※ 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....
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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
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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
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◆ From: 140.113.141.151
※ 編輯: doom8199 來自: 140.113.141.151 (09/08 19:28)