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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 ))
letp>1calculate∫02πdt(p+cost)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 )
letputAp=∫02πdt(p+cost)2changementeit=zgiveAp=∫∣z∣=11(p+z+z−12)2dziz=∫∣z∣=1−4idzz(2p+z+z−1)2=∫∣z∣=1−4idzz(2p+z+1z)2=∫∣z∣=1−4izdz(2pz+z2+1)2=∫∣z∣=1−4izdz(z2+2pz+1)2letφ(z)=−4iz(z2+2pz+1)2polesofφ?rootsofz2+2pz+1Δ′=p2−1>0⇒z1=−p+p2−1z2=−p−p2−1φ(z)=−4iz(z−z1)2(z−z2)2∣z1∣−1=p−p2−1−1=p−1−p2−1(p−1)2−(p2−1)=p2−2p+1−p2+1=−2p+2=−2(p−1)<0⇒∣z1∣<1∣z2∣>1⇒∫∣z∣=1φ(z)dz=2iπRes(φ,z1)
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(φ,z1)=limz→z1?1(2−1)!{(z−z1)2φ(z)}(1)=limz→z1{−4iz(z−z2)2}(1)=−4ilimz→z1(z−z2)2−2(z−z2)z(z−z2)4=−4ilimz→z1(z−z2)−2z(z−z2)3=4ilimz→z1z+z2(z−z2)2=4iz1+z2(z1−z2)2=4i−2p(2p2−1)2=−2ip(p2−1)∫−∞+∞φ(z)dz=2iπ(−2ipp2−1)=4pπp2−1⇒Ap=4pπp2−1.

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