4
$\begingroup$

Given a definite integral $$I=\int_a^b f(x)\text{d}x$$ for some function $f$, we can make any number of substitutions $x=u(x)$ which produce symbolically new integrals, all of which evaluate to the same value.

My main question is this: is it possible to have two definite integrals both of which give the same value yet each of which can not be transformed (i.e. made to look the same symbolically) into the other by substitution?

I get the impression the answer to my question is YES, but I'm not sure. Is there any information or theory out there that deals with this? How can we know when the transformation can not be done (if indeed it cannot be done)?

  • 0
    What are the rules of the game here? That is, what types of algebraic manipulations do you want to consider?2012-12-12
  • 0
    @orlandpm I'm just thinking of any algebraic substitution for the variable of integration.2012-12-12
  • 2
    Ok here's one that definitely provides a counterexample. Take $\rho$ to be a smooth function that vanishes in some interval $[0,a)$, with $0< a< 1$, and such that $\int_{0}^1\rho(x)\,dx = 1$. Now assume that we can find a substitution $u(x)$ such that $\rho(u(x)) \,du(x) = dx$ and $u(0) = 0$, $u(1) = 1$. This means that $\rho(u(x))u'(x)$ is identically $1$. But $\rho(u(x))$ must vanish in a neighborhood of $0$, so $\rho(u(x))u'(x)$ must vanish there too; this is a contradiction.2012-12-12
  • 1
    Given any two definite integrals, you can multiply by constants to produce integrals that are equal.2012-12-12
  • 0
    @DavidMitra +1, although I wanted to consider "non-trivial" substitutions which exclude multiplication by constants.2012-12-12
  • 0
    That's not what I meant to suggest. If it were true that equal integrals can be transformed into each other via substitution, then any two integrals could be transformed into each other, after multiplying one of them by an appropriate constant.2012-12-12

3 Answers 3