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Question Number 167393 by DrHZ last updated on 15/Mar/22

	Answered by mindispower last updated on 15/Mar/22
	$∫_0 ^(2π) (((1+e^(ix) +e^(−ix) )^n e^(inx) )/(3+e^(ix) +e^(−ix) ))dx=I_n we want Real I_n =∫_0 ^(2π) (((e^(2ix) +1+e^(ix) )^n e^(ix) )/(e^(2ix) +1+3e^(ix) ))dx u=e^(ix) =∫_C_1 (((x^2 +x+1)^n )/(x^2 +3x+1))dx=−i∫_C_1 f(x)dx C_1 ={z∈C;∣z∣≤1} x^2 +3x+1=0⇒x=((−3−^+ (√5))/2) ∣((−3+(√5))/2)=a∣<1 I_n =2iπRes(−if,a) =2π((((a^2 +a+1)^n )/(2a+3)))=2π((((−2a)^n )/(2a+3))) Re(I_n )=((2π)/(2a+3))(−2a)^n =((2π)/( (√5)))((√5)−3)^n$
	$\int_{0}^{2 π} \frac{{(1 + e^{i x} + e^{- i x})}^{n} e^{i n x}}{3 + e^{i x} + e^{- i x}} d x = I_{n}$ $w e w a n t R e a l I_{n}$ $= \int_{0}^{2 π} \frac{{(e^{2 i x} + 1 + e^{i x})}^{n} e^{i x}}{e^{2 i x} + 1 + 3 e^{i x}} d x$ $u = e^{i x}$ $= \int_{C_{1}} \frac{{(x^{2} + x + 1)}^{n}}{x^{2} + 3 x + 1} d x = - i \int_{C_{1}} f (x) d x$ $C_{1} = {z \in C; ∣ z ∣⩽ 1}$ $x^{2} + 3 x + 1 = 0 \Rightarrow x = \frac{- 3 \overset{+}{-} \sqrt{5}}{2}$ $∣ \frac{- 3 + \sqrt{5}}{2} = a ∣< 1$ $I_{n} = 2 i π R e s (- i f, a)$ $= 2 π (\frac{{(a^{2} + a + 1)}^{n}}{2 a + 3}) = 2 π (\frac{{(- 2 a)}^{n}}{2 a + 3})$ $R e (I_{n}) = \frac{2 π}{2 a + 3} {(- 2 a)}^{n} = \frac{2 π}{\sqrt{5}} {(\sqrt{5} - 3)}^{n}$
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