I can't understand some facts on the substitutions in indefinite integrals.

On my textbook is reported only the "standard" case (integral of a composed function and the derivative of the inner function):

Considered two intervals $I$ and $J$ let

.$f: I\rightarrow \mathbb{R}$ be a function that has an antiderivative $F(x)$ on $I$

. $\phi : J\rightarrow I$ be a function differentiable on $J$.

Then the function $f(\phi(x))\phi’(x)$ is integrable and we have

$\int f(\phi(x))\phi’(x)dx=F(\phi(x))+c$

The theorem is clear, but, reading other books, I saw that there are at least two other cases of substitutions in which there are different (and more) conditions to impose, since it is necessary to use the inverse of the function $\phi$.

I report the theorem(s) about these two cases

Considered two intervals $I$ and $J$, let

.$f: I\rightarrow \mathbb{R}$ be a function continuous $I$

.$\phi : J\rightarrow I$ be a function differentiable with continuous derivative on $J$ with $\phi’(x)\ne 0 \forall x \in J$

Some of the versions of the theorem now add the following condition:

.$f(J)=I$ (i.e. $\phi$ surjective)


$\int f(x) dx= \int f(\phi(t))\phi’(t)dt$

(using substitution $x=\phi(t)$, from which, since $\phi$ is invertible, we get back the $x$ with $t=\phi^{-1}(x)$)


$\int f(\phi(x)) dx= \int f(t)[\phi^{-1}(t)]’dt$

(using substitution $t=\phi(x)$, from which, since $\phi$ is invertible, we get the $dx$ to substitute in the integral)

The condition of $f$ continuous implies (Fundamental Theorem of Calculus) that it has an antiderivative and $\phi’(x)\ne 0$ in $J$ means that $\phi$ is injective (which means invertible).

Firstly I can't understand why it is required for $\phi(x)$ to have continuous derivative, instead of just being continuous (as in the first theorem, taken from my textbook).

I see that in the same theorems regarding definite integrals the same condition ($\phi \in C^{1}$)is imposed, but I think that in that case it is strictly necessary to have the function that we want to integrate continuous, so that we can apply the fundamental theorem of calculus. But is it necessary to impose it also for indefinite integrals, or maybe just in the second and in the third case I listed?

Secondly is it strictly necessary for $\phi(x)$ to be surjective besides being injective (so invertible)?. In other terms is it necessary that $\phi$ is a bijection between $I$ and $J$ or is it enough for it to be invertible (i.e. $\phi^{-1}$ exists)?

Thanks a lot in advice

  1. Let me clear up a misunderstanding / misuse of a word: "integrable" does not mean "has an antiderivative". In means that a certain definite integrals exists and is finite. In particular: Not every continuous function is integrable.
  2. $\phi$ being injective is not the same as being invertible. Without bijectivity there is no $\phi^{-1}$. And yes, if you want to say anything about the inverse of $\phi$ (that it satisfies certain integral equations for example) then $\phi^{-1}$ should exist. Otherwise you're making vacous statements.
  3. $\phi$ being continuously differentiable and bijective means that the inverse is continuously differentiable too. If you just assume continuity the inverse might not be continuous (although in your case you get that one for free since you're only considering intervals) and there is even less reason to think that $\phi^{-1}$ is differentiable. So again: Without such a condition you will not be able to even write down $(\phi^{-1})'$ or the integral equation meaningfully.
  • $\begingroup$ Concerning 3. One can surely do with less than continuous differentiability, but some assumptions are needed, in order to have all integrals make sense. After all, derivatives might be almost anything and in particular they can be non integrable. If one wants to relax the continuity assumption, however, one needs to deal with more complicated scenarios that belong to geometric measure theory more than they belong to calculus. $\endgroup$ – Giuseppe Negro Jan 13 '16 at 13:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.