let I_n = ∫_0 ^(2π) (dx/((p +cost)^n )) with p>1 find the value of I


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Question Number 38114 by maxmathsup by imad last updated on 21/Jun/18

$l e t I_{n} = \int_{0}^{2 π} \frac{d x}{{(p + c o s t)}^{n}} w i t h p > 1$ $f i n d t h e v a l u e o f I_{n}$
Commented by abdo mathsup 649 cc last updated on 08/Jul/18
$changement e^(it) =z give I_n = ∫_(∣z∣=1) (1/((p +((z+z^(−1) )/2))^n )) (dz/(iz)) = ∫_(∣z∣=1) ((−i 2^n )/((2p +z+z^(−1) )^n )) (dz/z) = ∫_(∣z∣=1) ((−i 2^n )/(z{2p +z +(1/z)}^n ))dz = ∫_(∣z∣=1) ((−i2^n z^(n−1) )/({2pz +z^2 +1}^n ))dz let ϕ(z) = ((−i 2^n z^(n−1) )/((z^2 +2pz +1)^n )) poles of ϕ? z^2 +2pz +1=0 Δ^′ =p^2 −1>0 ⇒ real roots z_1 =−p +(√(p^2 −1 )) and z_2 =−p−(√(p^2 −1)) we have proved that ∣z_1 ∣<1 and ∣z_2 ∣>1 so ∫_(∣z∣=1) ^ ϕ(z)dz =2iπ Res(ϕ,z_1 ) we?have ϕ(z) = ((−i 2^n z^(n−1) )/((z−z_1 )^n (z−z_2 )^n )) ⇒ Res(ϕ,z_1 ) =lim_(z→z_1 ) (1/((n−1)!)){ (z−z_1 )^n ϕ(z)}^((n−1)) =−i 2^n lim_(z→z_1 ) (1/((n−1)!)){ (z^(n−1) /((z−z_2 )^n ))}^()n−1)) but { z^(n−1) (z−z_2 )^(−n) }^((n−1)) = Σ_(k=0) ^(n−1 ) C_(n−1) ^(k ) {(z−z_2 )^(−n) }^((k)) (z^(n−1) )^((n−1−k)) we have (z−z_2 )^(−n) }^((1)) =−n(z−z_2 )^(−n−1) {(z−z_2 )^(−n) }^((2)) =(−1)^2 n(n+1)(z−z_2 )^(−n−2) { (z−z_2 )^(−n) }^((k)) =(−1)^k n(n+1)...(n+k−1)(z−z_2 )^(−n−k) also (z^n )^((p)) =n(n−1)...(n−p+1)z^(n−p) if p≤n⇒ (z^(n−1) )^((n−1−k)) =(n−1)(n−2)...(n−1−n+1+k +1)z^(n−1−n+1+k) =(n−1)(n−2).....(k+1) z^k =(((n−1)!)/(k!)) z^k ⇒ {z^(n−1) (z−z_2 )^(−n) }^((n−1)) = =Σ_(k=0) ^(n−1) C_(n−1) ^k (−1)^k n(n+1)...(n+k−1)(z−z_2 )^(−n−k) (((n−1)!)/(k!)) z^k Res(ϕ,z_1 )= −i 2^n Σ_(k=0) ^(n−1) C_(n−1) ^k (−1)^k n(n+1)...(n+k−1)(z_1 −z_2 )^(−n−k) (z_1 ^k /(k!)) ∫_(∣z∣=1) ϕ(z)dz =π 2^(n+1) A_n with A_n =Σ_(k=0) ^(n−1) C_(n−1) ^k (−1)^k n(n+1)...(n+k−1)(2(√(p^2 −1)))^(−n−k) (((−p+(√(p^2 −1)))^k )/(k!)) = I_p$
