- #1
WilcoRogers
- 9
- 0
Hello,
I am trying to show that:
[itex]\delta(x) = \lim_{\epsilon \to 0} \frac{\sin(\frac{x}{\epsilon})}{\pi x}[/itex]
Is a viable representation of the dirac delta function. More specifically, it has to satisfy:
[itex] \int_{-\infty}^{\infty} \delta(x) f(x) dx = f(0)[/itex]
I know that the integral of sin(x)/x over the reals is [itex]\pi[/itex], and so far as I can tell, it doesn't depend on epsilon. What I've tried so far is integration by parts, which leads me to:
[itex] f(x) - \int_{-\infty}^{\infty} f'(x) dx[/itex]
Which isn't really getting me somewhere, and the limit drops off due to the integral not caring what epsilon is. Is there another way of approaching this? Or am I on the right track, I just can't pull out an f(0) from this.
Any help is appreciated,
Thanks.
I am trying to show that:
[itex]\delta(x) = \lim_{\epsilon \to 0} \frac{\sin(\frac{x}{\epsilon})}{\pi x}[/itex]
Is a viable representation of the dirac delta function. More specifically, it has to satisfy:
[itex] \int_{-\infty}^{\infty} \delta(x) f(x) dx = f(0)[/itex]
I know that the integral of sin(x)/x over the reals is [itex]\pi[/itex], and so far as I can tell, it doesn't depend on epsilon. What I've tried so far is integration by parts, which leads me to:
[itex] f(x) - \int_{-\infty}^{\infty} f'(x) dx[/itex]
Which isn't really getting me somewhere, and the limit drops off due to the integral not caring what epsilon is. Is there another way of approaching this? Or am I on the right track, I just can't pull out an f(0) from this.
Any help is appreciated,
Thanks.