Re: [理工] [工數] 101南交光電
繞半圓
∞ ln(z+i) ln(z+i) 0 ln(z+i)
∮=∫ ---------dz + ∫ + --------dz + ∫ ---------dz = 2πiRes(i)
0 2 c 2 -∞ 2
z +1 z +1 z +1
∞ ln(z+i) ∞ ln(-z+i)
=∫ ---------dz +0+ ∫ ---------- dz = 2πiRes(i)
0 2 0 2
z +1 z +1
∞ ln(z+i) ∞ ln(-1)(z-i)
=∫ ---------dz +0+ ∫ ------------- dz = 2πiRes(i)
0 2 0 2
z +1 z +1
∞ ln(z+i) ∞ ln(z-i) ∞ 1
=∫ ---------dz +∫ --------- + πi∫ ------dz = 2πiRes(i)
0 2 0 2 0 2
z +1 z +1 z +1
2 2
∞ ln(z +1) iπ
=∫ ----------dz + ----- = 2πiRes(i)
0 2 2
z +1
ln(z+i) ln(2i)
Res(i)=lim -------- = --------
z→i 2z 2i
2 2 2
∞ ln(z +1) iπ iπ iπ
∫ ----------dz = πln(2i) - ----- = πln(2) + π--- - ----- = πln(2)
0 2 2 2 2
z +1
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 114.25.182.106
推
02/18 19:07, , 1F
02/18 19:07, 1F
推
02/18 19:40, , 2F
02/18 19:40, 2F
推
02/19 10:54, , 3F
02/19 10:54, 3F
討論串 (同標題文章)
完整討論串 (本文為第 2 之 2 篇)：
理工
7
24