- #1
arierreF
- 78
- 0
Prove that.
\int_a^b f(x)g' (x)\, dx = -f(0)
This is supposed to be a delta Dirac function property. But i can not prove it.
I thought using integration by parts.
\int_a^b f(x)g' (x)\, dx = f(x)g(x) - \int_a^b f(x)'g (x)\, dx
But what now?
Some properties:
\delta [g(x)] = \sum \frac{1}{|g'(xi)|}
\int_a^b f(x)\delta(x-xi)\, dx =
f(x_{0}) if a<x_{0}<b
0, other cases.
I just need a tip please.
\int_a^b f(x)g' (x)\, dx = -f(0)
This is supposed to be a delta Dirac function property. But i can not prove it.
I thought using integration by parts.
\int_a^b f(x)g' (x)\, dx = f(x)g(x) - \int_a^b f(x)'g (x)\, dx
But what now?
Some properties:
\delta [g(x)] = \sum \frac{1}{|g'(xi)|}
\int_a^b f(x)\delta(x-xi)\, dx =
f(x_{0}) if a<x_{0}<b
0, other cases.
I just need a tip please.
