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Mathematically speaking, when I try to use some signal to disprove a system is invertible, can I use the signal like $\delta(t+\infty)$ ($\delta$ representing the Dirac distribution)? For example, the two inputs $0$ and $\delta(t+\infty)$ are distinct, but their outputs from the system $\int_{-\infty}^{t}e^{\tau}x(\tau)d\tau$ will then both be $0$. In this way, can I say the system is invertible?

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    $\begingroup$ Personally, I'd say that you can't do that. In any case, it wouldn't be useful, since the system might be invertible for all signals that one actually cares about. $\endgroup$ – MBaz Feb 9 at 0:49
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It's possible, in mathematics, to complete real numbers with "infinite" values, with sound topological properties; for instance non-standard analysis or the Extended real number line (discussion at Math StackExchange).

However, in this context, for any standard real number $t$, the rule is to set $t\pm \infty = \pm \infty$ (see Arithmetic operations). So, using your notations, for every $t$, you would have [CAUTION ADVISED] $\delta(t+\infty)=\delta(\infty)$, to which I am not able to give a meaning, else than something being a constant evererywhere equal to $\delta(\infty)$. Of course, $\delta$ is a distribution, and should not be treated as a function, but defined as a functional operator, or via limits of functions.

However, so far, I have never seen

  • a (serious and useful) use of the extended real system in signal processing,
  • a reference trying to define such a Dirac at infinity.

For the latter, I fear (just intuition) that constructing this with test functions can be troublesome, since two different limits are required (location and amplitude). This is just an analogy, but $\delta(0)\times \delta(0)$ is, as far as I know, not defined.

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    $\begingroup$ The only sensible use I could see of this would be to use it as a test function, e.g. to have something to convolve a system with to get the value it converges to at $\infty$; but that would essentially require us to define that the convolution with $\delta(t+\infty)$ does that; it would feel consistent, but exactly as you state, there's nothing inherently meaningful about $\delta(\infty)$ $\endgroup$ – Marcus Müller Feb 9 at 13:30
  • $\begingroup$ Yes, I feel that one would need a double limit, one classical on the amplitude, and one on the ordinal variable, which could lead to ambiguities $\endgroup$ – Laurent Duval Feb 9 at 14:43
  • $\begingroup$ As far as I can remember when I used then, that mostly simplifies proofs and notationt $\endgroup$ – Laurent Duval Feb 9 at 20:14
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The Dirac delta generalized function is only defined under an integral

$$ Dirac(t-a) = \int_{-\infty}^\infty f(t)\delta(t-a)dt = f(a)\, $$

Typically there's also a function under the integral.

Hence

$$ Dirac(t-\infty) = f(-\infty)\, $$

if one considers $-\infty$ to be a point - not to mention problems evaluating the $Dirac$ at -$\infty$.

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