[工程數學]兩題積分(研究所入學考題)

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?qi...

or http://tw.group.knowledge.yahoo.com/math-etm/artic...

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)

先說謝謝,也請保重,祝早日康復!

1. 2 pi

You can use the complex variable to help you solve the proble.

The other way is to show integration of sin(x^2)/x^2 on R is the same as the integration of sinx /x on R.

The proof is following

\int sin(x^2)/x^2 dx = -sin(x^2)/x + \int (2/x)* sinx cos x dx (integration by part)

first terb will equal 0 when put x= -oo and oo

2sinx cos x = sin 2x

so second tern will be \int (sin 2x ) /x dx

and you do change variable this definitely = \int sin x /x

2009-12-17 04:16:16 補充：

consider \int e^2iz / z^2 dz on upper semi big circle dig of the semi circle around origin

e^2iz = 1+ 2iz + 4i^2z^2 /2

residue will be coefficient of 1/z so will be 2iz / z^2 .tern.

the Res(e^2iz / z^2 ; 0 ) = 2i

2 pi i * 2i = -4 pi

but only semi circle so, the answer is 2 pi.