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傅立叶变换的微分性质,减出来为什么不是零?

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abababa Posted at 2017-8-24 20:33:18 |Read mode
用Mathematical输入:
FourierTransform[D[x^3,{x,1}],x,s]-(I s)^1FourierTransform[x^3,x,s]
函数是$f(x)=x^3$,用傅立叶变换的微分性质:
\[\mathcal{F}(f'(x))=(is)^1\mathcal{F}(f(x))\]
上式应该是零,但实际减出来不是零,这是什么原因?

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战巡 Posted at 2018-12-6 02:54:20
你这玩意最后给你啥结果?是这个吧?
-6sqrt(2π)DirackDelta’’[s]

这玩意其实就是0,DirackDelta是狄拉克函数,DirackDelta[x]=0, x≠0,在x=0处不可导,其他地方导数当然是0,整个还是0

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