兩題積分求值

Q1:請參考http://tw.knowledge.yahoo.com/question/question?qi...

Q2: (設m>=0)

∫[-∞~∞] sin(mx)/(x^2+x+1) dx (配方,平移)

=∫[-∞~∞] sin[m(x- 1/2)] /(x^2 + w^2) dx ( w^2 = 3/4, w>0)

=∫[-∞,∞] [sin(mx)cos(m/2)-cos(mx)sin(m/2)]/(x^2+w^2) dx

= - sin(m/2)∫[-∞~∞] cos(mx)/(x^2+w^2) dx (奇函數的積分=0)

= - sin(m/2)∫[-∞~∞] exp(imx)/(x^2+w^2) dx 的實數部分

(counter integral 上半平面)

= - sin(m/2)*2πi*[Residue of exp(imx)/(x^2+w^2) at x=wi ] 取實數

= - sin(m/2)*2πi* exp(-mw)/(2wi) 取實數

= - 2πsin(m/2) exp(-√3 m/2) /√3

註:

若m<0, 則求 - sin(m/2)∫[-∞~∞] exp(-imx)/(x^2+w^2) dx 的實部

得 - 2πsin(m/2) exp(√3 m/2) /√3

故Q2 Ans= - 2πsin(m/2) exp(-√3 |m|/2) /√3

麻煩寫下一點過程,感恩!

1. Sqrt( 2 pi )

2. - (2 pi / Sqrt(3)) Exp( - |m| Sqrt(3)/2 ) sin( m / 2 )

The first one is by changing complex coordinate and closing a contour. The second is by residue theorem.