let-give-from-R-and-2-1-and-I-n-0-pi-cos-nt-1-2-cost-2-dt-calculate-I-n-

Question Number 32341 by abdo imad last updated on 23/Mar/18

l e t g i v e λ f r o m R a n d λ^{2} \neq 1 a n d

I_{n} (λ) = \int_{0}^{π} \frac{c o s (n t)}{1 - 2 λ c o s t + λ^{2}} d t . c a l c u l a t e I_{n} (λ) .

Commented by abdo imad last updated on 01/Apr/18

c h . t = 2 x g i v e I_{n} (λ) = 2 \int_{0}^{2 π} \frac{c o s (2 n x)}{1 - 2 λ c o s (2 x) + λ^{2}} d x a f t e r

w e u s e t h e c h . e^{i x} = z

I_{n} (λ) = 2 \int_{∣ z ∣= 1} \frac{\frac{z^{2 n} + z^{- 2 n}}{2}}{1 - 2 λ \frac{z^{2} + z^{- 2}}{2}} \frac{d z}{i z}

= 2 \int_{∣ z ∣= 1} \frac{z^{2 n} + z^{- 2 n}}{i z (2 - 2 λ (z^{2} + z^{- 2}))} d z

= \int_{∣ z ∣= 1} \frac{z^{2 n} + z^{- 2 n}}{i z (1 - λ z^{2} - λ z^{- 2})} = \int_{∣ z ∣= 1} \frac{- i (z^{2 n} + z^{- 2 n})}{z (1 - λ z^{2} - \frac{λ}{z^{2}})} d z

= \int_{∣ z ∣= 1} \frac{- i z (z^{2 n} + z^{- 2 n})}{z^{2} - λ z^{4} - λ} d z

= \int_{∣ z ∣ = 1} \frac{i (z^{2 n + 1} + z^{- 2 n + 1})}{λ z^{4} - z^{2} + λ} d z l e t p u t

φ (z) = \frac{i (z^{2 n + 1} + z^{- 2 n + 1})}{λ z^{4} - z^{2} + λ} . p o l e s o f φ ?

λ z^{4} - z^{2} + λ = 0 \Rightarrow Δ = 1 - 4 λ^{2} i f Δ ⩾ 0 t h e r o o t s a r e r e a l s

z_{1}^{2} = \frac{1 + \sqrt{1 - 4 λ^{2}}}{2} a n d z_{2}^{2} = \frac{1 - \sqrt{1 - 4 λ^{2}}}{2} \Rightarrow

z_{1} = ξ \sqrt{\frac{1 + \sqrt{1 - 4 λ^{2}}}{2}} a n d z_{2} = ξ \sqrt{\frac{1 - \sqrt{1 - 4 λ^{2}}}{2}}

w i t h ξ^{2} = 1 a n d w e g e t \int_{R} φ (z) d x = 2 i π \sum_{i} R e s (φ, x_{i})

i f Δ < 0 Δ = {(i \sqrt{4 λ^{2} - 1})}^{2} \Rightarrow

z_{1}^{2} = \frac{1 + i \sqrt{4 λ^{2} - 1}}{2} \Rightarrow z_{1} = ξ \sqrt{\frac{1 + i \sqrt{4 λ^{2} - 1}}{2}}

z_{2}^{2} = \frac{1 - i \sqrt{4 λ^{2} - 1}}{2} \Rightarrow z_{2} = ξ \sqrt{\frac{1 - i \sqrt{4 λ^{2} - 1}}{2}}

a n d w e c h o o s e t h e r o o t s w i c h v e r i f y ∣ z_{i} ∣⩽ 1 \dots . b e c o n t i n u e d