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Galileo
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#12
Aug28-05, 06:33 AM
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The first endpoint at t is a bit of a problem, because of the singularity from the delta function. We do have: [itex]\delta(t-u) \to 0[/itex] as [itex]u\to t[/itex].
At the other endpoint, it is simply [itex]\delta(t)[/itex]. So:

[tex]-\delta(t-u)f(u)|^{u=t}_{u=0}=-(0)f(t)+\delta(t)f(0)=f(0)\delta(t)[/tex]