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I am wondering what is the derivative of the following function with respect to $x(t)$ in sense of distributions. $$ I\left(\int_0^t x(\tau)d\tau \leq c\right) $$ where $I$ is the indicator function and $c$ is a constant.

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I don't think one can interpret such a derivative in the sense of distributions. Distributions generalize locally integrable functions; they require an underlying measure. The space of functions $x(t)$ is infinite-dimensional and does not carry a natural measure. A functional on this space is not a distribution, and neither is its derivative.

Formally, replacing $x(\tau)$ with $x(\tau)+\epsilon \phi(\tau)$ and taking the derivative with respect to $\epsilon$ at $\epsilon=0$ taking the difference of indicator functions, and passing to the limit we arrive at something that can be written as $$ \delta\left(\int_0^t x(\tau)\,d\tau = c\right)\,\operatorname{sign} \phi \tag1$$ The meaning of (1) is that integrating $$ \delta\left(\int_0^t (x(\tau)+\epsilon\phi(\tau))\,d\tau = c\right)\,\operatorname{sign} \phi $$ over $\epsilon \in (a,b)$ gives $$ I\left(\int_0^t (x(\tau)+b\phi(\tau))\,d\tau \ge c\right)- I\left(\int_0^t (x(\tau)+a\phi(\tau))\,d\tau \ge c\right)$$

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