换元积分的函数φ需要为单调的?

hbghlyj

stackexchange数学教师版块:

⋯⋯most high school teachers I know and most school textbooks warn students that you should not use integration by substitution if φ is not monotone.

In AnalysisIII-2022-notes.pdf Proposition 4.2, monotonicity of $g$ is not required

PROPOSITION 4.2 (Substitution rule). Suppose that $f:[a, b] → \mathbb{R}$ is continuous and that $\phi:[c, d] →[a, b]$ is continuous on $[c, d]$, has $\phi(c)=a$ and $\phi(d)=b$, and maps $(c, d)$ to $(a, b)$. Suppose moreover that $\phi$ is differentiable on $(c, d)$ and that its derivative $\phi'$ is integrable on this interval. Then$$
∫_{a}^{b} f=∫_{c}^{d}(f ∘ \phi) \phi'
$$Remark. It may help to see the statement written out in full:$$
∫_{a}^{b} f(x) d x=∫_{c}^{d} f(\phi(t)) \phi'(t) d t
$$Proof. Let us first remark that $f ∘ \phi$ is continuous and hence integrable on $[c, d]$. It therefore follows from Proposition 1.5 that $(f ∘ \phi) \phi'$ is integrable on $[c, d]$, so the statement does at least make sense.

Since $f$ is continuous on $[a, b]$, it is integrable. The first fundamental theorem of calculus implies that its antiderivative$$F(x):=∫_{a}^{x} f$$is continuous on $[a, b]$, differentiable on $(a, b)$ and that $F'=f$.

By the chain rule and the fact that $\phi((c, d)) \subset(a, b), F ∘ \phi$ is differentiable on $(c, d)$, and$$
(F ∘ \phi)'=\left(F' ∘ \phi\right) \phi'=(f ∘ \phi) \phi'
$$By the remarks at the start of the proof, it follows that $(F ∘ \phi)'$ is an integrable function. By the second form of the fundamental theorem,\begin{aligned}∫_{c}^{d}(f ∘ \phi) \phi' &=∫_{c}^{d}(F ∘ \phi)' \\&=(F ∘ \phi)(d)-(F ∘ \phi)(c) \\&=F(b)-F(a) \\&=F(b)=∫_{a}^{b} f\end{aligned}

Theorem 7.1.44 in chapter7.pdf assumes that $\phi$ is a strictly increasing continuous function

Theorem 7.1.44 (Change of Variables) Suppose that $\phi$ is a strictly increasing continuous function that maps an interval $[A, B]$ onto $[a, b], \alpha$ is monotonically increasing on $[a, b]$, and $f \in \Re(\alpha)$ on $[a, b]$. For $y \in[A, B]$, let $\beta(y)=\alpha(\phi(y))$ and $g(y)=f(\phi(y))$. Then $g \in \Re(\beta)$ and
$$
\int_{A}^{B} g(y) d \beta(y)=\int_{a}^{b} f(x) d \alpha(x) .
$$
Proof. Because $\phi$ is strictly increasing and continuous, each partition $\mathcal{P}=$ $\left\{x_{0}, x_{1}, \ldots, x_{n}\right\} \in \wp[a, b]$ if and only if $\mathcal{Q}=\left\{y_{0}, y_{1}, \ldots, y_{n}\right\} \in \wp[A, B]$ where $x_{j}=\phi\left(y_{j}\right)$ for each $j \in\{0,1, \ldots, n\}$. Since $f\left(\left[x_{j-1}, x_{j}\right]\right)=g\left(\left[y_{j-1}, y_{j}\right]\right)$ for each $j$, it follows that
$$
U(\mathcal{Q}, g, \beta)=U(\mathcal{P}, f, \alpha) \quad \text { and } \quad L(\mathcal{Q}, g, \beta)=L(\mathcal{P}, f, \alpha)
$$
The result follows immediately from the Integrability Criterion.

hbghlyj

What is the importance of functions of class $C^k$?
You can certainly consider $k$-times differentiable functions on, say, $[a,b]\subset \mathbb R$ and give them a notation, like $D^k[a,b]$.  
The point however is that many well-known and interesting theorems, true for $C^k[a,b]$, will fail for $D^k[a,b]$or won't even make sense. Here are three examples from elementary calculus:      
     
a) Integration by parts for $u,v\in C^1[a,b]$: $$\int_{a}^{b}u(x)v^{\prime }(x)dx=\left(u({b})v(b)-u(a)v(a)\right) -\int_{a}^{b}u^{\prime}(x)v(x)dx$$ The integrals don't even make sense a priori if $u', v'$ are not continuous.  

b) Taylor's formula
$$f(x)=\sum_{i=0}^k\frac{f^{(i)}(a)}{i!}(x-a)^i+\int_a^x\frac{f^{(k+1)}(t)}{k!}(x-t)^kdt\quad (x\in [a,b])$$ is valid for $f\in C^{k+1}([a,b])$ and again doesn't make sense for $f\in D^{k+1}([a,b])$   

c) The change of variables formula $$  \int_a^b f(\phi (t))\phi'(t)dt=\int_{\phi(a)}^{\phi(b)}  f(x)dx            $$ again necessitates that $\phi$ be a $C^1$-function and not merely a differentiable one.

hbghlyj

If $g$ is not monotone, then $g$ may be not injective, then $g^{-1}$ is not well-defined in$$\int_If(x)dx=\int_{g^{-1}(I)}f(g(t))g'(t)dt$$After proving the theorem when $g$ is monotone, we can generalize the theorem by dividing the interval into sub-intervals so that $g$ is monotone in each sub-interval, then the integral is multiplied by number of inverse images. For example
$$\int_0^11dx=1$$
Applying the theorem with $f(x)=1,g(t)=t^2$, we have $g^{-1}([0,1])=[-1,1],f(g(t))=1,g'(t)=2t,$
$$\int_0^{-1}2tdt+\int_0^12tdt=2$$Note the interval should be in the correct direction: $g(-1)=1$, so $[-1,0]$ is reversed: $\int_0^{-1}$