Is there an opposite of a dirac delta function, a function that is infinitely wide and infinitesimally high?
The dirac delta function is defined as:
$$
\delta(x) =
\begin{cases}
0, & \text{if } x \neq 0\\
\infty, & \text{if } x = 0
\end{cases}
$$
where
$$
\int_{-\infty}^{\infty} \delta(x) \, dx = 1
\\
\int_{-\infty}^{\infty} f(x) \delta(x – a) \, dx = f(a)
$$
Is there a function:
$$
\epsilon(x) = \lim_{\epsilon \to 0} \epsilon
$$
where
$$
\int_{-\infty}^{\infty} \epsilon(x) \, dx = 1
$$
The dirac delta function $\delta(x)$ is infinitely high but infinitesimally wide, this new function $\epsilon(x)$ is infinitesimally high but infinitely wide. Im guessing theres no such function/distribution given how limits work and it probably would not be useful if it did, but is there any credence to this idea?
Best Answer
As others have said, the delta function is not a function but a distribution. I won't complain about your definition but point out it really is the limit of a tall peaky function where you have to be integrating over another function and you take the limit outside the integral. I would suggest a better definition is $\delta(x)=\lim_{\epsilon \to 0} k(\epsilon)$ with the integral of $\delta(x)$ over any interval including $0$ to be $1$ and $k(\epsilon)$ being some tall peaky function, probably always positive and not doing strange things as $\epsilon$ changes.
In that spirit the obvious answer is to define a set of functions $g(\epsilon,x)$ as $$g(\epsilon,x)=\begin {cases} \epsilon /2 & |x| \le 1/\epsilon \\ 0& \text{otherwise} \end {cases}$$ It has the property that $\lim_{\epsilon \to 0} g(\epsilon,x)=0$ and the integral of $g$ over the real line is constant at $1$. If you take $$\lim_{\epsilon \to 0} \int_{-\infty }^\infty f(x)g(x,\epsilon) dx$$ (note I have taken the limit out of $g$ and put it in front of the integral) you get something you can call the average value of $f$ over the real line. You can get the same thing from $$\lim_{L \to \infty} \frac 1{2L}\int_{-L}^L f(x) dx$$ which just plugs in the definition of the $g$ I suggested and uses $L=1/\epsilon$. I suspect this is not as useful as picking out the value of a function at a point which is what the delta function lets you do.