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I've noticed that using integration by substitution blindly could lead to some strange results. For example: with $u = x^2$, we might naively follow the usual procedure to find $$ \int_{-1}^1 x^4\,dx = \int_{-1}^{1} x^3 \,x\,dx = \int_{u(-1)}^{u(1)} u^{3/2}\,du = \int_1^1 u^{3/2}\,du =0 $$ The incorrect step here is writing $x = u^{1/2}$, since we would have $x = -u^{-1/2}$ over $[-1,0)$. If we split the original integral into one over $[-1,0]$ and another over $[0,1]$, we of course get the correct answer. However, it seems impossible to make this particular substitution in this particular problem. Certainly, there is no function $f$ such that $\int_{u(-1)}^{u(1)}f(u)\,du$ produces the correct result. Interestingly, this does produce the correct result if the integrand is an odd power of $x$.

In a more advanced course, one might account for this by saying that substitution will only work correctly if the substitution map is injective (one to one) over the domain of interest. However, having been a student and teacher of integral calculus, I've never seen this addressed (in the context of intro calculus) by a teacher or textbook. That leads me to the following questions:

  • Why doesn't this come up more often? Is there a conspiracy to avoid problems where this situation arises, or is this a problem that only comes up in "pathological cases"?

  • Why, from a pedagogical standpoint, is this not addressed?

  • Is it a coincidence that in my toy example, we get the correct answer when the integrand is $x^n$ with $n$ odd?

EDIT: So apparently, some teachers/texts do choose to address the issue, which I guess is not all that surprising. Still, I would be interested in hearing an argument that "skipping it is not such a big deal".

Then again, perhaps that's a better question for the math-ed SE.

user84413
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Ben Grossmann
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  • I actually saw this a couple weeks ago. Someone was having quite a pain with the substitutions. To address the second question, in a course, the problems are probably designed so this never happens. Lastly, you could've split the answer over positive and negative parts. – Simply Beautiful Art Dec 22 '16 at 14:03
  • Thanks for asking this. I've encountered this situation before, but being a physics grad, I had unfortunately never been taught the rigor behind it. – zahbaz Dec 22 '16 at 19:41

4 Answers4

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with $u = x^2$, we might naively follow the usual procedure to find $$ \int_{-1}^1 x^4\,dx =\int_{-1}^{1} x^3 \,x\,dx = \int_{u(-1)}^{u(1)} u^{3/2}\,du = \int_1^1 u^{3/2}\,du =0 $$

  1. Here's a strong version of the integration-by-substitution theorem, which does not generally require injectivity, and which the above example does not counteract:

    • If $g'$ is integrable on $[a,b]$ and $f$ is integrable, and has an antiderivative, on $g[a,b],$ then $$\int_a^bf\big(\color{violet}{g(x)}\big)\,\color{cyan}{g'(x)}\,\mathrm{d}x=\int_{g(a)}^{g(b)}f(u)\,\mathrm{d}u.$$

    Making your steps explicit to exhibit that the mistake occurs before performing any integration/substitution: $$0.4=\int_{-1}^1 x^4\,\mathrm dx\color{red}=\int^1_{-1}\left(\frac12\color{violet}{(x^2)}^{\frac32}\right)\color{cyan}{(2x)}\,\mathrm dx= \int_{g(-1)}^{g(1)} \frac12u^{3/2}\,\mathrm du=0.$$ $\color{red}{\textbf{The red (second) equality is false}}$ because \begin{align}x<0&\implies x^3\ne-x^3=(x^2)^{\frac32},\\a,b\in\mathbb Z\;\text{ and }\;x<0&\implies x^{ab}=(x^a)^b.\end{align}

    $\quad$ $\quad$ $\quad$ enter image description here

The short of it is that because the integrand $x^4$ can be expressed throughout $[-1,1]$ as $f(\color{violet}{x^2})\,\color{cyan}{(2x)}$ only piecewise, it needs to be preprocessed as such for the above theorem, which is not inherently asking for injectivity, to even be applicable.

  1. We might well have made the same mistake while invoking the Fundamental Theorem of Calculus, then fallaciously concluded that the FTC is broken, when in fact the red equalities below are similarly false: $$\text{on }[-1,1],\;\;\frac{\mathrm d}{\mathrm dx}\left[\frac15(x^2)^{\frac52}\right]=x(x^2)^{\frac32}\color{red}=x^4,\\\therefore\;\;0.4=\int_{-1}^1x^4\,\mathrm dx\color{red}=\left[\frac15(x^2)^{\frac52}\right]_{x={-1}}^{x={1}}=0.$$
ryang
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You should take a look at Michael Spivak's Calculus, where the substitution theorem has exactly this requirement (injective substitution), and a discussion of the problem you ask about. (That discussion may take place in one or more of the many excellent problems in the integration chapter; I can't remember and don't have my copy with me.)

