Menu Close

calculate-arctan-x-2-3-x-2-x-1-2-dx-

Tinku Tara 0 Comments
Pin




Question Number 78625 by mathmax by abdo last updated on 19/Jan/20
calculate ∫_(−∞) ^(+∞) ((arctan(x^2 −3))/((x^2 +x+1)^2 ))dx
$${calculate}\:\:\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({x}^{\mathrm{2}} −\mathrm{3}\right)}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 19/Jan/20
let W(z)=((arctan(z^2 −3))/((z^2 +z+1)^2 )) poles of W? z^2 +z+1 =0 →Δ =1−4 =−3 ⇒z_1 =((−1+i(√3))/2) =e^(i((2π)/3)) z_2 =e^(−((2iπ)/3)) ⇒W(z) =((arctan(z^2 −3))/((z−e^((i2π)/3) )^2 (z−e^(−((2iπ)/3)) )^2 )) ∫_(−∞) ^(+∞) W(z)dz =2iπRes(W,e^(i((2π)/3)) ) Res(W,e^((i2π)/3) ) =lim_(z→e^((i2π)/3) ) (1/((2−1)!)){(z−e^((i2π)/3) )^2 W(z)}^((1)) =lim_(z→e^((i2π)/3) ) { ((arctan(z^2 −3))/((z−e^(−((2iπ)/3)) )^2 ))}^((1)) =lim_(z→e^((i2π)/3) ) ((((2z)/(1+(z^2 −3)^2 ))×(z−e^(−((2iπ)/3)) )^2 −2(z−e^(−((2iπ)/3)) )arctan(z^2 −3))/((z−e^(−((2iπ)/3)) )^4 )) =lim_(z→e^((i2π)/3) ) ((((2z(z−e^(−i((2π)/3)) ))/(1+(z^2 −3)^2 ))−2arctan(z^2 −3))/((z−e^(−((2iπ)/3)) )^3 )) =((((2e^(i((2π)/3)) (2isin(((2π)/3)))/(1+(e^((i4π)/3) −3)^2 ))−2arctan(e^((i4π)/3) −3))/((2isin(((2π)/3)))^3 )) =((((2i(√3)e^(i((2π)/3)) )/(1+(e^((i4π)/3) −3)^2 ))−2arctan(e^((i4π)/3) −3))/((i(√3))^3 )) ....be continued...
$${let}\:{W}\left({z}\right)=\frac{{arctan}\left({z}^{\mathrm{2}} −\mathrm{3}\right)}{\left({z}^{\mathrm{2}} \:+{z}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{poles}\:{of}\:{W}? \\ $$$${z}^{\mathrm{2}} +{z}+\mathrm{1}\:=\mathrm{0}\:\rightarrow\Delta\:=\mathrm{1}−\mathrm{4}\:=−\mathrm{3}\:\Rightarrow{z}_{\mathrm{1}} =\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$${z}_{\mathrm{2}} ={e}^{−\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} \:\Rightarrow{W}\left({z}\right)\:=\frac{{arctan}\left({z}^{\mathrm{2}} −\mathrm{3}\right)}{\left({z}−{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \left({z}−{e}^{−\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} } \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi{Res}\left({W},{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \right) \\ $$$${Res}\left({W},{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)\:={lim}_{{z}\rightarrow{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} } \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} {W}\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} } \:\:\:\left\{\:\frac{{arctan}\left({z}^{\mathrm{2}} −\mathrm{3}\right)}{\left({z}−{e}^{−\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} } \:\:\:\frac{\frac{\mathrm{2}{z}}{\mathrm{1}+\left({z}^{\mathrm{2}} −\mathrm{3}\right)^{\mathrm{2}} }×\left({z}−{e}^{−\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} −\mathrm{2}\left({z}−{e}^{−\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} \right){arctan}\left({z}^{\mathrm{2}} −\mathrm{3}\right)}{\left({z}−{e}^{−\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} \right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} } \:\:\:\frac{\frac{\mathrm{2}{z}\left({z}−{e}^{−{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \right)}{\mathrm{1}+\left({z}^{\mathrm{2}} −\mathrm{3}\right)^{\mathrm{2}} }−\mathrm{2}{arctan}\left({z}^{\mathrm{2}} −\mathrm{3}\right)}{\left({z}−{e}^{−\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} \right)^{\mathrm{3}} } \\ $$$$=\frac{\frac{\mathrm{2}{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \left(\mathrm{2}{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right.}{\mathrm{1}+\left({e}^{\frac{{i}\mathrm{4}\pi}{\mathrm{3}}} −\mathrm{3}\right)^{\mathrm{2}} }−\mathrm{2}{arctan}\left({e}^{\frac{{i}\mathrm{4}\pi}{\mathrm{3}}} −\mathrm{3}\right)}{\left(\mathrm{2}{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)^{\mathrm{3}} } \\ $$$$=\frac{\frac{\mathrm{2}{i}\sqrt{\mathrm{3}}{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} }{\mathrm{1}+\left({e}^{\frac{{i}\mathrm{4}\pi}{\mathrm{3}}} −\mathrm{3}\right)^{\mathrm{2}} }−\mathrm{2}{arctan}\left({e}^{\frac{{i}\mathrm{4}\pi}{\mathrm{3}}} −\mathrm{3}\right)}{\left({i}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }\:\:….{be}\:{continued}… \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *