Re: [理工] [工數]-中央95-光電
※ 引述《kagato (包)》之銘言:
: 求這個積分
: ∞ i2πβx
: I(β) = ∫ exp^ dx
: 0
: 答案: πδ(2πβ) + i/(2πβ)
: 後面那項不知道怎麼來的,拜託板上高手@@
∞ i2πβx
I(β) = ∫ exp^ dx
0
∞ ∞
∫e^-iwxdx=∫e^-iwxu(x)dx
0 -∞
1
=πδ(w) + ---
iw
w=-2πβ
------------------------------------------------
Let g(x)=e^-ax ,x>0
-e^ax ,x<0
∞
G(w)=F(g)=∫g(x)e^-iwxdx
-∞
∞
=-2i∫e^-axsinwxdx
0
e^-ax ∞
=-2i{------}{-asinwx-wcoswx}|
a^2+w^2 0
w
=-2i{-------}
a^2+w^2
2
limg(x)=1,x>0 limF(g(x))=----
a->0 -1,x<0 a->0 iw
1
and let u(x)=lim ---(g(x)+1)
a->0 2
1 1
F(u(x))=---G(w)| +πδ(w)=---- + πδ(w)
2 a=0 iw
--
為者常成.行者常至
--
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◆ From: 123.193.214.165
※ 編輯: iyenn 來自: 123.193.214.165 (11/12 19:35)
推
11/12 19:35, , 1F
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11/12 22:25, , 13F
11/12 22:25, 13F
※ 編輯: iyenn 來自: 123.193.214.165 (11/12 23:31)
※ 編輯: iyenn 來自: 123.193.214.165 (11/12 23:42)
推
11/12 23:42, , 14F
11/12 23:42, 14F
推
01/12 16:17, , 15F
01/12 16:17, 15F
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