John Hughes
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    neat! I don't see Spivak being used in classes, perhaps that should change. – Ben Grossmann Dec 22 '16 at 14:05
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    I used the integration chapter in a class I taught at Bryn Mawr back in the 1980s; there was nearly a revolution from the students saying "Why isn't THIS our textbook??? It's SOOOO much better!" Then they got to the problem section in the "techniques of integration" chapter, and said, "Oh, nevermind..." (I believe that the first integral in that chapter may be something like $\int x^2 e^x ~dx$, but I could be mis-remembering.) It was also used in the honors intro course at Berkeley in the late 1970s/early 80s. I got to TA that class, and what a pleasure it was... – John Hughes Dec 22 '16 at 14:07
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    ha, sounds about right. I'll have to take a look – Ben Grossmann Dec 22 '16 at 14:12
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In most cases a substitution rule is introduced for integrals of the specific form $$\int _{{a}}^{{b}}f(\varphi (t))\cdot \varphi '(t)\,{\mathrm {d}}t$$

where $f:I \to \Bbb R$ is continuous and $\varphi: [a,b] \to I$ is continuously differentiable. Then it holds $$\int _{{a}}^{{b}}f(\varphi (t))\cdot \varphi '(t)\,{\mathrm {d}}t=\int _{{\varphi (a)}}^{{\varphi (b)}}f(x)\,{\mathrm {d}}x$$

If your integrand is $t^n$ for $n$ even you wont find $f$ and $\varphi$ s.t.

$$f(\varphi (t))\cdot \varphi '(t) = t^n$$

But if $n$ is odd you can choose $\varphi(t) = t^2$ and $f(t) = \frac{1}{2}t^\frac{n-1}{2}$ and the substitution works… so there is no need to consider integrals of your form because you cannot use subsitution rule for them because they don't satisfy the given assumptions.

Gono
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    So, which assumption isn't satisfied in my case? – Ben Grossmann Dec 22 '16 at 15:26
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    In your case is $[a,b] = [-1,1]$ and $I = [0,1]$. So try to find $f$ and $\varphi$ such that $f(\varphi (t))\cdot \varphi '(t) = t^4$. Using $\varphi(t) = t^2$ will lead you to $$f(t) = \frac{1}{2}\sqrt{x}^3$$ but for this $f$ it holds $$f(\varphi (t))\cdot \varphi '(t) = |x|^3\cdot x \not= x^4$$ if $x<0$ so not on the whole space you integrate at… By splitting the integral in two different parts you can choose two different $f$ namely $$f_{[-1,0]}(x) = -\frac{1}{2}\sqrt{x}^3$$ and $$f_{[0,1]}(x) = \frac{1}{2}\sqrt{x}^3$$ and so you can use the substitution rule for each part seperately. – Gono Dec 22 '16 at 15:39
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    Well, then simply define $$ f(x) = \begin{cases} \frac 12 \sqrt{x^3} & x \geq 0\ -\frac 12 \sqrt{x^3} & x < 0 \end{cases} $$ which is continuous over $[-1,1]$. By your statement, the substitution should now work. However, we still end up with $\int_1^1 f(u),du$ – Ben Grossmann Dec 22 '16 at 15:53
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    This wont work either. You make a fine but hard mistake! $f$ has to be a continous function $f: I \to \Bbb R$ and if you want to use the substitution $\varphi(x) = x^2$ and $\varphi:[-1,1] \to I$ you get $I = [0,1]$. So f has to be given as a continous function from $[0,1] \to Bbb R$! And your given $f$ equals $\frac{1}{2}\sqrt{x^3}$ on $[0,1]$ but then $$\int_{-1}^1 x^4 dx \not= \int_{-1}^1 f(\varphi(x)) \varphi'(x) dx$$ and you cannot use your $f$ for substitution rule. Indeed there is no need to define your $f$ for $x\le 0$ because as an argument you plug in $\varphi(x)$ so elements of $I$. – Gono Dec 22 '16 at 16:00
  • It's much easier to describe it more general: If you want to solve the integral $$\int_a^b g(x) dx$$ by substitution you have to show that there exists a continously differentiable function $\varphi: [a,b] \to I$ and a continous function $f:I \to \Bbb R$ s.t. $$\int_a^b g(x) dx = \int_a^b f(\varphi(x))\varphi'(x) dx$$ holds. In your case $g(x) = x^4$ and you want to use $\varphi(x) = x^2$ and you claim it works… so it's an easy task to you: Give me your $f$! You will see: As hard as you look for it, you won't find it… Your $f$ given above doesn't fulfill the equality, so try a new one. – Gono Dec 22 '16 at 16:03
  • Interesting. I'll think about it and get back to you. – Ben Grossmann Dec 22 '16 at 16:26
  • you're welcome :-) – Gono Dec 22 '16 at 16:29
  • After thinking about it and asking another question, this answer has convinced me that there's a reasonable way to forego requiring an injective substitution, at least in the single-variable setting. Thanks so much. – Ben Grossmann Dec 22 '16 at 22:29
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Well they are adressed. See for an example http://faculty.swosu.edu/michael.dougherty/book/chapter07.pdf.

The easiest way to avoid this problem is to choose a bijective-substitution. Eg one that is one-to-one and has an inverse.

N3buchadnezzar
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