Define an operator on real valued functions analogous to integration, but with bounded contributions. Given a strictly increasing function $b(x)$ with $b(0) = 0$ I'm looking to define an operator $\Psi_{b}$ on (smooth if necessary, otherwise continuous or integrable) functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+} $ to  which acts like the integral operator with the exception that the value of $\Psi_{b}(f)(x)$ is less than $\int^x b(t) dt $
When we say $\Psi_{b}(f)$ "acts like" integration it should converge to $\int^x f(t) dt $ when  $\int^x f(t) dt $ << $\int^x b(t) dt $ for all functions $f$. It should converge to  $\int^x b(t) dt $ when $f(x)$ > $b(x)$ for all $x$.
It should also ideally not kill any gradients, so $\Psi_{b}(f)$ should not involve hard caps that make the result completely unresponsive to changes in $f$, i.e. for all $a$, the gradient of $\Psi_{b}(f)$ with respect to $f(a)$ should be nonzero.
Finally, $\frac{d}{dx} \Psi_{b}(f)(x)$ should converge to $f(x)$ when  $f(x)/b(x) \rightarrow 0$ - i.e. the "bounded contribution integral" should act like an ordinary integral whenever the bound is not a binding constraint.
The first obvious attempt to solve this was to bound the input function $f$ below $b$ pointwise. However pointwise bounds don't work nicely with integrals because a delta function (or some close approximation thereof) can pack all the area into a tiny interval.
So, is there an operator that satisfies these requirements, or a proof that it's not possible?
Any pointers to similar work or relevant fields is appreciated.
 A: One possible answer is to bound $\int^x f(t) dt $  below  $\int^x b(t) dt $  point by point.
Let $F(x) = \int^x f(t) dt $
Let  $B(x) = \int^x b(t) dt $
Let $\tilde{F}(x) = \Psi_{b}(f)(x)$ and $\tilde{f}(x) = \frac{d}{dx}\Psi_{b}(f)(x)$ as a shorthand when the function $b$ is fixed.
We then define a sigmoid-like function $\sigma()$ where $\sigma(0) = 0$ and  $\sigma(t) \rightarrow 1$  as $t \rightarrow \infty $, with  $\sigma(t) \approx t$ for small $t$ For example, $\sigma(t) = \frac{t}{\sqrt{1+t^{2} } }$
Then $\tilde{F}(x)$ =  $B(x) \times \sigma(\frac{F(x)}{B(x)})$
This seems to have the right properties:

*

*when $F()$ << $B()$ , the sigmoid function can be approximated as the identity function and $\tilde{F}(x) \approx B(x) \times \frac{F(x)}{B(x)} = F(X)$

*When $F()$ >> $B()$ , the sigmoid can be approximated as 1 and  $\tilde{F}(x) \approx B(x) \times 1 = B(X)$

*Since $\sigma()$ and $B()$ are differentiable we can differentiate to get $\tilde{f}(x)$.


Though, in coming up with this solution I see that there's a mistake in my list of desired properties. $\tilde{F}$ as defined here is highly nonlocal, taking area from one place and moving it a long way away. It's not so much a "bounded contribution integral" as a "deferred contribution integral".
A: A second attempt at a solution to this is to concede that the idea of "bounding" the contributions of a function to an integral requires a length scale below which contributions above the bound can be deferred and above which they cannot.
So a solution that captures the intuition is to convolve the function $f()$ with a low pass filter to get rid of fluctuations on a scale you don't want, and then apply a sigmoid function to bound the filtered version of $f()$ below $b()$, and then use the ordinary integral to add up these filtered, bounded contributions.
There are three free parameters in this operator - the lowpass filter, the sigmoid function and the bounding function. There isn't really a single canonical answer here.

