※ 引述《chihhungw (嘻嘻哈哈)》之銘言：
: 1. 從負無限大積到無限大
: x^/1+x^4 dx
: 2. 從零積到無限大
: 1/1+x^3
: 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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1、2 題同類型，我直接算 general case:
令 C : closed contour of 1/n circle with radius = R>1
C1: |z|=R>1 、 from Arg(z)=0 to Arg(z)=2π/n
C2: line curve from |z|=R to |z|=0 with Arg(z)=2π/n
set n-m >= 2 and n、m 屬於 {N,0}
z^m z^m
考慮 ∮ _______ dz = 2πi _________ ｜ iπ/n
C 1 + z^n n*z^(n-1) z=e
-2πi iπ(m+1)/n
= _____ e
n
z^m R x^m z^m z^m
由 ∮ _______ dz = ∫ _______ dx + ∫ _______ dz + ∫ _______ dz
C 1 + z^n 0 1 + x^n C1 1 + z^n C2 1 + z^n
z^m
當 R→∞ , ∫ _______ dz → 0 (by ML inequality)
C1 1 + z^n
-2πi iπ(m+1)/n ∞ x^m z^m
所以 _____ e = ∫ _______ dx + ∫ _______ dz
n 0 1 + x^n C2 1 + z^n
( change C2: let z=re^(i2π/n) → dz = e^(i2π/n) dr )
m i(2mπ/n)
-2πi iπ(m+1)/n ∞ x^m 0 r *e i(2π/n)
→ _____ e = ∫ _______ dx + ∫ _____________e dr
n 0 1 + x^n ∞ 1 + r^n
-2πi iπ(m+1)/n ∞ x^m i2π(m+1)/n
→ _____ e = ∫ _______ [1 - e ] dx
n 0 1 + x^n
∞ x^m -2πi e^[iπ(m+1)/n]
→ ∫ _______ dx = _____ * ___________________
0 1 + x^n n 1 - e^[i2π(m+1)/n]
-2πi 1
= _____ * ________________________________
n e^[-iπ(m+1)/n] - e^[iπ(m+1)/n]
-2πi 1
= _____ * ____________________
n -2i sin[π(m+1)/n]
π
= __________________
n*sin[π(m+1)/n]
例如第二題 n=3 , m=0
∞ 1 π 2π
∫ _______ dx = ___________ = ______
0 1 + x^3 3*sin[π/3] 3√3
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剩下四題我 po下一篇，不然篇幅有點長 XD
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