簡單的laplace transform

L{ ( e^-t^2 )。( sin^2t )。δ( e^-( t^2 - π^2／4 ) -1 ) }= int[t=0 to infinity] {e^(-st)*( e^-t^2 )*( sin^2t )*δ( e^-( t^2 - π^2／4 ) -1 ) dt , by definition.
Consider the change of variable: u= e^-( t^2 - π^2／4 ) = e^(π^2／4)* e^-t^2, thus 1. t= {ln[e^(π^2／4)/u}^(1/2); 2. du= u*(-2t) dt or dt = -(1/2)u/t du.
Then we may convert L{ ( e^-t^2 )*( sin^2t )*δ( e^-( t^2 - π^2／4 ) -1 ) }
= (1/2) int[u=0 to e^(π^2／4)] {{ e^[-s*{ln[e^(π^2／4)/u]}^(1/2) * [ e^(-π^2/4 )] * {ln[e^(π^2／4)/u]}^(-1/2) * sin^2{ln[e^(π^2／4)/u]^(1/2)} * δ(u-1) }} du . Since 1 is in between 0 and e^(π^2／4), according to the property of delta function, this integral is to be the function (1/2) * e^[-s*{ln[e^(π^2／4)/u]}^(1/2) * [ e^(-π^2/4 )] * sin^2{ln[e^(π^2／4)/u]^(1/2)} evaluated at u=1. So it equals to (1/2) * e^(-sπ/2) * e^(-π^2/4) * (2/ π ) * sin^2{π/2} = (1/ π) * e^[-(π/2)*(1+s)] .

2010-03-28 09:54:20 補充：
更正最後一段
So it equals to (1/2) * e^(-sπ/2) * e^(-π^2/4) * (2/ π ) * sin^2{π/2} = (1/ π) * e^[-(π/2)*(1+s)] .成為
So it equals to (1/2) * e^(-sπ/2) * e^(-π^2/4) * (2/ π ) * sin^2{π/2} = (1/ π) * e^[-(π/2)*(π/2 +s)] .

多謝各位指正

變數代換即可!
教書大最後一步錯了!

這我知道

2010-03-27 12:52:22 補充：
為什麼

( 1／π )。e^-[ ( 2πs+π^2 )／4 ] = ( 1／ π ) 。e^[ - ( π + πs ／2 ) ]?

最重要的是要知道何時的t值令δ(f(t))中的f(t)等於0