$c h a n g e m e n t e^{i t} = z g i v e$ $I_{n} = \int_{∣ z ∣= 1} \frac{1}{{(p + \frac{z + z^{- 1}}{2})}^{n}} \frac{d z}{i z}$ $= \int_{∣ z ∣= 1} \frac{- i 2^{n}}{{(2 p + z + z^{- 1})}^{n}} \frac{d z}{z}$ $= \int_{∣ z ∣= 1} \frac{- i 2^{n}}{z {2 p + z + \frac{1}{z}}^{n}} d z$ $= \int_{∣ z ∣= 1} \frac{- i 2^{n} z^{n - 1}}{{2 p z + z^{2} + 1}^{n}} d z l e t$ $φ (z) = \frac{- i 2^{n} z^{n - 1}}{{(z^{2} + 2 p z + 1)}^{n}} p o l e s o f φ ?$ $z^{2} + 2 p z + 1 = 0$ $Δ^{^{'}} = p^{2} - 1 > 0 \Rightarrow r e a l r o o t s$ $z_{1} = - p + \sqrt{p^{2} - 1} a n d z_{2} = - p - \sqrt{p^{2} - 1}$ $w e h a v e p r o v e d t h a t ∣ z_{1} ∣< 1 a n d ∣ z_{2} ∣> 1 s o$ $\int_{∣ z ∣= 1}^{} φ (z) d z = 2 i π R e s (φ, z_{1}) w e ? h a v e$ $φ (z) = \frac{- i 2^{n} z^{n - 1}}{{(z - z_{1})}^{n} {(z - z_{2})}^{n}} \Rightarrow$ $R e s (φ, z_{1}) = {l i m}_{z \to z_{1}} \frac{1}{(n - 1)!} {{(z - z_{1})}^{n} φ (z)}^{(n - 1)}$ $= - i 2^{n} {l i m}_{z \to z_{1}} \frac{1}{(n - 1)!} {\frac{z^{n - 1}}{{(z - z_{2})}^{n}}}^{) n - 1)} b u t$ ${z^{n - 1} {(z - z_{2})}^{- n}}^{(n - 1)}$ $= \sum_{k = 0}^{n - 1} C_{n - 1}^{k} {{(z - z_{2})}^{- n}}^{(k)} {(z^{n - 1})}^{(n - 1 - k)}$ $w e h a v e$ ${(z - z_{2})}^{- n}}^{(1)} = - n {(z - z_{2})}^{- n - 1}$ ${{(z - z_{2})}^{- n}}^{(2)} = {(- 1)}^{2} n (n + 1) {(z - z_{2})}^{- n - 2}$ ${{(z - z_{2})}^{- n}}^{(k)} = {(- 1)}^{k} n (n + 1) . . . (n + k - 1) {(z - z_{2})}^{- n - k}$ $a l s o$ ${(z^{n})}^{(p)} = n (n - 1) . . . (n - p + 1) z^{n - p} i f p ⩽ n \Rightarrow$ ${(z^{n - 1})}^{(n - 1 - k)} = (n - 1) (n - 2) . . . (n - 1 - n + 1 + k + 1) z^{n - 1 - n + 1 + k}$ $= (n - 1) (n - 2) . . . . . (k + 1) z^{k}$ $= \frac{(n - 1)!}{k!} z^{k} \Rightarrow$ ${z^{n - 1} {(z - z_{2})}^{- n}}^{(n - 1)} =$ $= \sum_{k = 0}^{n - 1} C_{n - 1}^{k} {(- 1)}^{k} n (n + 1) . . . (n + k - 1) {(z - z_{2})}^{- n - k} \frac{(n - 1)!}{k!} z^{k}$ $R e s (φ, z_{1}) =$ $- i 2^{n} \sum_{k = 0}^{n - 1} C_{n - 1}^{k} {(- 1)}^{k} n (n + 1) . . . (n + k - 1) {(z_{1} - z_{2})}^{- n - k} \frac{z_{1}^{k}}{k!}$ $\int_{∣ z ∣= 1} φ (z) d z = π 2^{n + 1} A_{n} w i t h$ $A_{n} = \sum_{k = 0}^{n - 1} C_{n - 1}^{k} {(- 1)}^{k} n (n + 1) . . . (n + k - 1) {(2 \sqrt{p^{2} - 1})}^{- n - k} \frac{{(- p + \sqrt{p^{2} - 1})}^{k}}{k!}$ $= I_{p}$
Commented by abdo mathsup 649 cc last updated on 08/Jul/18

$I_{n} = \int_{∣ z ∣= 1} φ (z) d z = π 2^{n + 1} A_{n}$
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