- #1
maverick280857
- 1,774
- 5
Hi everyone,
I was wondering how to deal with delta functions of functions that have double zeros.
For instance, how does one compute an integral of the form
\int_{-\infty}^{\infty}dx g(x)\delta(x^2)
where g(x) is a well behaved continuous everywhere function?
In general how does one find
\int_{-\infty}^{\infty}dx g(x)\delta(f(x))
where f(x) has a finite number of multiple zeros along with some simple zeros. I know that
\delta(f(x)) = \sum_{i=1}^{N}\frac{\delta(x-x_{i})}{|f'(x_{i})|}
where x_{i}'s (for i = 1 to N) are simple zeros of f(x) and it is known that f(x) has no zeros of multiplicitiy > 1.
but this is of course not valid here. Using this, however I could write
\delta(x^2-a^2) = \frac{1}{2|a|}\left(\delta(x-a) + \delta(x+a)\right)
But the limit of this as a \rightarrow 0 tends to infinity.
Any ideas?
Thanks in advance.
Cheers.
I was wondering how to deal with delta functions of functions that have double zeros.
For instance, how does one compute an integral of the form
\int_{-\infty}^{\infty}dx g(x)\delta(x^2)
where g(x) is a well behaved continuous everywhere function?
In general how does one find
\int_{-\infty}^{\infty}dx g(x)\delta(f(x))
where f(x) has a finite number of multiple zeros along with some simple zeros. I know that
\delta(f(x)) = \sum_{i=1}^{N}\frac{\delta(x-x_{i})}{|f'(x_{i})|}
where x_{i}'s (for i = 1 to N) are simple zeros of f(x) and it is known that f(x) has no zeros of multiplicitiy > 1.
but this is of course not valid here. Using this, however I could write
\delta(x^2-a^2) = \frac{1}{2|a|}\left(\delta(x-a) + \delta(x+a)\right)
But the limit of this as a \rightarrow 0 tends to infinity.
Any ideas?
Thanks in advance.
Cheers.
