- #1
Hi, if I have an interval on the x-axis, defined by the parameter L, can this, interval be transformed to a Dirac delta function instead, on the x-axis?
Thanks!
Thanks!
Wouldn't it simply become a point ? Like in: the interval ##[x,x+\epsilon]## becomes the point ##x## for ##\epsilon \downarrow 0##
For an interval (a,b) where a<b, that would make L negative. If 'L' stands for length, do you mean L = b-a?Thank you! This is indeed a great suggestion. However, L is included in a function I have, f(x), and currently it is a parameter. It would be useful to use your form somehow, as you write:
\begin{equation}
\int_{-\infty}^x \delta(t-a)dt - \int_{-\infty}^x \delta(t-b)dt
\end{equation}
Is L actually a-b here,
t is just a dummy variable for the integration, so it does not appear as an input or parameter of the indicator function.and t is the variable of my function, f(x)?
I'm confused about what you are trying to do here.If e^(ix)+L is my function, would the equivalent be :
\begin{equation}
\int_{-\infty}^x \delta(e^{it}-a)dt - \int_{-\infty}^x \delta(e^{it}-b)dt
\end{equation} ?
'It just says'? Where are you seeing this reference to the delta function? You need to back up and explain more about what your original problem is and why you want to use the delta function.It just says \delta and it gives therefore no idea to me on how it looks like a function.
Who said that it is a 'dummy' function? I never said that.Nevertheless, you say the dirac f is a dummy function.
Why would you want to replace L with a delta function? Is L a constant?The thing I am not sure about is that because L is part of the existing function, f(x), say:
\begin{equation}
f(x) = e^{-ipk}/L
\end{equation}
how would the Dirac variant of this look like, where L is represented by the integral?
Why would you want to replace L with a delta function? Is L a constant?