2
$\begingroup$

I know that the substitution rule works like this:

By the chain rule: $$F(g(x))' = f(g(x))g'(x)$$

Then $$\begin{align} \int_a^b F(g(x))'dx &= \int_a^b f(g(x))g'(x)dt\\ F(g(b)) - F(g(a)) &= \int_a^b f(g(x))g'(x)dt\\ \int_{g(a)}^{g(b)} f(u)du &= \int_a^b f(g(x))g'(x)dt \end{align}$$ Perfectly fine. The problem with this proof is that it uses the fact that $F(g(b))-F(g(a))$ is the same as the integral of a function $f(u)$ from $g(a)$ to $g(b)$. However, if we just want to find the antiderivative, that's a problem.

In this proof, he says:

$$\displaystyle \int f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \ \mathrm d u = \int f \left({x}\right) \ \mathrm d x$$ where $x = \phi \left({u}\right)$

I can't accept this. It is basically saying that we should integrate with respect to $\phi(u)$. Not saying that this is not possible, but it requires further theory about integration that is not on the scope of this proof. I realy really can't understand this proof as valid. The other text books I've read also say the same thing...

$\endgroup$
4
+50
$\begingroup$

In this proof, he says: $$ \int f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \ \mathrm d u = \int f \left({x}\right) \ \mathrm d x $$ where $x = \phi \left({u}\right)$.

Maybe I've misunderstood your objection, but the linked proof asserts that if $\phi$ is differentiable on $[a, b]$, $I$ is an open interval containing the image $\phi\bigl([a, b]\bigr)$, the function $f$ is continuous on $I$, and $F$ is an antiderivative of $f$ on $I$, then $$ \int (f\circ \phi)\, \phi' = F\circ \phi. $$ This merely re-expresses the chain rule, provided an indefinite integral is a synonym for a general antiderivative.

I'll grant that the notation with dummy variables of integration is less than ideal in this situation, but the assertion itself doesn't look controversial.

$\endgroup$
4
$\begingroup$

As stated in many calculus textbooks (and ProofWiki),† the Substitution Rule (for the indefinite integral) is wrong. It is usually stated as:

$$\int f \left({g \left({x}\right)}\right) g' \left({x}\right) \ \mathrm d x = \int f \left({u}\right) \ \mathrm d u. \tag{1}$$

But as noted by David Gale in "Teaching Integration by Substitution" (1994, PDF):

Of course the [above] equation is false. The expression $\int f(x) \ \mathrm dx$ stands for antiderivative, as in a table of integrals, and the variable, be it $x$, $t$, $u$ or anything else is a dummy. Clearly the antiderivatives on the left and right above are not equal. What the books mean, no doubt, is that if you substitute $g(x)$ for $u$ after taking the antiderivative on the right you get the antiderivative on the left. I expect some readers will say I am being pedantic or that there is no need to be so rigorous at the freshman level, but I think this kind of lapse is symptomatic of a rather strange set of standards and perhaps it sheds light on why none of the books proves the inverse substitution theorem. It is because none of them formulates it.

Following Gale's remarks, here is one way we can correct $(1)$ without modifying it too much:

$$\int f \left({g \left({x}\right)}\right) g' \left({x}\right) \ \mathrm d x = \int f \left({u}\right) \ \mathrm d u \Bigg|_{u=g(x)}.$$

Or alternatively, if $F'=f$, then:

$$\int f \left({g \left({x}\right)}\right) g' \left({x}\right) \ \mathrm d x = F\left(g(x)\right)+C.$$

Or:

$$(f\circ g)\cdot g'=(F\circ g)' \iff F'=f.$$

Proof. By the Chain Rule, $(F\circ g)'=(F'\circ g)\cdot g'$.

Thus, $(F\circ g)'= (f\circ g)\cdot g' \iff F'=f$. ∎

†See e.g. ProofWiki, Stewart (2011), Thomas & Finney (1996, p. 294).

$\endgroup$
2
$\begingroup$

I think the books should write:

$$\int_a^b f(g(t))g'(t)dt = \int_{g(a)}^{g(b)} f(x)dx$$

And: $$\int f(g(t))g'(t)dt = F(g(x))$$

But not

$$\color{Red}{\displaystyle \int f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \ \mathrm d u = \int f \left({x}\right) \ \mathrm d x}$$ $\color{Red}{\mbox{where }x = \phi \left({u}\right)}$

But the problem of finding the antiderivative remains unsolved. At least, it can't be proven this way rigorously.

$\endgroup$
1
$\begingroup$

Firstly, please note: This would really be better suited as a comment, but it's too long, so I've made it an answer. Please don't downvote just saying "should be a comment" - it's too long! :)

Here goes. I see below you have mentioned about integrating with respect to other functions and needing further integration theory to do so. It may pay to look into some measure theory. In particular, image measures and (of course) integration by substitution. Here is a link to the lecture notes for a lecture course that I'm doing, given last term: Probability and Measure, Lecture Notes.

