Is there an opposite of a dirac delta function, a function that is infinitely wide and infinitesimally high?

Question

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?

Answer 1

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.

Answer 2

I suppose you could consider the family of functions

$$\epsilon(x;s)=\frac{1}{s^2}\exp\left(-\pi~\frac{x^2}{s^2}\right)$$

We remark the following properties: $$\int_\Bbb R \epsilon(x;s)\mathrm dx=1 \\ \forall s\in\mathbb R_+$$ and $$\lim_{s\to\infty}\epsilon(x;s)=0 \\ \forall x\in\mathbb R$$

However unlike the limit definitions of the Dirac delta distribution, the pointwise limit of this family is a well-defined function - that is, the zero function.