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In my physics book I saw the following math snippet:

Let $$y(t)=\sin(t)\int_{-\epsilon}^{\epsilon}x(\tau)\cos(t-\tau)d\tau$$ be the output signal for input signal $x(t)$.

So, as a mathematician, I'm interested in how we can describe this from mathematical point of view. What is $y$ here? Apparently, $y$ is a function which maps elements from one set (domain) to elements of another set (range). The range is obviously $\mathbb{R}$ (or some subset of it). But, what is the domain?

As I understood, $y$ has the above form for all $x$ and for all $t$. So, it doesn't really mean that $y$ depends only on $t$, it also depends on $x$. But, it means that $y$ can be written as $y(x, t)$. However, $x$ is a function (in the book it is earlier described that $x:\mathbb{R}\mapsto\mathbb{R}$), so what is then the domain of $y$?

For example, consider the following function $f:\mathbb{R^2}\mapsto\mathbb{R}$. This function takes two real numbers and returns one real number (actually it takes an ordered pair of real numbers). From the ZFC point of view, we can say that $\forall a,b,c\in\mathbb{R}\left(f(a,b)=c\iff\left((a,b),c\right)\in f\right)$.

Now, let's back to $y$. We have that $y$ takes two arguments, the first one is $x$ (input signal) and the other one is $t$ (time variable). Time variable is a real number, so we have $t\in\mathbb{R}$, that's obvious. But, what about $x$? Again, according to ZFC, $x$ is some sort of set. It can be represented as $x=\left\{(a,b)\mid x(a)=b\right\}$. So, to determine the domain of $y$, I need to know the set which contains exactly all possible values of set $x$, right? But, what is that set?

Let's call $X$ the set that contains all possible values of set $x$ and nothing more. First of all, notice that $\left|x\right|=\left|\mathbb R\right|$ because $x$'s argument $a$ (the first element of ordered pair) can contain only real numbers and can appear exactly once. How many different functions $x$ can exist? Because $a$ is fixed (we know that there will be $a$ for all reals and that it will appear only once), so how many different $b$'s can exist for every $a$? My understanding is that there are $\left|\mathcal{P}\left(\mathbb R\right)\right|$ (power set of reals) different functions $x$. Therefore, what is the domain of $y$?

My opinion is that the domain of $y$ is (according to the above paragraphs) actually $\mathcal{P}\left(\mathbb R\right)\times\mathbb R$. Is it correct? If no, then why not? If yes, then ok.

Thanks, I appreciate your work in advance to anyone who decide to help.

Edit

This is answer to @PaulSinclair's comment. My question is not related to the exact same formula from my book. Instead, my question is related to finding domain of a function which takes another function as an argument. If it is easier to you, forget about the snippet I posted. Just consider a simplified example:

Let $f:D\mapsto\mathbb R$ be a function which takes two arguments: $t\in\mathbb R$ and arbitrary function $g:\mathbb R\mapsto\mathbb R$. Find the domain $D$ (domain of function $f$).

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  • $\begingroup$ Gosh! $y$ is also function of $\epsilon$. And whoah! If you give a different meaning to $\int$, then it gives a different answer to $y$! So $y$ is also a function of integration! You should definitely think of it as $y(t, x, \epsilon, \int \cdot d\tau)$ with domain $\Bbb R \times \scr F(\Bbb R, \Bbb R)\times (0, \infty) \times \mathcal I$, where $\scr F(\Bbb R)$ is the set of integrable functions on $\Bbb R$, and $\mathcal I$ is a set of definitions for integration! Or, you could be sensible and accept everything but $t$ as a given constant and just take $\operatorname{Dom}(y) = \Bbb R$. $\endgroup$
    – Paul Sinclair
    Commented Nov 16, 2017 at 0:35
  • $\begingroup$ @PaulSinclair. Added an edit, PTAL. $\endgroup$
    – user503399
    Commented Nov 16, 2017 at 0:58
  • $\begingroup$ The domain of $f$ in your edit is $\Bbb R \times \Bbb R^{\Bbb R}$. (If you've not seen it before, $A^B$ is a common notation for the set of all maps from $B$ into $A$, which comes from the fact that if $A$ and $B$ are finite, then $|A^B| = |A|^{|B|}$) $\endgroup$
    – Paul Sinclair
    Commented Nov 16, 2017 at 3:18
  • $\begingroup$ @PaulSinclair. "If you've not seen it before" - I've actually seen it before, but my question is: 1) Is $\mathcal P\left(\mathbb R\right)={\mathbb R}^{\mathbb R}$ (or just their cardinality?) and 2) Is ${\mathbb R}^{\mathbb R}\times\mathbb{R}={\mathbb R}^{\mathbb R}$? $\endgroup$
    – user503399
    Commented Nov 16, 2017 at 9:26
  • $\begingroup$ No: $\Bbb R^{\Bbb R} = \{f \subseteq \Bbb R \times \Bbb R \mid \pi_1(f) = \Bbb R \text{ and }\forall x,y \in f, \pi_1(x) = \pi_1(y) \implies x = y\} \subsetneq \mathcal P(\Bbb R^2)$ (not $\mathcal P(\Bbb R))$. And while $\Bbb R^{\Bbb R}$ is equinumerous with $\Bbb R^{\Bbb R}\times \Bbb R$, they are not the same set. $\endgroup$
    – Paul Sinclair
    Commented Nov 16, 2017 at 18:27

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