Fourier transform問題

2010-03-24 6:04 am
F{ e^-( a。x^2 ) }

a > 0

回答 (4)

2010-03-24 6:32 pm
✔ 最佳答案
By definition F(w)=F{ e^-( a。x^2 ) } = integral[x=-infinity to infinity] {e^(iwx)e^(-ax^2)} dx = integral[x=-infinity to infinity]{e^(-ax^2+iwx)} dx .

the exponent of the integrand: (-ax^2+iwx)=-a[x-(iw/(2a)]^2+[-w^2/(4a^2)]
thus F(w)= e^[-w^2/(4a^2)] * integral[x=-infinity to infinity]{e^(-a[x-(iw/(2a)]^2} dx. Using the famous integral integral[x=-infinity to infinity]{e^(-x^2} dx =sqrt(pi) , we have F(w)=e^[-w^2/(4a^2)] *sqrt(pi/a).

2010-03-25 02:36:08 補充:
更正(-ax^2+iwx)=-a[x-(iw/(2a)]^2+[-w^2/(4a^2)] --->(-ax^2+iwx)=-a[x-(iw/(2a)]^2+[-w^2/(4a)]
故 F(w)=e^[-w^2/(4a^2)] *sqrt(pi/a) 應為F(w)=e^[-w^2/(4a)] *sqrt(pi/a)
有勞天助提醒
2010-03-25 12:17 am
To:教大
配方請小心
2010-03-24 10:27 am
您的方法就是學校老師教的方法

我其實是想問問看有沒有法二

2010-03-24 23:10:35 補充:
我想用contour integral可以嗎?
2010-03-24 10:22 am
這是高微的題目,十分技巧,重點在一致收斂性
為簡化問題 取a=1

令I(w)=F{ e^-(x^2 ) } ..化簡..= S(-oo,oo) e^-(x^2 )*coswx dx = 2 S(0,oo) e^-( x^2 )*coswx dx

將I(w)對w微分

I'(w)= d/dw 2S(0,oo) e^-(x^2 )*coswx dx

2010-03-24 02:22:26 補充:
瑕積分號與微分號換序.....要求 (e^-(x^2 )*coswx)的微分為一致收斂
請自行利用 weierstrass test 判別其為一致收斂

=2S(0,oo) e^-( x^2 )*x *-sinwx dx = -2 S(0,oo) e^-( x^2 )*x *sinwx dx
現在可以積分了!! 用分部積分 = -w S(0,oo) e^-( x^2 ) *coswx dx = -wI(w)

I'(w)=-w I(w) 解此一階ODE 得 I(w)=sqrt(pi) *e^-( (w/2)^2 )


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