))ln(−1)+(1/(2i))ln(((2(√3)−1)/(2(√3)+1))) =(1/(2i))(iπ) +(1/(2i))ln(((2(√3)−1)/(2(√3)+1))) =(π/2) +(1/(2i))ln(((2(√3)−1)/(2(√3)+1))) ⇒cos(arctan(2(√3)i))=−sin(−(i/2)ln(((2(√3)−1)/(2(√3)+1))) =sin((i/2)ln(((2(√3)−1)/(2(√3)+1)))) =sh(−(1/2)ln(((2(√3)−1)/(2(√3)+1)))) =((e^(−(1/2)ln(((2(√3)−1)/(2(√3)+1)))) −e^((1/2)ln(((2(√3)−1)/(2(√3)+1)))) )/2) =−(1/2){(√((2(√3)−1)/(2(√3)+1)))−(1/( (√((2(√3)−1)/(2(√3)+1)))))} ⇒ I=((π(√3))/(18×22))(2(√3)+11){(((2(√3)−1)/(2(√3)+1)))^(−(1/2)) −(((2(√3)−1)/(2(√3)+1)))^(1/2) }
$${we}\:{have}\:{arctan}\left({z}\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{1}+{iz}}{\mathrm{1}−{iz}}\right)\:\Rightarrow{arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{i}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(−\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(−\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left({i}\pi\right)\:+\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}}\right)\:\Rightarrow{cos}\left({arctan}\left(\mathrm{2}\sqrt{\mathrm{3}}{i}\right)\right)=−{sin}\left(−\frac{{i}}{\mathrm{2}}{ln}\left(\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}}\right)\right. \\ $$$$={sin}\left(\frac{{i}}{\mathrm{2}}{ln}\left(\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}}\right)\right)\:={sh}\left(−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}}\right)\right) \\ $$$$=\frac{{e}^{−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}}\right)} −{e}^{\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}}\right)} }{\mathrm{2}}\:=−\frac{\mathrm{1}}{\mathrm{2}}\left\{\sqrt{\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}}}−\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}}}}\right\}\:\Rightarrow \\ $$$${I}=\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{18}×\mathrm{22}}\left(\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{11}\right)\left\{\left(\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} −\left(\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right\} \\ $$

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