let-I-n-0-2pi-dx-p-cost-n-with-p-gt-1-find-the-value-of-I-n-

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

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) \dots (n + k - 1) {(z - z_{2})}^{- n - k}

a l s o

{(z^{n})}^{(p)} = n (n - 1) \dots (n - p + 1) z^{n - p} i f p ⩽ n \Rightarrow

{(z^{n - 1})}^{(n - 1 - k)} = (n - 1) (n - 2) \dots (n - 1 - n + 1 + k + 1) z^{n - 1 - n + 1 + k}

= (n - 1) (n - 2) \dots . . (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) \dots (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) \dots (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) \dots (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}