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Question Number 53113 by maxmathsup by imad last updated on 17/Jan/19
let I =∫_(−∞) ^(+∞) ((t+1)/((t^2 −t+1)^2 ))dt find value of I .
$${let}\:{I}\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{t}+\mathrm{1}}{\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$${find}\:{value}\:{of}\:{I}\:. \\ $$
Commented by maxmathsup by imad last updated on 18/Jan/19
we have t^2 −t+1 =(t−(1/2))^2 +(3/4) so changement t−(1/2) =((√3)/2) x give I =∫_(−∞) ^(+∞) (((1/2)+((√3)/2)x +1)/(((3/4))^2 (x^2 +1)^2 )) ((√3)/2) dx =((16)/9) ((√3)/4) ∫_(−∞) ^(+∞) ((3+(√3)x)/((x^2 +1)^2 ))dx =((4(√3))/9) ∫_(−∞) ^(+∞) ((3+(√3)x)/((x^2 +1)^2 ))dx let ϕ(z) =((3 +(√3)z)/((z^2 +1)^2 )) we have ϕ(z) =((3+(√3)z)/((z−i)^2 (z+i)^2 )) so the ples of ϕ are +^− i(doubles) and residus theorem give ∫_(−∞) ^(+∞) 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) { ((3+(√3)z)/((z+i)^2 ))}^((1)) =lim_(z→i) (((√3)(z+i)^2 −2(z+i)(3+(√3)z))/((z+i)^4 )) =lim_(z→i) (((√3)(z+i)−2(3+(√3)z))/((z+i)^3 )) =((2(√3)i−2(3+(√3)i))/(−8i)) =((−6)/(−8i)) =(3/(4i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (3/(4i)) =((3π)/2) ⇒ I =((4(√3))/9) ((3π)/2) =((2π(√3))/3) .
$${we}\:{have}\:{t}^{\mathrm{2}} −{t}+\mathrm{1}\:=\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\:\:{so}\:{changement}\:{t}−\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{x}\:{give} \\ $$$${I}\:=\int_{−\infty} ^{+\infty} \:\:\frac{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}\:+\mathrm{1}}{\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{dx}\:=\frac{\mathrm{16}}{\mathrm{9}}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{\mathrm{3}+\sqrt{\mathrm{3}}{x}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\:\int_{−\infty} ^{+\infty} \:\:\frac{\mathrm{3}+\sqrt{\mathrm{3}}{x}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:\:{let}\:\varphi\left({z}\right)\:=\frac{\mathrm{3}\:+\sqrt{\mathrm{3}}{z}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\:{we}\:{have}\: \\ $$$$\varphi\left({z}\right)\:=\frac{\mathrm{3}+\sqrt{\mathrm{3}}{z}}{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} }\:\:{so}\:{the}\:{ples}\:{of}\:\varphi\:{are}\:\overset{−} {+}{i}\left({doubles}\right)\:{and}\:{residus}\:{theorem} \\ $$$${give}\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} ={lim}_{{z}\rightarrow{i}} \left\{\:\frac{\mathrm{3}+\sqrt{\mathrm{3}}{z}}{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \frac{\sqrt{\mathrm{3}}\left({z}+{i}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{i}\right)\left(\mathrm{3}+\sqrt{\mathrm{3}}{z}\right)}{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\sqrt{\mathrm{3}}\left({z}+{i}\right)−\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{3}}{z}\right)}{\left({z}+{i}\right)^{\mathrm{3}} }\:=\frac{\mathrm{2}\sqrt{\mathrm{3}}{i}−\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{3}}{i}\right)}{−\mathrm{8}{i}}\:=\frac{−\mathrm{6}}{−\mathrm{8}{i}}\:=\frac{\mathrm{3}}{\mathrm{4}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\mathrm{3}}{\mathrm{4}{i}}\:=\frac{\mathrm{3}\pi}{\mathrm{2}}\:\Rightarrow\:{I}\:=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\:\frac{\mathrm{3}\pi}{\mathrm{2}}\:=\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{3}}\:. \\ $$

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