※ 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.... --- 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 --- 剩下四題我 po下一篇，不然篇幅有點長 XD --

※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.113.141.151