求证f(t)δ'(t)=f(0)δ'(t)-f(0)δ(t)
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我觉得应该是,f(t)δ'(t)=f(0)δ'(t)-f'(0)δ(t)吧!这个式子怎么推出来的比较难,证明的话,两边对t从(-∞,+∞)积分 左边积分∫f(t)δ'(t)dt=∫f(t)dδ(t)=f(t)δ(t)|(t=∞)-f(t)δ(t)|(t=-∞)-∫f'(t)δ(t)dt=-=-∫f'(0)δ(t)dt= -f'(0).右边积分∫【f(0)δ'(t)-f'(0)δ(t)】dt=∫f(0)δ'(t)dt-∫f'(0)δ(t)dt =∫f(0)dδ(t)-f'(0) =f(0)δ(t)|(t=∞)-f(0)δ(t)|(t=-∞)-∫0*δ(t)dt-f'(0) =-f'(0) 所以等式两边对t从(-∞,+∞)积分相等,于是f(t)δ'(t)=f(0)δ'(t)-f'(0)δ(t)