arierreF
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Prove that.
[itex]\int_a^b f(x)g' (x)\, dx = -f(0)[/itex]
This is supposed to be a delta Dirac function property. But i can not prove it.
I thought using integration by parts.
[itex]\int_a^b f(x)g' (x)\, dx = f(x)g(x) - \int_a^b f(x)'g (x)\, dx[/itex]
But what now?
Some properties:
[itex]\delta [g(x)] = \sum \frac{1}{|g'(xi)|}[/itex]
[itex]\int_a^b f(x)\delta(x-xi)\, dx =[/itex]
[itex]f(x_{0})[/itex] if [itex]a<x_{0}<b[/itex]
0, other cases.
I just need a tip please.
[itex]\int_a^b f(x)g' (x)\, dx = -f(0)[/itex]
This is supposed to be a delta Dirac function property. But i can not prove it.
I thought using integration by parts.
[itex]\int_a^b f(x)g' (x)\, dx = f(x)g(x) - \int_a^b f(x)'g (x)\, dx[/itex]
But what now?
Some properties:
[itex]\delta [g(x)] = \sum \frac{1}{|g'(xi)|}[/itex]
[itex]\int_a^b f(x)\delta(x-xi)\, dx =[/itex]
[itex]f(x_{0})[/itex] if [itex]a<x_{0}<b[/itex]
0, other cases.
I just need a tip please.