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I have a statement that leads to a paradox, but I'm incapable of finding the part where I'm wrong.

The integrator system

$$x(t) \mapsto y(t)=\int_{-\infty}^{t}{x(\tau) \, {\rm d} \tau}$$

is a linear time-invariant (LTI) system. Also, the differentiator

$$x(t) \mapsto y(t)=\frac{dx(t)}{dt}$$

is a linear time-invariant (LTI) system due to the fact that the system acts as the $D$ operator and the $D$ operator is LTI.

The inverse system of the integrator is the differentiator, and due to the commutative property of convolution, the integrator must be the inverse function of the differentiator but this is wrong since giving inputs $x_1(t)=t$ and $x_2(t)=t+5$ to the cascade of differentiator and integrator produces the same output. In other words, the integrator is incapable of retrieving the constant term of the input.

Well, what is wrong in this reasoning?

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    $\begingroup$ Exhaustive answer over on math.SE. $\endgroup$
    – Peter K. ♦
    Commented Jun 29, 2021 at 19:13
  • $\begingroup$ @PeterK. What I'm looking for is more of a system view of the problem, namely the part that the flaw occurs, this should be in either linearity and time-invariance of differentiator and integrator or in properties of convolution. Now I know that some informality such as generalized functions is introduced in signals and systems, but unless we find the part where this informality is causing this paradox, we can't be sure of the other conclusions we draw using these properties. $\endgroup$
    – Amir Soleimani
    Commented Jun 29, 2021 at 19:31

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The flaw is in the assumption that convolution is associative. This is only true if it is allowed to interchange the order of the corresponding integrals, which is not the case here because the ideal integrator is an unstable ("marginally stable") system with a pole on the imaginary axis.

On a more intuitive note, a differentiator has a zero at DC, so applying a differentiator first will remove any DC component of the input signal, which clearly can't be recovered.

Also take a look at this answer to a very related question.

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  • $\begingroup$ It is reasonable to think of this part as the part that creates a bug, but unlike associativity, the proof for the commutative property of convolution uses only one integral and a simple variable change. When using Laplace transform for proof the Fubini's theorem is violated and what you are saying is correct but I don't see how a simple variable change in the simpler proof can cause this! $\endgroup$
    – Amir Soleimani
    Commented Jun 29, 2021 at 20:03
  • $\begingroup$ @AmirSoleimani: At first I wrongly wrote "commutative" but it should have been "associative", because it involves not only a single integral but two integrals, because it is about interchanging two systems, so we have the convolution of an input signal with two impulse responses, hence two integrals. $\endgroup$
    – Matt L.
    Commented Jun 29, 2021 at 20:51

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