※ 引述《chihhungw (嘻嘻哈哈)》之銘言： : 3. 從零積到pi : dx/2+cosx : 4. 從零積到2pi : 1+4cosx/17-8cosx dx : 5. 從零積到2pi : e^(cosx) cos(sin(x)) dx : 另外一個 find all solutions of sin(z)=i : 非常感恩阿.... --- 3. set z=e^(ix) → dz = i*e^(ix) dx = iz dx cos(x) = [z + z^(-1)]/2 sin(x) = [z - z^(-1)]/2 π 1 2π 1/2 則 ∫ ________ dx = ∫ ________ dx (注意 cosx 對稱於 x=π) 0 2 + cosx 0 2 + cosx 1/2 1 = ∮ __________________ * ___ dz (where C: |z|=1) C 2 + [z + z^(-1)]/2 iz -i = ∮ ____________ dz C z^2 + 4z + 1 -i = ∮ ____________________________ dz C [z - (-2+√3)][z - (-2-√3)] 2π = _____ ( one pole z = (-2+√3) ) 2√3 4. set z=e^(ix) → dz = i*e^(ix) dx = iz dx cos(x) = [z + z^(-1)]/2 sin(x) = [z - z^(-1)]/2 2π 1 + 4cosx 1 + 2[z + z^(-1)] 1 ∫ __________ dx = ∮ __________________ * __ dz (where C: |z|=1) 0 17 - 8cosx C 17 - 4[z + z^(-1)] iz 2z^2 + z + 2 = i/4 *∮ _____________ dz C (z-1/4)(z-4)z 1/8 + 1/4 + 2 2 = -π/2*[ ______________ + __________ ] (-15/4)(1/4) (-1/4)(-4) = 4π/15 (two pole z=1/4 、 0) 5. e^z 考慮 ∮ ____ dz = 2πi (where C: |z|=1) (one pole z=0) C z 2π (cosx + isinx) → ∫ e idx = 2πi 0 2π cosx → ∫ e *[cos(sinx) + isin(sinx)] dx = 2π 0 2π cosx 2π cosx 　→ ∫ e *cos(sinx) dx = 2π and ∫ e *sin(sinx) dx = 0 0 0 --- sin(z)=i → sinx*coshy + icosx*sinhy = i , where z=x+yi → sinx*coshy = 0 ____(1) cosx*sinhy = 1 ____(2) from (1) , coshy > 0 for all y 屬於R then sinx = 0 → x = kπ , k屬於Z by (2) → (-1)^k*sinhy = 1 → [e^y - e^(-y)]/2 = (-1)^k → (e^y)^2 - 2(-1)^k*(e^y) - 1 = 0 → e^y = (-1)^k ± √2 (observe e^y > 0 ) → y = ln[ √2 + (-1)^k] hence z = kπ + i*ln[ √2 + (-1)^k] , k屬於Z --

※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.113.141.151 ※ 編輯: doom8199 來自: 140.113.141.151 (09/08 19:28)