calculate-F-x-0-2pi-cos-4t-x-2-2x-cost-1-dt-

Question Number 39037 by maxmathsup by imad last updated on 01/Jul/18

c a l c u l a t e F (x) = \int_{0}^{2 π} \frac{c o s (4 t)}{x^{2} - 2 x c o s t + 1} d t

Commented by math khazana by abdo last updated on 04/Jul/18

F (x) = R e (\int_{0}^{2 π} \frac{e^{i 4 t}}{x^{2} - 2 x c o s t + 1} d t) c h a n g e m e n t

e^{i t} = z g i v e

\int_{0}^{2 π} \frac{e^{i 4 t}}{x^{2} - 2 x c o s t + 1} d t = \int_{∣ z ∣= 1} \frac{z^{4}}{x^{2} - 2 x \frac{z + z^{- 1}}{2} + 1} \frac{d z}{i z}

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

= \int_{∣ z ∣= 1} \frac{- {i z}^{4}}{x^{2} z - {x z}^{2} - x + z} d z

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

= \int_{∣ z ∣= 1} \frac{i z^{4}}{x z^{2} - (1 + x^{2}) z + x} l e t c o n s i d e r

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

Δ = {(1 + x^{2})}^{2} - 4 x^{2} = 1 + 2 x^{2} + x^{4} - 4 x^{2}

= {(1 - x^{2})}^{2} \Rightarrow \sqrt{Δ} = ∣ 1 - x^{2} ∣

c a s e 1 ∣ x ∣< 1 \Rightarrow \sqrt{Δ} = 1 - x^{2} \Rightarrow

z_{1} = \frac{1 + x^{2} + 1 - x^{2}}{2 x} = \frac{1}{x} (x \neq 0)

z_{2} = \frac{1 + x^{2} - 1 + x^{2}}{2 x} = x

∣ z_{1} ∣ = \frac{1}{∣ x ∣} > 1 a n d ∣ z_{2} ∣ =∣ x ∣< 1 s o

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, z_{2}) b u t

φ (z) = \frac{i z^{4}}{x (z - z_{1}) (z - z_{2})} \Rightarrow

R e s (φ, z_{2}) = \frac{i z_{2}^{4}}{x (z_{2} - z_{1})} = \frac{i x^{4}}{x (x - \frac{1}{x})} = \frac{i x^{4}}{x^{2} - 1}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{{i x}^{4}}{x^{2} - 1} = \frac{- 2 π x^{4}}{x^{2} - 1} = \frac{2 π x^{4}}{1 - x^{2}} \Rightarrow

F (x) = R e (\int_{- \infty}^{+ \infty} φ (z) d z) = \frac{2 π x^{4}}{1 - x^{2}} .

Commented by math khazana by abdo last updated on 04/Jul/18

c a s e 2 i f ∣ x ∣< 1 w e f o l l o w t h e s a m e r o a d \dots