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I'm wondering if there is an analytical form for a $C^\infty$ approximation with a compact support of a Heaviside step function $f(x) = I_{x \geq 0}$. In attempting to construct one, I'm taking a bump function $$g(x) = \exp\left(\frac{1}{x^2-1}\right)I_{|x| \leq 1}$$ and trying to compute the convolution $$g*f(t) = \int_\mathbb{R}f(x)g(t-x)dx = \int_{-1}^t \exp\left(\frac{1}{x^2-1}\right)dx$$

Does someone know how to take this integral, or at least construct its analytical or finite series approximation?

Hope
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    No, you cannot compute this integral explicitly. – Kavi Rama Murthy Oct 03 '18 at 10:25
  • @Hope There is no explicit form for this integral but you can obtain an explicit formulation with another choice of $g$. As $(g *f)(t)=\int_{-\infty}^t g(x) dx$ a "good" choice is any bump function with a known antiderivative. For example if you do not require compact support $1/(1+x^2)$ leads tot he well known $\arctan$. – Delta-u Oct 03 '18 at 14:20
  • @Delta-u Another $g(x)$ would be fine as long as it has compact support and all zero derivatives on the border. – Hope Oct 05 '18 at 08:04
  • Did you tried something like modifing $1/(1+\exp((2x-1)/(x^2-x)))$ as a piecewise function? – Joako Nov 05 '23 at 04:15

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An approximation for the Heaviside step function could be made through a smooth transition function like $1 \leq m \to \infty $: $$\begin{align}f(x) &= \begin{cases} {\displaystyle \frac{1}{2}\left(1+\tanh\left(m\frac{2x}{1-x^2}\right)\right)}, & |x| < 1 \\ \\ 1, & x \geq 1 \\ 0, & x \leq -1 \end{cases}\end{align}$$ which could be made as close as desire by increasing the value of $m$. Check it in Desmos.

smooth approximation

Joako
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