推 ntust661:推^^ 02/18 19:07
推 iohan:感謝cg大 02/18 19:40
推 justyju:感謝 02/19 10:54
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
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