Earlier today I was having a little fun with Catalan's constant and its various integral representations: showing that they all do indeed evaluate to the same thing. This got me wondering whether this is always possible, if we are given several integral representations of the same real number. I then thought of a counter-example:


But I partly put this down to the fact that the integral on the left is non-elementary, whereas the one on the right is not.

What I am more interested in, is: if we consider two elementary integrals such that:

$$\int_a^b f(x)\;dx=\int_c^d g(x)\;dx$$ Does there exist a chain of manipulations which will lead us from one to the other?

One could also ask the same question about non-elementary integrals (edit: it was recently pointed out to me something like this might be a counter-example to this second case).

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    Note that there always exist real numers $u,v$, not both zero, such that$$\int_a^buf(x)\,dx=\int_c^dvg(x)\,dx.$$ Thus, such an equality of integrals can hardly be more than condiered "accidental". – Hagen von Eitzen Jan 26 '13 at 17:46
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    If, in addition, the integrands are positive, then there is a substitution that converts one integral into the other. This is a special case of [Moser's theorem](http://mathoverflow.net/questions/82581/): if two volume forms give the same mass to a compact manifold, one is the pushforward of the other. –  Jan 26 '13 at 17:59
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    I won’t downvote, but it seems to me, in agreement with @Hagen, that the rules of this game are so lax that it’s much too easy to win. – Lubin Feb 06 '13 at 21:59
  • I hate to say it, but if A=B and you want to manipulate A to obtain B, then... you don't have to do anything, just scratch out A and write B, whereas they are equal. Alternatively, multiply A by 1=B/A (assuming A≄0), cancel the A's, and you have B... So clearly, yes, there are manipulations you can do to one definite integral to obtain another, if you know they are equal. – Douglas B. Staple Mar 24 '13 at 03:36
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    What happens if you restrict the space that the functions exist in? – yiyi Mar 28 '13 at 07:31

3 Answers3


Consider the integrals $$ I=\int_a^b f(x)\;\mathrm dx\qquad J=\int_c^d g(y)\;\mathrm dy $$ and assume that $f\gt0$ everywhere and that $I=J$. Then the following change of variable transforms $I$ into $J$. Consider the primitives $F$ and $G$ of $f$ and $g$ defined by $$ F(x)=\int_a^x f(t)\;\mathrm dt\qquad G(y)=\int_c^y g(s)\;\mathrm ds $$ and the change of variables $x=u(y)$ defined on $(c,d)$ by $$ u(y)=F^{-1}(G(y)) $$ Then $F'(u)\mathrm du=G'(y)\mathrm dy$, that is, $f(u(y))u'(y)=g(y)$, thus, this change of variables transforms $I$ into $J$. If $f$ is not positive everywhere, consider $f_c=f+c$ with $c$ large enough and apply the above to $f_c$ and to the corresponding $g_c$.

Is the change of variable $u$ admissible? This depends on your definition of "admissible" but if the functions $f$ and $g$ are elementary and if the property of being elementary is preserved by taking a primitive and by taking the inverse, then the change of variable $u$ is indeed elementary.

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To make this a reasonable question you need to impose some more constraints. A good constrained version of this problem is to consider only algebraic functions and integrate them over domains which are given by polynomial inequalities, where all of the coefficients are algebraic. The values of such integrals are called periods. A simple example is $\pi$ because you can get it from integrating $3/4$ over the region $x^2+y^2+z^2<1$ in $\mathbb{R}^3$.

There is a conjecture by Kontsevich and Zagier that there is an algorithm that can convert one of these integrals into another in a finite number of elementary steps, just like you ask for. However, it is just a conjecture at this point. You might enjoy reading what they have to say on this subject: Periods.

Dan Piponi
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Maybe consult the material on "decreasing rearrangement" of integrals. Say, in Hardy, Littlewood, Polya, INEQUALITIES.

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