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# HW5 - (1 i x dx = R π e x cos xdx i R π e x sin xdx =-1 e...

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MATH 185 SOLUTIONS OF HW V (1) (37. 3) If m = n then k = m - n is a nonzero integer and therefore 2 π 0 e miθ e - niθ = 2 π 0 e ( m - n ) = 2 π 0 e kiθ = 1 ki [ e kiθ ] 2 π 0 = 0 since e ki is a periodic function of period 2 π . If m = n then e miθ e - niθ = 1 so the integral is just 2 π . (2) (37. 4) Evaluating π 0 e (1+ i ) x dx by thinking of it as a real-valued function. π 0 e (1+ i ) x dx = 1 (1 + i ) [ e (1+ i ) x ] π 0 = 1 (1 + i ) ( e (1+ i ) π - 1) = 1 (1 + i ) ( - e π - 1) = - 1 + e 2 + i 1 + e 2 Meanwhile e (1+ i ) x = e x cos x + ie x sin
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Unformatted text preview: (1+ i ) x dx = R π e x cos xdx + i R π e x sin xdx =-1+ e 2 + i 1+ e 2 . Since e x cos x and e x sin x are real-valued functions, by comparing the real and imaginary part of the equation we have R π e x cos xdx =-1+ e 2 and R π e x sin xdx = 1+ e 2 . Date : today. 1...
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