Approximating Delta distribution with an Exponential

Question

Is it alright to approximate a Delta distribution with an exponential like this:

$$\delta(x-1) = \omega\,e^{-\omega (x-1)}, \hspace{1cm} x \geq 1,$$

where, $\omega \gg 1 $, and, $$\int_1^{\infty} f(x)\,\delta(x-1) = f(1).$$ Also, what are the caveats of differentiating this distribution, e.g., can I write, $$\frac{\partial}{\partial x}\delta(x-1) = -\omega\,\delta(x-1).$$

Thanks.

Answer 1

The Dirac Delta has a standard regularization by the family of normed Gaussians $$\delta_\varepsilon (x) = \frac{1} {\sqrt{2\pi \varepsilon^2}} \ e^{- \frac{x^2}{2 \varepsilon^2}}$$ that has unit 1-norm and falls off to zero sudenly for $|x| > 3 \varepsilon$. The central points of a regularization of the $\delta$-distribution:

1. Approximate point support at $x=0$,
2. approximate unit step function of its indefinite integral,
3. indefinite differentiability (smoothness) to implement the unlimited algebra of partial differentiation of products with smooth functions of compact support.

Check your approach against the thre points.

Answer 2

The approximation you use for $\delta(x)$ is $D_\omega(x) = \omega e^{-\omega x} H(x),$ where $H$ is the Heaviside step function. The derivative of this is $$ D_\omega'(x) = -\omega^2 e^{-\omega x} H(x) + \omega e^{-\omega x} \delta(x) = -\omega D_\omega(x) + \omega \delta(x) . $$