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Question Number 38113 by maxmathsup by imad last updated on 21/Jun/18
let p>1 calculate ∫_0 ^(2π) (dt/((p +cost)^2 ))
$${let}\:{p}>\mathrm{1}\:{calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\:\frac{{dt}}{\left({p}\:+{cost}\right)^{\mathrm{2}} } \\ $$
Commented by math khazana by abdo last updated on 08/Jul/18
let put A_p = ∫_0 ^(2π) (dt/((p +cost)^2 )) changement e^(it) =z give A_p = ∫_(∣z∣=1) (1/((p +((z+z^(−1) )/2))^2 )) (dz/(iz)) =∫_(∣z∣=1) ((−4i dz)/(z(2p +z +z^(−1) )^2 )) = ∫_(∣z∣=1) ((−4idz)/(z( 2p +z +(1/z))^2 )) =∫_(∣z∣=1) ((−4izdz)/((2pz +z^2 +1)^2 )) =∫_(∣z∣=1) ((−4izdz)/((z^2 +2pz +1)^2 )) let ϕ(z) = ((−4iz)/((z^2 +2pz +1)^2 )) poles of ϕ? roots of z^2 +2pz +1 Δ^′ =p^2 −1 >0 ⇒ z_1 =−p +(√(p^2 −1)) z_2 =−p −(√(p^2 −1)) ϕ(z) = ((−4iz)/((z−z_1 )^2 (z−z_2 )^2 )) ∣z_1 ∣ −1 = p−(√(p^2 −1))−1 =p−1 −(√(p^2 −1)) (p−1)^2 −(p^2 −1) =p^2 −2p +1−p^2 +1 =−2p+2 =−2(p−1)<0 ⇒∣z_1 ∣<1 ∣z_2 ∣>1 ⇒ ∫_(∣z∣=1) ϕ(z)dz =2iπ Res(ϕ,z_1 )
$${let}\:{put}\:{A}_{{p}} =\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\frac{{dt}}{\left({p}\:+{cost}\right)^{\mathrm{2}} }\:\:{changement} \\ $$$${e}^{{it}} \:={z}\:{give} \\ $$$${A}_{{p}} =\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{\mathrm{1}}{\left({p}\:+\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}\right)^{\mathrm{2}} }\:\frac{{dz}}{{iz}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\frac{−\mathrm{4}{i}\:{dz}}{{z}\left(\mathrm{2}{p}\:+{z}\:+{z}^{−\mathrm{1}} \right)^{\mathrm{2}} } \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{−\mathrm{4}{idz}}{{z}\left(\:\mathrm{2}{p}\:+{z}\:+\frac{\mathrm{1}}{{z}}\right)^{\mathrm{2}} } \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{−\mathrm{4}{izdz}}{\left(\mathrm{2}{pz}\:+{z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{−\mathrm{4}{izdz}}{\left({z}^{\mathrm{2}} \:+\mathrm{2}{pz}\:+\mathrm{1}\right)^{\mathrm{2}} }\:\:{let} \\ $$$$\varphi\left({z}\right)\:=\:\frac{−\mathrm{4}{iz}}{\left({z}^{\mathrm{2}} \:+\mathrm{2}{pz}\:+\mathrm{1}\right)^{\mathrm{2}} }\:{poles}\:{of}\:\varphi? \\ $$$${roots}\:{of}\:{z}^{\mathrm{2}} \:+\mathrm{2}{pz}\:+\mathrm{1} \\ $$$$\Delta^{'} \:={p}^{\mathrm{2}} −\mathrm{1}\:\:>\mathrm{0}\:\Rightarrow\:{z}_{\mathrm{1}} =−{p}\:+\sqrt{{p}^{\mathrm{2}} \:−\mathrm{1}} \\ $$$${z}_{\mathrm{2}} =−{p}\:−\sqrt{{p}^{\mathrm{2}} \:−\mathrm{1}} \\ $$$$\varphi\left({z}\right)\:\:\:=\:\:\frac{−\mathrm{4}{iz}}{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} \left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\mid{z}_{\mathrm{1}} \mid\:−\mathrm{1}\:=\:{p}−\sqrt{{p}^{\mathrm{2}} −\mathrm{1}}−\mathrm{1}\:={p}−\mathrm{1}\:−\sqrt{{p}^{\mathrm{2}} \:−\mathrm{1}} \\ $$$$\left({p}−\mathrm{1}\right)^{\mathrm{2}} −\left({p}^{\mathrm{2}} −\mathrm{1}\right)\:={p}^{\mathrm{2}} \:−\mathrm{2}{p}\:+\mathrm{1}−{p}^{\mathrm{2}} \:+\mathrm{1} \\ $$$$=−\mathrm{2}{p}+\mathrm{2}\:=−\mathrm{2}\left({p}−\mathrm{1}\right)<\mathrm{0}\:\Rightarrow\mid{z}_{\mathrm{1}} \mid<\mathrm{1} \\ $$$$\mid{z}_{\mathrm{2}} \mid>\mathrm{1}\:\Rightarrow\:\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right) \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 08/Jul/18
Res(ϕ,z_1 ) =lim_(z→z_1 ) ?(1/((2−1)!)) {(z−z_1 )^2 ϕ(z)}^((1)) =lim_(z→z_1 ) { ((−4iz)/((z−z_2 )^2 ))}^((1)) =−4i lim_(z→z_1 ) (((z−z_2 )^2 −2(z−z_2 )z)/((z−z_2 )^4 )) =−4i lim_(z→z_1 ) (((z−z_2 ) −2z)/((z−z_2 )^3 )) =4i lim_(z→z_1 ) ((z +z_2 )/((z−z_2 )^2 )) =4i ((z_1 +z_2 )/((z_(1 ) −z_2 )^2 )) =4i ((−2p)/((2(√(p^2 −1)))^2 )) = ((−2ip)/((p^2 −1))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (((−2ip)/(p^2 −1)))= ((4pπ)/(p^2 −1)) ⇒ A_p = ((4pπ)/(p^2 −1)) .
$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:?\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\:\left\{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} \:\varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\:\left\{\:\frac{−\mathrm{4}{iz}}{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=−\mathrm{4}{i}\:{lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \frac{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} \:−\mathrm{2}\left({z}−{z}_{\mathrm{2}} \right){z}}{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$$$=−\mathrm{4}{i}\:{lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\frac{\left({z}−{z}_{\mathrm{2}} \right)\:−\mathrm{2}{z}}{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$=\mathrm{4}{i}\:{lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\frac{{z}\:+{z}_{\mathrm{2}} }{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\mathrm{4}{i}\:\frac{{z}_{\mathrm{1}} \:+{z}_{\mathrm{2}} }{\left({z}_{\mathrm{1}\:} −{z}_{\mathrm{2}} \right)^{\mathrm{2}} }\:=\mathrm{4}{i}\:\:\frac{−\mathrm{2}{p}}{\left(\mathrm{2}\sqrt{{p}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} }\:=\:\frac{−\mathrm{2}{ip}}{\left({p}^{\mathrm{2}} \:−\mathrm{1}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\:\left(\frac{−\mathrm{2}{ip}}{{p}^{\mathrm{2}} −\mathrm{1}}\right)=\:\frac{\mathrm{4}{p}\pi}{{p}^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow \\ $$$${A}_{{p}} =\:\frac{\mathrm{4}{p}\pi}{{p}^{\mathrm{2}} \:−\mathrm{1}}\:. \\ $$

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