If one has a step function with an infinite number of intervals $$s(x) = \sum\limits_{i=0}^\infty \alpha_i \chi_{A_i}(x)$$ and its integral $$I_s(t)=\int_0^ts(x)dx,$$ does there exist a smooth, continuous function $c(x)$ with integral $$I_c(t)=\int_0^tc(x)dx$$ such that $$ \lim_{t\to\infty} \frac{I_s(t)}{I_s(t)} = 1 $$ and if so, what might be a method of finding $c(x)$? Does it have a closed form? Is it unique? If not, why? I've looked at sum smoothing and asymptotic expansions of said smoothed sums, but no source I've read thus far has commented on the integral of these smooth sums, and I haven't the slightest idea of where to look for more information or how to solve this problem. Incidentally, if anyone has any further reading on the topic I'd be much obliged.
2026-02-23 07:51:59.1771833119
Can a step function be "smoothed" yet have the same definite integral?
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