Are there  functions satisfying the following integral  condition? Can we find two functions $f$ and $g$ that are reasonably defined nontrivial(not everywhere zero, $f\neq g$, not linear polynomials) functions such that the following condition is satisfied?
$$ f( \left(\int_{0}^{t} g(x) \ \text{d}x\right))  = g( \left(\int_{0}^{t} f(x) \ \text{d}x\right)) $$
P.S.: I migrated this question from here on   Math.SE. I am sure this site hosts very knowledgeable mathematicians that keeping on migrating to another site is foolish. I felt a really good feeling for some time as nobody answered my question. But is usually the case that: "There is a general principle that a stupid man can ask such questions to which one hundred wise men would not be able to answer. In accordance with this principle I shall formulate some problems." Vladimir Arnold
Motivation:  The equation that I wrote out was not random. At least, the symmetry I find in it and the absence of an iota of clue at proceeding with any method makes me fall in love with finding a solution. Part of the motivation was to find a function that in some way resembles the exponential function. The exponential map is invariant under differentiation. So, the natural curiosity to find a nontrivial map invariant under integration. For obvious reasons, such map does not exist because of the presence of the constant of integration in indefinite integrals. Hence, I added an extra condition that would make the would-be function more nontrivial and more appealing.
 A: Take, e.g., two (distinct, non-trivial) bump functions $F$ and $G$ s.t. $supp\: F\cap G\left(\mathbb{R}\right)=supp\: G\cap F\left(\mathbb{R}\right)=\emptyset$
. Then their derivatives $f=F^{\prime}$, and $g=G^{\prime}$ are clearly satisfying the required identity.
A: I'm not sure I understand your quotation. It could equally be "A small child can find a rock so heavy that a hundred strong men could not lift it. In accordance with this principle, I shall go around pointing out heavy rocks." 
Part of the skill in mathematics is knowing which problems from the millions available are likely to be solvable. You should probably try to give some explanation as to why you expect a solution - or at least, why you want one - so people have a reason to think about this particular question over any other.
(This is a comment, not an answer, but I don't have the power)
A: For real analytic functions, just looking at the power series gives information. For example, it seems that if all the coefficients of the power series expansion are nonzero, then the only solution (class) is $f=g.$
A: Following Igor's comment(or his hunch), I started out with finding polynomial counterexamples of the type: $f(x)= ax^n $ and  $ g(x)= bx^m $ on $(0,\infty)$. It is easy to see that such polynomials satisfy the integral condition only if $m=n$, and $a=b$ if $n$ is even or $a=\pm b$ if $n$ is odd. Then a class of counterexamples on $(0,\infty)$ can be constructed by letting $a= -b$ with odd $n\geq3$ .
A: The problem does not seem as far from the main stream as it might first appear. Pick any given function g and let G denote the integral of g. Let F denote the integral of f. Then the equation becomes F'(G(x))=g(F(x)), with initial condition F(0)=0. In other words, for any given g, the problem reduces to a functional differential equation. 
