How does the substitution theorem fail in this case?

Question

I have a very basic question on the substitution rule for definite integrals. Suppose I want to evaluate \begin{equation} \int_{-1}^{1}x\ \text{d}x \end{equation} with the substitution $x=\varphi(u)=u^2$, so that $\text{d}x = 2u\ \text{d}u$. The real problem here are the integration extremes, since the function $\varphi$ is not injective. I am anyway unable to see which hypothesis of the substitution theorem is not satisfied, could please someone help me?

Answer 1

The version of the substitution theorem that I know and love is this: $$ \int_{g(a)}^{g(b)} f(t) ~ dt = \int_a^b f(g(x)) g'(x) ~dx $$ where $g$ must be differentiable on the interval $[a, b]$, and all terms must be defined (i.e. the domain of $f$ must contain $g([a, b])$, for instance).

I don't see how this theorem applies to your integral at all.

You're probably applying the "inverse substitution theorem", which you can state by replacing $g$, in the theorem above, with $h^{-1}$. But that theorem only applies when the function $h$ is invertible, and when (once again) all terms are defined (i.e., there are restrictions on the relations between the upper and lower limits of the two integrals and the domain/codomain of $h$).

Post-comment addition

Let me try to work from what I've said above to the revised theorem for you.

Start from $$ \int_{g(a)}^{g(b)} f(t) ~ dt = \int_a^b f(g(x)) g'(x) ~dx $$ Assume $g$ is invertible, with inverse $h$. If $g(a) = A$ and $g(b) = B$, then $h(A) = a$ and $h(B) = b$. So we can rewrite $$ \int_{A}^{B} f(t) ~ dt = \int_{h(A)}^{h(B)} f(g(x)) g'(x) ~dx $$ Now let's do a little more, and substitute $h^{-1}$ for $g$:

$$ \int_{A}^{B} f(t) ~ dt = \int_{h(A)}^{h(B)} f(h^{-1}(x) (h^{-1})'(x) ~dx $$

Now in doing these substitutions, I've assumed two things in addition to the assumptions of the ordinary substitution rule:

1. $g$, restricted to the interval $[a, b]$, is invertible
2. The numbers $A$ and $B$ are in the image $g([a, b])$.

If we apply this theorem to compute, say $$ \int_0^8 t^\frac{2}{3} ~ dt $$ by letting $t = s^3$, the "just substitute" algebra looks like this: \begin{align} dt &= 3s^2 ds\\ \int_0^8 t^\frac{2}{3} ~ dt &= \int_0^2 (s^3)^\frac{2}{3} 3 s^2 ~ ds\\ &= \int_0^2 (s^2) 3 s^2 ~ ds\\ &= \int_0^2 3 s^4 ~ ds\\ &= \left. \frac{3}{5} s^5 \right|_0^2\\ &= \frac{3}{5} 2^5. \end{align} [where I may have gotten the algebra wrong, of course!]

What we're really doing is saying that $$ h^{-1}(s) = t $$ so that $$ h(s) = s^\frac{1}{3} $$ and $$ h^{-1}(s) = s^3. $$

Now applying the theorem as written in the displayed box above, we have $A = 0, B = 8, a = 0, b = h(8) = 2$, and $f(t) = t^\frac{2}{3}$. So we get \begin{align} \int_0^8 t^\frac{2}{3} ~ dt &= \int_{h(A)}^{h(B)} f(h^{-1}(x)) (h^{-1})'(x) ~dx \\ &= \int_{h(A)}^{h(B)} (h^{-1}(x))^\frac{2}{3} (h^{-1})'(x) ~dx \end{align} Recalling that $h^{-1}(s) = s^3$, we have that its derivative is $3s^2$, so \begin{align} \int_0^8 t^\frac{2}{3} ~ dt &= \int_{h(A)}^{h(B)} f(h^{-1}(x)) (h^{-1})'(x) ~dx \\ &= \int_{h(A)}^{h(B)} (x^3)^\frac{2}{3} \cdot 3x^2 ~dx\\ &= \int_{0}^{2} x^2 \cdot 3x^2 ~dx \end{align} which is what we got with the "just do the substitution" thing before.

Now there's actually something subtle: it turns out that there's a more general version of the theorem that doesn't actually require $h$ to be invertible, so that you can sometimes do substitutions like $x = u^3 - u, dx = 3u^2 - 1$. But I don't have the time or energy to write out the details. Fortunately for me, Michael Spivak has already done so, beautifully, in his book Calculus. I encourage you to take a look at the chapter on methods of integration for the details.

Answer 2

There are to problems here:

• your function $\varphi$ is not injective;
• not all points of $[-1,1]$ are of the form $\varphi(x)$ for some $x$.

Note that if you do the some thing with the integral$$\int_0^1x\,\mathrm dx,\tag1$$then none of these problems occur and all goes well. In other words, the integral $(1)$ is indeed equal to$$\int_0^12u^3\,\mathrm du.$$

Answer 3

Integration by substitution: $$\int_{\varphi(a)}^{\varphi(b)} f(x) \ \mathrm dx = \int_a^b f(\varphi(t))\varphi'(t) \ \mathrm dt$$

The substitution is "$x = \varphi(t)$".

In this case, you want $x = \varphi(t) = t^2$.

Then, you need to find $a$ such that $\varphi(a) = -1$ and $b$ such that $\varphi(b) = 1$, but the former is clearly impossible.