Will the rules of calculus stay the same when a real-valued function is defined over infinite number of variables? So the question would be:


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*Can we ever talk about a real-valued function that is defined over infinite number of variables?

*Will the rules of calculus remain the same for such functions described in 1.?  
 A: There is a theory of "calculus on Banach spaces", which comes close to what you are asking for. See for example Wikipedia:Fréchet derivatives for some information about Fréchet and Gateaux derivatives for functions $f : U \to V$ where $U$ and $V$ are Banach spaces.
Some things remain the same in the infinite dimensional setting, but some things are fundamentally different. For example the closed unit ball in a Banach space is compact if and only if the space is finite dimensional.
A: Regarding 1, of course we can. Consider $X:= \mathbb R^{\mathbb N} = \{f: \mathbb R\to \mathbb N\}$, the set of all sequences of real numbers. It's perfectly reasonable to want to talk about functions from $X$ to $\mathbb R$. Or if you don't want to have to deal with the fact that a function $f: X \to \mathbb R$ depends on "infinitely many variables", you can specialise to $c_c = \{(x_n)_{n\geq1} \in X \mid x_n$ is eventually $0\}$.
Unfortunately $X$ (and $c_c$ for that matter) is not all that well-behaved. While it does have a natural topology which is even metrisable, it's missing many of the nice properties that makes the set of real numbers so useful—for example it's not locally compact and it admits linear maps into the reals that are not continuous.
A: There is no equivalent to Lebesgue Measure, which is used to define volume and hence integration on $\mathbb R ^n$ for $n \lt \infty$, for infinite dimensional spaces. Formally this means that there is no measure on any infinite-dimensional Banach $B$ space which is:
1) Locally finite: Each point hasan open neighborhood with finite measure.
2) Strictly positive: Every nontrivial open subset has strictly positive measure.
3) Translation Invariance: If $E \subseteq B$ and $\mu(E) = L$ then $\mu(E+x) = L  \ \forall x \in B$.
This is proved by showing that we can exploit the infinite-dimensionality of the unit ball to pack infinitely many balls of radius $\frac{1}{4}$ inside it. Which leads to the conclusion that the unit ball has infinite measure.
So there is no directly analogous way to define definite integration for functions with infinite-dimensional domain.
