let p>1 calculate ∫_0 ^(2π) (dt/((p +cost)^2 ))


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Question Number 38113 by maxmathsup by imad last updated on 21/Jun/18

$l e t p > 1 c a l c u l a t e \int_{0}^{2 π} \frac{d t}{{(p + c o s t)}^{2}}$
Commented by math khazana by abdo last updated on 08/Jul/18

$l e t p u t A_{p} = \int_{0}^{2 π} \frac{d t}{{(p + c o s t)}^{2}} c h a n g e m e n t$ $e^{i t} = z g i v e$ $A_{p} = \int_{∣ z ∣= 1} \frac{1}{{(p + \frac{z + z^{- 1}}{2})}^{2}} \frac{d z}{i z}$ $= \int_{∣ z ∣= 1} \frac{- 4 i d z}{z {(2 p + z + z^{- 1})}^{2}}$ $= \int_{∣ z ∣= 1} \frac{- 4 i d z}{z {(2 p + z + \frac{1}{z})}^{2}}$ $= \int_{∣ z ∣= 1} \frac{- 4 i z d z}{{(2 p z + z^{2} + 1)}^{2}}$ $= \int_{∣ z ∣= 1} \frac{- 4 i z d z}{{(z^{2} + 2 p z + 1)}^{2}} l e t$ $φ (z) = \frac{- 4 i z}{{(z^{2} + 2 p z + 1)}^{2}} p o l e s o f φ ?$ $r o o t s o f z^{2} + 2 p z + 1$ $Δ^{^{'}} = p^{2} - 1 > 0 \Rightarrow z_{1} = - p + \sqrt{p^{2} - 1}$ $z_{2} = - p - \sqrt{p^{2} - 1}$ $φ (z) = \frac{- 4 i z}{{(z - z_{1})}^{2} {(z - z_{2})}^{2}}$ $∣ z_{1} ∣ - 1 = p - \sqrt{p^{2} - 1} - 1 = p - 1 - \sqrt{p^{2} - 1}$ ${(p - 1)}^{2} - (p^{2} - 1) = p^{2} - 2 p + 1 - p^{2} + 1$ $= - 2 p + 2 = - 2 (p - 1) < 0 \Rightarrow∣ z_{1} ∣< 1$ $∣ z_{2} ∣> 1 \Rightarrow \int_{∣ z ∣= 1} φ (z) d z = 2 i π R e s (φ, z_{1})$
Commented by math khazana by abdo last updated on 08/Jul/18
$Res(ϕ,z_1 ) =lim_(z→z_1 ) ?(1/((2−1)!)) {(z−z_1 )^2 ϕ(z)}^((1)) =lim_(z→z_1 ) { ((−4iz)/((z−z_2 )^2 ))}^((1)) =−4i lim_(z→z_1 ) (((z−z_2 )^2 −2(z−z_2 )z)/((z−z_2 )^4 )) =−4i lim_(z→z_1 ) (((z−z_2 ) −2z)/((z−z_2 )^3 )) =4i lim_(z→z_1 ) ((z +z_2 )/((z−z_2 )^2 )) =4i ((z_1 +z_2 )/((z_(1 ) −z_2 )^2 )) =4i ((−2p)/((2(√(p^2 −1)))^2 )) = ((−2ip)/((p^2 −1))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (((−2ip)/(p^2 −1)))= ((4pπ)/(p^2 −1)) ⇒ A_p = ((4pπ)/(p^2 −1)) .$
$R e s (φ, z_{1}) = {l i m}_{z \to z_{1}} ? \frac{1}{(2 - 1)!} {{(z - z_{1})}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to z_{1}} {\frac{- 4 i z}{{(z - z_{2})}^{2}}}^{(1)}$ $= - 4 i {l i m}_{z \to z_{1}} \frac{{(z - z_{2})}^{2} - 2 (z - z_{2}) z}{{(z - z_{2})}^{4}}$ $= - 4 i {l i m}_{z \to z_{1}} \frac{(z - z_{2}) - 2 z}{{(z - z_{2})}^{3}}$ $= 4 i {l i m}_{z \to z_{1}} \frac{z + z_{2}}{{(z - z_{2})}^{2}}$ $= 4 i \frac{z_{1} + z_{2}}{{(z_{1} - z_{2})}^{2}} = 4 i \frac{- 2 p}{{(2 \sqrt{p^{2} - 1})}^{2}} = \frac{- 2 i p}{(p^{2} - 1)}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π (\frac{- 2 i p}{p^{2} - 1}) = \frac{4 p π}{p^{2} - 1} \Rightarrow$ $A_{p} = \frac{4 p π}{p^{2} - 1} .$
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