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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @TedShifrin $\sin t , 0\le t \le \pi$ is not a parametrization of $[0,1]$. Of course it depends of the definition used, but a parametrization $\phi : [a,b] \rightarrow [0,1] $ should have $\phi(a)=0, \phi(b)=1$. It's a nice, illustrative example, but it doesn't contradict anything and it doesn't adress the original question. The fact that the image of $[0,\pi]$ by $\sin t$ is $[0,1]$ doesn't prove anything $\endgroup$ – Emilio Jan 27 '18 at 3:54

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\ $".

  • $\begingroup$ 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
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    $\begingroup$ So to be explicit: is the message here, "no, substitutions need not be injective"? $\endgroup$ – Radon Rosborough Apr 8 '15 at 2:09
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    $\begingroup$ @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

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