Often I see the technique change of variables applied to the domain of integration where one replaces the integration symbol with a derivative:

$$ dx = dx\frac{du}{du} = \frac{dx}{du}du = \left(\frac{du}{dx}\right)^{-1}du $$

where $u(x)$ denotes the replacement variable. This apparently relies on the interpretation of $dx$ and $du$ as infinitesimally "small pieces" which seems to be disputed.

I know that when it comes to integration the correct way of applying this change of variables is via integration by substitution. Let's consider the following example:

$$ \int_a^bx^3dx \stackrel{u(x) \equiv x^2}{=} \int_{u(a)}^{u(b)}\sqrt{u}\cdot u \underbrace{\left(\frac{du}{dx}\right)^{-1}}_{\frac{1}{2x} = \frac{1}{2\sqrt{u}}}du = \frac{1}{2}\int_{a^2}^{b^2}u\,du = \left[\frac{u^2}{4}\right]_{u=a^2}^{u=b^2} $$

Actually this is obtained via integration by substitution:

$$ \int_a^bx^3dx = \frac{1}{2}\int_a^b2x\cdot x^2dx\stackrel{u(x) \equiv x^2}{=} \frac{1}{2}\int_{u(a)}^{u(b)}u\,du = \left[\frac{u^2}{4}\right]_{u=a^2}^{u=b^2} $$

Now for this example the latter decomposition into substitute $u(x)$ and its derivative $u'(x)$ is obvious but often, I find, this is not the case. Consider for example:

$$ \int_a^b J_0\left(\exp(\sigma\cdot x)\right)dx $$

where $J_0$ denotes the Bessel function of first kind and order. Finding such a decomposition here is non-obvious but on the other hand it's tempting to apply a change of variables $u(x) \equiv \exp(\sigma\cdot x)$ which leads to:

$$ \frac{1}{\sigma}\int_{\exp(\sigma a)}^{\exp(\sigma b)} \frac{J_0(u)}{u}du $$

which can be solved using Meijer G-functions. Now in reverse order it's possible to deduce the decomposition from the integrand but carrying out the integration by substitution is way more complicated.

So this eventually brings me to my question. Since often it is easier (more obvious and more convenient) to apply a change of variables than to perform integration by substitution, are there any limitations to this method (in terms of applicability) I should be aware of? Can I always perform a change of variables or do I need to crosscheck its validity by deducing the corresponding substitution rule? Or are the two methods actually (inherently) the same?


Change of variable and substitution are the same.

The difference is only in names. In general whenever the substitution makes an integral easier to evaluate we change variable.

  • $\begingroup$ Actually my question is not about terminology. I understand that integration by substitution formally involves a "change of variable" but by the latter I referred to the trick of treating $dx$ as an infinitesimal distance and hence converting it into a derivative. For me it is not clear if this trick can be applied under any circumstances and whether or how it is related to the formally correct integration by substitution. $\endgroup$ – a_guest Jul 25 '18 at 22:21
  • $\begingroup$ @a_guest du=u'dx is the key, dx is not a derivative, it is a symbol which stands for the linearized change in the variable $x$ $\endgroup$ – Mohammad Riazi-Kermani Jul 25 '18 at 22:57

I think what you're asking about is sometimes called the inverse substitution theorem. A few months ago I found that someone had actually written a paper taking issue with how integration by substitution is taught, that you may find enlightening: Gale, D. Teaching Integration by Substitution, The American Mathematical Monthly, Vol. 101, No. 6 (Jun.–Jul., 1994), pp. 520–526 .

Since that's probably still paywalled, I'll summarise the particularly relevant bit (page 4 of the PDF copy):

Inverse Substitution Theorem If $h=(f \circ g) \cdot g'$, $H'=h$ and $g$ has an inverse, $$ (H \circ g^{-1})' = f. $$

The paper offers two proofs of this, one that uses the Inverse Function Theorem but concludes that $f$ has an antiderivative, and one that assumes $f$ has an antiderivative $F$, but does not need the Inverse Function Theorem. The former is surely a more useful set of hypotheses, and that proof goes as follows:

$$ (H \circ g^{-1})' = (H' \circ g^{-1}) \cdot (g^{-1})' = ([(f \circ g) \cdot g'] \circ g^{-1} ) \cdot (g^{-1})', $$ by the chain rule and the definition of $h$. The first bracket expands to $$ [(f \circ g) \circ g^{-1}] \cdot ( g' \circ g^{-1} ), $$ and then the first bracket here simplifies to $f$ by associativity of function composition, so we have $$ f \cdot (g' \circ g^{-1}) \cdot (g^{-1})', $$ and the last two terms are the derivative of $g \circ g^{-1}$, i.e. the identity function, which has derivative $1$, and hence the result.


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.