# Why should the substitution be injective when integrating by substitution?

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.

• You are correct. The formula only requires that $\phi$ be continuously differentiable.
– wj32
Aug 18, 2013 at 12:23
• @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? Aug 18, 2013 at 12:46
• 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. Aug 18, 2013 at 13:21
• @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? Aug 18, 2013 at 13:32
• Referring to your first formula, I'm considering the case where $g\in \mathscr R ([a,b])$, that is: $g$ is Riemann-integrable in $[a,b]$. Aug 18, 2013 at 13:41

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 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\$$".

• 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. Aug 18, 2013 at 23:35
• @Pteromys: That's exactly the "second kind of substitution" referred to above. Only LHS and RHS are interchanged. Aug 19, 2013 at 10:10
• So to be explicit: is the message here, "no, substitutions need not be injective"? Apr 8, 2015 at 2:09
• @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. Apr 8, 2015 at 8:48
• Christian, would you mind taking a look at this answer of mine? It's precisely about this... Thanks math.stackexchange.com/questions/1911892/… Sep 2, 2016 at 16:17

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.

• 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 $\pi$ Aug 18, 2013 at 13:08
• 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. Aug 18, 2013 at 13:28
• 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. Aug 18, 2013 at 13:28
• 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. Jun 19, 2015 at 16:15
• @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 Jan 27, 2018 at 3:54