22
$\begingroup$

I made a silly mistake in evaluating some integral by using a non-injective $u$-substitution. But why should $u$-substitutions be injective in the first place?

I reasoned in the following way: the formula $$ \int_{\phi(a)}^{\phi(b)}g(x)\ dx = \int_a^b g(\phi(t))\phi^\prime(t)\ dt $$ holds for a general $C^1$ function $\phi$, even if it is not injective. When you calculate an integral of the form $\int_a^b f(\phi(t))\ dt$, to use the formula above from right to left, you should find a function $f$ such that $$ f(\phi(t)) = g(\phi(t))\phi^\prime(t), $$ which do not exist if $\phi$ is not injective, i.e., $\phi(t) = 0$ for some $t$. This is why substitutions should be injective.

Is my reasoning correct? If so, I believe that if $\phi^\prime(t) = 0 \Rightarrow f(\phi(t)) = 0$, a function $g$ that satisfies the formula above may exist and $\phi$ should not necessarily be injective. Is this right?

I am often confused about the fact $\phi$ should be injective. Is there an intuitive way to interpret this fact, so that I always remember to take a $\phi$ that is injective?

I would be grateful if you could help me understand this matter.

$\endgroup$
  • $\begingroup$ You are correct. The formula only requires that $\phi$ be continuously differentiable. $\endgroup$ – wj32 Aug 18 '13 at 12:23
  • $\begingroup$ @wj32 Thank you. Could I also ask you about the possibility of $\phi$ not being injective, and the intuition about the issue, as I stated above? $\endgroup$ – Pteromys Aug 18 '13 at 12:46
  • $\begingroup$ Correct me if I'm wrong, but I believe that the formula is true for a general $\phi \in C^1$ only if $$\frac{\text d}{\text d x} \int _{\phi(a)} ^{\phi(x)} g(t)\,\text d t=g(\phi(x))\phi '(x)=\frac{\text d}{\text d x}\int _a ^x g(t) \phi '(t) \text d t,$$ which is the case if $g\in C^0.$ I don't have a counterexample at hand (if there exist one), but I think the general case requires $\phi$ to be monotonous. $\endgroup$ – pppqqq Aug 18 '13 at 13:21
  • $\begingroup$ @pppqqq My textbook also states the formula for $f\in C^0$ and $g\in C^1$. You state that the general case requires $\phi$ to be monotonous, but what is the general case you are talking about? $\endgroup$ – Pteromys Aug 18 '13 at 13:32
  • $\begingroup$ Referring to your first formula, I'm considering the case where $g\in \mathscr R ([a,b])$, that is: $g$ is Riemann-integrable in $[a,b]$. $\endgroup$ – pppqqq Aug 18 '13 at 13:41
1
$\begingroup$

Well, imagine the substitution as tracing a path (along the $x$-axis in this case). If you go from $a$ to $b$ and then back from $b$ to $a$ you will cancel out the integral and not compute the integral on $[a,b]$ as you intended. And all sorts of intermediate things can happen.

Try "parametrizing" $[0,1]$ by $x=\sin t$, $0\le t\le\pi$, and computing $\displaystyle\int_0^1 x\,dx$, for example. Of course, if you do the official substitution, you end up with $\int_0^0 x\,dx = 0$. But the function has "covered" the interval $[0,1]$ and then "uncovered" it.

