int[1/(sinh(z)*z^2)]dz=?
積分路徑:lzl=1
int[1/(sinh(z)*z^2)]dz=?
2009-12-18 1:48 pm
回答 (1)
2009-12-18 7:57 pm
✔ 最佳答案
把 1/[z^2 sinh(z)] 展開為 Mac-Laurin's series找 1/z項係數,再積分(或用Residue)即可sinh(z)=z+z^3/3!+z^5/5!+...= z*(1+w), w=z^2/6+z^4/120+...
1/sinh(z)=(1/z)*(1-w+w^2+...)=(1/z)[-z^2/6+(1/36- 1/120)z^4+...]
1/[z^2 sinh(z)]= -1/(6z)+....
so, int 1/[z^2 sinh(z)] dz= -(1/6) *2 π i= - iπ/3
參考: Myself
收錄日期: 2021-04-30 14:13:44
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091218000016KK01169