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 Sci Advisor HW Helper P: 2,002 The first endpoint at t is a bit of a problem, because of the singularity from the delta function. We do have: $\delta(t-u) \to 0$ as $u\to t$. At the other endpoint, it is simply $\delta(t)$. So: $$-\delta(t-u)f(u)|^{u=t}_{u=0}=-(0)f(t)+\delta(t)f(0)=f(0)\delta(t)$$