$\endgroup$
  • $\begingroup$ Are you sure about the example you gave? For that integral, $x$ runs from $0$ to $1$, so $t$ runs from $0$ to $\pi/2$ and not $\pi$ $\endgroup$ – Parth Thakkar Aug 18 '13 at 13:08
  • $\begingroup$ But if you run from $0$ to $\pi$, you hit every $x$ twice, and look what happens to the integral. That's exactly my point. $\endgroup$ – Ted Shifrin Aug 18 '13 at 13:28
  • $\begingroup$ What is actually relevant is a notion of degree of a mapping ... which you can learn about in a differential topology course. If the map has degree $1$, even if it hits some values more than once, then it'll work out fine. $\endgroup$ – Ted Shifrin Aug 18 '13 at 13:28
  • $\begingroup$ Hmmm, so all this isn't all that simple as it seems. I don't know even know what topology is, and I am dealing with something like differential topology. I'll rather keep quiet :) $\endgroup$ – Parth Thakkar Aug 18 '13 at 13:30
  • 2
    $\begingroup$ This answer is true but not relevant, since you wouldn't change the bounds to $0$ and $\pi$ when making the substitution $x = \sin t$. The lesson here is just that it's the values of the substitution function $\phi$ that are important, not the endpoints of the image on an interval. $\endgroup$ – Jim Belk Jun 19 '15 at 16:15
17
$\begingroup$

When $f:\ I\to{\mathbb R}$ has a primitive $F$ on the interval $I$, then by definition $$\int_a^b f(t)\ dt =F(b)-F(a)$$ for any $a$, $b\in I$; in particular $b<a$ is allowed.

When $\phi$ is differentiable on $[a,b]$ and $g$ has a primitive $G$ on an interval $I$ containing $\phi\bigl([a,b]\bigr)$, then by the chain rule $G \circ \phi$ is a primitive of $(G\circ\phi)\cdot\phi'$ on $[a,b]$. It follows that $$\int_{\phi(a)}^{\phi(b)} g(x)\ dx =G\bigl(\phi(b)\bigr)-G\bigl(\phi(a)\bigr)=\int_a^bg\bigl(\phi(t)\bigr)\phi'(t)\ dt\ .\tag{1}$$ No question of injectivity here.

Now there is a second kind of substitution. Here we are given an integral $$\int_a^b f(x)\ dx$$ without any $\phi$ visible neither in the boundaries nor in the integrand. It is up to us to choose a clever $\phi$ defined on some interval $J$ such that (i) $a$, $b\in \phi(J)$ and (ii) $f\circ\phi$ is defined on $J$. Assume that $\phi(a')=a$, $\>\phi(b')=b$. Then according to $(1)$ we have $$\int_a^b f(x)\ dx=\int_{a'}^{b'}f\bigl(\phi(t)\bigr)\>\phi'(t)\ dt\ .$$ No question of injectivity here, either. Consider the following example: $$\int_0^{1/2} x^2\ dx=\int_{-\pi}^{25\pi/6}\sin^2 t\>\cos t\ dt.$$ It is true that for this second kind of substitution one usually chooses an injective $\phi$ so that one can immediately write $\phi^{-1}(a)$ and $\phi^{-1}(b)$ instead of "take an $a'$ such that $\phi(a')=a\ $".

$\endgroup$
  • $\begingroup$ I was most interested in the cases where the formula for integration by substitution is used from right to left, when there is no apparent $\phi^\prime(t)$ factor in the integrand on the RHS, but thank you any way. $\endgroup$ – Pteromys Aug 18 '13 at 23:35
  • $\begingroup$ @Pteromys: That's exactly the "second kind of substitution" referred to above. Only LHS and RHS are interchanged. $\endgroup$ – Christian Blatter Aug 19 '13 at 10:10
  • 2
    $\begingroup$ So to be explicit: is the message here, "no, substitutions need not be injective"? $\endgroup$ – Radon Rosborough Apr 8 '15 at 2:09
  • 1
    $\begingroup$ @raxod502: True. On the other hand it would be difficult to cook up an example for which some noninjective substitution would produce a computational advantage. $\endgroup$ – Christian Blatter Apr 8 '15 at 8:48
  • $\begingroup$ Christian, would you mind taking a look at this answer of mine? It's precisely about this... Thanks math.stackexchange.com/questions/1911892/… $\endgroup$ – An old man in the sea. Sep 2 '16 at 16:17

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.