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By the fundemental theorem calculus we are able to say that:

The Substition Formula: If $f$ and $g'$ are continuous, then $$ \int_{g(a)}^{g(b)}f(u)du= \int_a^b f\big( g(x) \big)\cdot g'(x)dx $$

This formula is often stated for indefinite integrals by:

$$ \int f(u)du= \int f\big( g(x) \big)\cdot g'(x)dx \quad \text{where} \quad u=g(x) $$

I was trying to find a rigorous proof to the second formulation, but have come up short so far. I was wondering whether it can indeed be rigorously proved, since it is often taught in early calculus class. I found a remark in Spivak's book on calculus saying that:

"This formula cannot be taken literally (after all, $\int f(u)du$ should mean a primitive of $f$ and the symbol $\int f\big( g(x) \big) g'(x)dx$ should mean a primitive of $ (f\circ g)\cdot g'$; these are certainly not equal)."

Is there a rigorous sense in which the second formulation has a proof, or is just a symbolic manipulation?

Keen-ameteur
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    It's understood that $u:=g(x)$, in which case the two primitives are exactly equal (after substituting back $u$). – Alex R. Oct 29 '19 at 20:03

1 Answers1

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As stated by Spivak:

"This formula cannot be taken literally (after all, $\int f(u)du$ should mean a primitive of $f$ and the symbol $\int f\big( g(x) \big) g'(x)dx$ should mean a primitive of $ (f\circ g)\cdot g'$; these are certainly not equal)."

Using the chain rule, if $F$ is a primitive of $f$ then $F \circ g$ is a primitive of $(f \circ g) g'$ since

$$\frac{d}{dx}F(g(x)) = F'(g(x)) g'(x) =f(g(x))g'(x)$$

If we interpret primitive or antiderivative precisely as a function whose derivative is the integrand, then the equality is false. This is what Spivak is saying. If as in the comment we interpret the operation $\int f(u) \, du$ to mean write $F(u)$ with $g(x)$ substituted into that expression for $u$ then it has some validity.

RRL
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  • Do you mean by this that the formula for indefinite is not true and cannot be proven? – Keen-ameteur Oct 29 '19 at 19:54
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    Well $F\circ g \neq F$ as they are different functions. The "second formulation" is a kind of shorthand that implies the change of variables result for definite integrals. – RRL Oct 29 '19 at 19:56
  • I would think that it does. – Keen-ameteur Oct 29 '19 at 20:20
  • Only up to an arbitrary constant. Maybe its not the best analogy. Nevertheless I think the second formulation is just an imprecise way (not to be taken literally as said by Spivak himself) to convey the first formulation without being careful about integration limits. You asked for a proof and I gave you the proof that they are not equal. – RRL Oct 29 '19 at 20:25
  • ... if we interpret primitive or antiderivative as a function whose derivative is the integrand. – RRL Oct 29 '19 at 20:32
  • My problem is that I am currently involved with a course which teaches the second formulation prior to the first, and as such I am having problem "rigourising" indefinite integral computations using it. – Keen-ameteur Oct 29 '19 at 20:33
  • Please see my edit. I am sticking to the precise meaning of a term like anti-derivative. One can interpret the phrase " where $u = g(x)$" as one sees fit. I am agreeing with Spivak. – RRL Oct 29 '19 at 20:38