L[f(t)]=∫(0->∞)e^(-st)*f(t)dt
L[f(t)]'=∫(0->∞)e^(-st)*(-t)f(t)dt
L[tf(t)]=-L[f(t)]'
L[f(t)/t]'= - ∫e^(-st)*f(t)dt= -L[f(t)]
L[f(t)/t]=∫ (s'->∞)L[f(t)] ds
cost是
y''+y=0的解
s^2*L[cost]-s*cos0+sin0+L[cost]=0
L[cost]=s/(s^2+1)
那么
L[(1-cost)/t]=L[1/t]-L[cost/t]=
=∫1/s-s/(s^+1)=
=lns-1/2ln(s^2+1) | s'->∞
=ln(s/sqrt(s^2+1)) | s'->∞
=-1/2*ln(1+1/s^2) | s'->∞
=1/2*ln(1+1/s'^2)
s换回s'
L[(1-cost)/t]=1/2*ln(1+1/s^2)
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