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Bobdemaths
#2
Jan23-13, 12:36 PM
P: 3
OK I found the answer. To get rid of the derivatives of Dirac deltas, I used the identity [itex]f(x) \delta ' (x-y) = f(y) \delta ' (x-y) - f'(x) \delta (x-y)[/itex] which follows from the convolution between delta and [itex](f \times \varphi)'[/itex] ([itex]\varphi[/itex] is a test function).
This also produces the remaining wanted term.