In particular, chapter 2. Theorem 2.22 is "Change of Variables". It is only a sketch proof, however (unfortunately!).

Basically, the way I would recommend that you try to prove your statement is by considering it for so-called simple functions (ie (finite) sums of indicator functions) initially. In fact, you can probably show quite straightforwardly that the set of functions for which you claim holds is a vector space. Then try to prove that it holds for just an individual indicator function. Given this, any simple function has the property (as in a vector space). Also show that it holds for constant functions.

Then, either you know, or you (one) can show that any measurable function is the limit of simple functions (Monotone Class Theorem, Theorem 1.33). Then apply the Monotone Convergence Theorem (Theorem 2.7).

This is certainly how I would attempt it.

Hopefully this helps! :)

$\endgroup$
  • $\begingroup$ link is broken, you still have it? $\endgroup$ – Guerlando OCs Nov 9 '15 at 5:57
  • $\begingroup$ The lecturer has moved away, and his links have been taken down. I've updated with a new link -- same course, taught by the bloke before (who's now teaching me Advanced Probability). $\endgroup$ – Sam T Nov 9 '15 at 13:58
1
$\begingroup$

I do not have a reputation of 50 of more so I cannot comment directly on Lucas Zanella's original post dated Sep 15 2014. That said, I too feel like the proof in [1] for indefinite integrals is incomplete.

From the `Proof for Indefinite Integrals' in [1], I recap as follows:

Let $\displaystyle F \left({u}\right) = \int f \left({u}\right) \ \mathrm d u$ (*).

By definition $F \left({u}\right)$ is an antiderivative of $f\left({u}\right)$.

Further, by the Chain Rule

$ \frac {\mathrm d F \left({\phi \left({u}\right)}\right)} {\mathrm d u} = f \left({\phi \left({u}\right)}\right) \frac{d {\phi \left({u}\right)} }{du} $

So $F \left({\phi \left({u}\right)}\right)$ is an antiderivative of $f \left({\phi \left({u}\right)}\right) \frac{d {\phi \left({u}\right)} }{du}$.

Therefore:

$\displaystyle \int f \left({\phi \left({u}\right)}\right) \frac{d {\phi \left({u}\right)} }{du} \ \mathrm d u = F \left({\phi \left({u}\right)}\right) + C$, where $C$ is a constant.

Further, by assumption (*) above,

$\displaystyle \int f \left({\phi \left({u}\right)}\right) \frac{d {\phi \left({u}\right)} }{du} \ \mathrm d u = \int f \left({\phi(u) }\right) \ \mathrm d \phi(u) + C$.

Next, dropping the compound argument

$\displaystyle \int f \left({\phi \left({u}\right)}\right) \frac{d {\phi \left({u}\right)} }{du} \ \mathrm d u = \int f \left({\phi }\right) \ \mathrm d \phi + C$.

Please note that it has NOT be shown that

$\displaystyle \int f \left({\phi \left({u}\right)}\right) \frac{d {\phi \left({u}\right)} }{du} \ \mathrm d u = \int f \left({\phi }\right) \ \mathrm d \phi + 0$.

References.

[1] https://proofwiki.org/wiki/Integration_by_Substitution#Proof_for_Indefinite_Integrals

$\endgroup$
0
$\begingroup$

Using the differential of $\phi$

$$ \phi'(u)\ \mathrm du = \mathrm d(\phi(u)) \implies \int f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \ \mathrm d u = \int f \left({\phi \left({u}\right)}\right) \mathrm d(\phi(u)) $$

now let: $\mu=\phi(u)$

$$ \int f \left({\phi \left({u}\right)}\right) \mathrm d(\phi(u)) = \int f \left(\mu\right)\ \mathrm d\mu $$

this is not formally rigorous, but it should be able to convince you (if not, i'll try to find a more formally correct argument)

$\endgroup$
  • $\begingroup$ That's what I understood from the article, but is is not rigorous. Integrating with respect to another function requires more theory about integration. $\endgroup$ – Lucas Zanella Sep 15 '14 at 8:12
  • $\begingroup$ @LucasZanella you could try to fix your proof for definite integrals to make it work for the indefinite one's hint: substitute $b$ with $x$ $\endgroup$ – Francesco Alem. Sep 15 '14 at 8:14
  • 1
    $\begingroup$ Ive though of somethinf in the form $\int^{g(u)} f(x)dx = \int^u f(g(x))g'(x)dt$ but such notation (just the upper bound) does not exist $\endgroup$ – Lucas Zanella Sep 15 '14 at 8:18
  • $\begingroup$ okay but you can put an arbitrary lower bound, then add a constant of integration and you're almost done. $\endgroup$ – Francesco Alem. Sep 15 '14 at 8:20

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.