✔ 最佳答案
第一題:√(2π)
第二題:2πcos(m/2) exp(-√3 |m|/2) /√3
明天有空再寫
2009-12-24 11:29:59 補充:
Q1:
1. 原積分=2∫[0~∞] sin(x^2)/x^2 dx (by parts)
=4∫[0~∞] cos(x^2) dx
=√(2π)
此積分求法請參考:
http://tw.knowledge.yahoo.com/question/question?qid=1009081308379
or
http://tw.group.knowledge.yahoo.com/math-etm/article/view?aid=102
Q2: (設m>=0)
∫[-∞~∞] cos(mx)/(x^2+x+1) dx (配方,平移)
=∫[-∞~∞] cos[m(x- 1/2)] /(x^2 + w^2) dx ( w^2 = 3/4, w>0)
=∫[-∞,∞] [cos(mx)cos(m/2)+sin(mx)sin(m/2)]/(x^2+w^2) dx
= cos(m/2)∫[-∞~∞] cos(mx)/(x^2+w^2) dx (奇函數的積分=0)
= cos(m/2)∫[-∞~∞] exp(imx)/(x^2+w^2) dx 的實數部分
(counter integral 上半平面)
= cos(m/2)*2πi*[Residue of exp(imx)/(x^2+w^2) at x=wi ]
= cos(m/2)*2πi* exp(-mw)/(-2wi)
=2πcos(m/2) exp(-√3 m/2) /√3
註:
若m<0, 則求cos(m/2)∫[-∞~∞] exp(-imx)/(x^2+w^2) dx 的實部
得2πcos(m/2) exp(√3 m/2) /√3
故Q2 Ans=2πcos(m/2) exp(-√3 |m|/2) /√3
2009-12-24 11:32:18 補充:
更正:倒數第6行為= cos(m/2)*2πi* exp(-mw)/(2wi) (差一個負號)