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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?

$$ f \left(\int_{0}^{t} g(x) \ \text{d}x\right) = g \left(\int_{0}^{t} f(x) \ \text{d}x\right) $$

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    you might want to remove the "quote" code so that LaTeX shows up.2011-01-18
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    and check your title for typos...2011-01-18
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    Why don't you take $f=g=e^x$?2011-01-18
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    Using antiderivatives does not remove the problem caused by constants of integration. Antiderivatives are still only unique up to constants. Are you using "ad" to refer to _all_ of the antiderivatives?2011-01-21
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    @Theo, thanks for your observation. I have now edited the question.2011-01-21
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    @Qiaochu, by ad I meant only one particular antiderivative.2011-01-21
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    @Chulumba: which one?2011-01-21
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    @Chulumba: There is no well-defined notion of having "zero as the constant of integration". The closest thing that makes sense is in fact $\int_0^x f(t) dt$ (try it for your example and see that it does give what you expect), and so Moron's answer is valid.2011-01-21
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    Okay, for preciseness, let's take the constants of integration to be $0$ in all cases.2011-01-21
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    Rahul is right. And it's important to understand that, that "There is no well-defined notion of having 'zero as the constant of integration'"2011-01-21
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    What I meant was "antiderivatives with zero as their constants". Isn't this well-defined? Your proposition is also okay. I agree.2011-01-21
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    But in the answer Moron gave below, the integrands are products of functions not compositions of functions in the sense I asked.2011-01-21
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    I have made an edit that I think makes my question as clear as can be. Thanks for everybody's participation and I look forward to hearing your response for the now-clear question.2011-01-21
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    The upper limit of the outer integrals probably should be something like $y$, not $x$ which you have already used in the inner integral...2011-01-21
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    Are the outer integrals necessary at all? If two definite integrals from 0 to $x$ are equal for all $x$, then the integrands are equal except on a set of measure zero, right?2011-01-21
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    @Chulumba: $f = C_1 x, \ g = C_2 x$ is not diallowed by the current problem.2011-01-21
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    @Willie, $x$ is a dummy variable. Once the inner integral is evaluated, $x$ is history, so can be reused as an upper limit of the outer integrals.2011-01-22
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    @mjqxxxx, assume that I am interested over sets of measure 0, too. So no need to remove the outer integrals.2011-01-22
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    @ Moron, thanks. I had it in my first statement of the question but forgot to exclude it while editing. I have now edited again.2011-01-22
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    @Chulumba: I doubt that you _are_ interested in sets of measure zero. Otherwise let $f(x)$ be any function, and let $g(x) = f(x) + I_A(x)h(x)$, where $I_A$ is the indicator function over $A$, and $A$ has measure zero, and $h(x)$ is any function.2011-01-22
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    @mjqxxxx, sorry. My knowledge of measures is only superficial, hence missed to see that possibility. Thanks. So, you have proved me that I need to give my question a finishing touch!2011-01-22
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    @mjqxxxx, corrected thanks.2011-01-24
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    You can write this as $$F'\circ G = G' \circ F,$$ $$F(0)=G(0)=0$$ to analyze it more functionally.2011-08-16
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    Since for many elementary functions there is (sort of) an "obvious" antiderivative (e.g. sin(x) makes more sense than sin(x)+7 as an antiderivative of cos(x)), I just want to point out a situation in which there is no reasonably consistent way to "set $C=0$": Consider $I=\int 2\sin(x)\cos(x)dx$. The substitution $u=\sin(x)$ gives $I=\int 2udu = u^2+C_1=\sin(x)^2+C_1$. On the other hand, using the double angle identity for sine gives $I = \int\sin(2x)dx = -\frac{1}{2}cos(2x)+C_2$. These two forms are connected by the double angle identity for cosine but give different answers when $C_1=C_2$2011-08-16
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    Can x be complex?2011-08-29

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If there is such a function, it can't be represented by Tailor series:

Given: $f( \left(\int_{0}^{t} g(x) \ \text{d}x\right)) = g( \left(\int_{0}^{t} f(x) \ \text{d}x\right))$

$f( G(t) $ -$ G(0)$ ) = g( F(t) -$ F(0)$ )

$g(x)=g_0+g_1x+g_2x^2+...$

$f(x)=f_0+f_1x+f_2x^2+...$

$G(x)=g_0x+g_1\frac{x^2}2+g_2\frac{x^3}3+...$

$F(x)=f_0x+f_1\frac{x^2}2+f_2\frac{x^3}3+...$

$f_0+f_1(g_0x+...)+f_2(g_0x+...)^2+...=g_0+g_1(f_0x+...)+g_2(f_0x+...)^2+...$

Since there's only one term of $kx^0$ and $kx^1$ on each side, we can conclude their respective constants are equal:

$f_0=g_0,f_1=g_1$

For every power afterward it is possible to substitute every previous set of constants in order to obtain another.

$f_2=g_2, f_3=g_3, ...$ (Hence $f(x)=g(x)$)

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    Bummer, dude...2011-08-16
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    I'm confused. Theo, were you commenting on a different version of the answer than what is there now?2011-08-16
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    @Robert: Yes, my comment addressed the first few versions of this post but I removed it shortly after the re-write to the current version happened, as it had no longer anything to do with the answer. Apparently I must have done so just an instant before you posted your comment, because I see that comment only now.2011-08-17