- #1
arierreF
- 79
- 0
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.