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If $f$ is a continuous function defined on $[a, b],$ and $\phi(t)$ a continuously differentiable function from $[\alpha, \beta]$ into $[a, b]$. Then

$\int_\alpha^\beta{f(\phi(t)) \phi\prime (t)dt}=\int_{a}^{b}f(x)dx$

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This is proof.Question I have is that why we need $\phi(t)$ to be continuously differentiable function?Why we need that derivative of $\phi(t)$ to be continuous?I can't see where we used that.

unit 1991
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  • Continuity is the simplest way to make sure the integral exists. This could be relaxed a little bit. – Randall Jan 13 '22 at 17:09
  • @Randall If function is not continuous at finitely many points it still can be integrable.But inside integral we have $f((ϕ(t))ϕ′(t)$ not only $\phi(t)$ – unit 1991 Jan 13 '22 at 17:11
  • Meaning that even if your statement is correct inside integral is $f((ϕ(t))ϕ'(t)$ why if $\phi'(t)$ is continuous follows that integral exists? – unit 1991 Jan 13 '22 at 17:12
  • Because you're integrating a product of continuous functions, which is continuous. – Randall Jan 13 '22 at 17:13
  • But if $f=0$ we can have that $ϕ′(t) $ to be discontinuous or not? – unit 1991 Jan 13 '22 at 17:15
  • I don't understand what $f$ has to do with $\phi'(t)$. – Randall Jan 13 '22 at 17:16
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    There's no claim that the condition is strictly necessary. You could for instance split the integral wherever $\phi'$ is discontinuous and then apply this result on each piece, and then you'd get a slight generalization. – Ian Jan 13 '22 at 17:18
  • All that said, it is rarely necessary to pay much attention to functions which are differentiable but not continuously differentiable. They don't come up much in practice (partially because a function that is the derivative of something can only have a "type II" discontinuity, which aren't that common in practice either). So I would suggest not dwelling on the "continuously" part of "continuously differentiable" hypotheses in theorems. – Ian Jan 13 '22 at 17:20
  • @Randall Inside integral it is $f(\phi(t))ϕ′(t) $ product of two functions.You are saying that we need $ϕ′(t)$ to be continuous so that product of two functions to be also continuous.But if $f=0$ then $f(\phi(t))ϕ′(t)=0$ which means that $ϕ′(t)$ can be discontinuous? – unit 1991 Jan 13 '22 at 17:20
  • @unit1991 OK, sure, but then you're integrating the $0$ function, and we don't need any fancy theorems at all to do that. – Randall Jan 13 '22 at 17:27
  • @Ian is saying it best. No one is stating that the condition is necessary. – Randall Jan 13 '22 at 17:28
  • @unit1991. If $f$ is continuous, then it is sufficient only that $\phi'$ be Riemann integrable for the rule to hold. On the other hand, if we weaken the condition on $f$ -- for example assume only that $f$ is Riemann integrable -- then in some proofs, it is necessary to assume that $\phi'$ is continuous. See here. In your case what really matters is that $f \circ \phi \phi'$ is integrable. – RRL Jan 13 '22 at 17:28
  • @RRL Thank you.I will think more about that and read your linked question. – unit 1991 Jan 13 '22 at 17:32
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    @Ian $ϕ′(t)$ can be discontinuous only in finitely many points yes?So only that way we can apply on each piece. – unit 1991 Jan 13 '22 at 17:33
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    @unit1991: Also see this -- particularly the comments. – RRL Jan 13 '22 at 17:36
  • @RRL in here https://math.stackexchange.com/questions/3746099/understanding-the-substitution-theorem-of-riemann-integration/3747622#3747622 how you got $F'(t) = f(\phi(t)) \phi'(t)$? – unit 1991 Jan 13 '22 at 18:15
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    @unit1991: Leibniz rule. In future you should ask a question about another answer by commenting there. – RRL Jan 13 '22 at 18:24
  • @unit1991 Doing this splitting without getting into further technicalities requires only finitely many discontinuities yes. – Ian Jan 13 '22 at 18:28
  • Apply chain rule to differentiate $F(t) = G(\phi(t))$ where $G(x) = \int_{\phi(a)}^x f(u), du$. SInce $f$ is continuous $G'(x) = f(x)$ and by the chain rule $F'(t) = G'(\phi(t))\phi'(t)$. – RRL Jan 13 '22 at 18:30
  • @RRL Now I understand thank you. – unit 1991 Jan 13 '22 at 18:32

1 Answers1

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This question has generated quite a few comments. In view of this interest you might want to know a more definitive answer. Preiss and Uher published this in their days as students in Prague.

Preiss, David; Uher, Jaromír A remark on the substitution for the Riemann integral. (Czech. English summary) Časopis Pěst. Mat. 95 1970 345–347.

Theorem: Let $g $ be a Riemann integrable function on $[a,b]$; for $s∈[a,b]$ put $G(t)=∫_s^t g(w)dw$ $(t∈[a,b])$; let $f$ be a function bounded on $[c,d]=G([a,b])$; then if either one of the Riemann integrals $∫_a^b f(G(t))g(t)dt$, $∫_{G(a)}^{G(b)}f(x)dx$ exists, the other exists also and these integrals are equal.

The proof is based on an earlier result of R. O. Davies [Math. Gaz. 45 (1961), 23–25].

An online proof for this (in English, not Czech) can be found here: http://www.teaching.math.rs/vol/tm1813.pdf

B. S. Thomson
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