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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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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? – Qiaochu Yuan Jan 21 '11 at 9:49
@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. – Rahul Jan 21 '11 at 13:38
But in the answer Moron gave below, the integrands are products of functions not compositions of functions in the sense I asked. – Chulumba Jan 21 '11 at 13:52
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? – mjqxxxx Jan 21 '11 at 17:26
@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. – mjqxxxx Jan 22 '11 at 12:41

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)$ )






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


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... – The Chaz 2.0 Aug 16 '11 at 21:58
I'm confused. Theo, were you commenting on a different version of the answer than what is there now? – Robert Israel Aug 16 '11 at 23:27
@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. – t.b. Aug 17 '11 at 0:16

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