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As long as we don't switch the orientation, does $\int_cf\:dx$ depend on the parameterization of $C$ or no?

I have a feeling that it does not depend. However, can someone give me a rigorous proof as to why it does not depend?

Noah Nader
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1 Answers1

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Hint: Let $\gamma: [a,b] \to D(f)$ and $\sigma:[c,d] \to D(f)$ be two distinct parametrization of $C$ with the same orientation. Hence there exist a continuous function $\varphi: [c,d] \to [a,b]$, strictly increasing with $\varphi(c)=a, \ \varphi(d)=b$, such that $\sigma = \gamma \circ \varphi$. Now use the definition of $\int_C f$ and the fact that $\varphi$ is uniformly continuous on $[c,d]$ to prove that for any $\varepsilon >0$ $$ \left| \int_{\gamma}f - \int_{\gamma \circ \varphi} f \right| < \varepsilon $$ Thus, indeed the value $\int_C f$ does not depend on the parametrization chosen.

Alonso Delfín
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  • shouldn't $\varphi$ be also piecewise continuously differentiable to make that work? Curve integral use the derivative of the parametrization. I don't think that uniform continuity of $\varphi$ will be enough. – user251257 Jul 29 '15 at 22:19
  • @user251257 Curve Integral does not necessary use the derivative of the integral, it suffices for $\gamma$ to be of bounded variation. – Alonso Delfín Jul 29 '15 at 22:33
  • @user251257 For completeness I state here the definition: Let $f$ be a continuos function in $D(f)$ and $\gamma:[a,b] \to D(f)$ be of bounded variation, then $I:=\int_\gamma f :=\int_a^b f( \gamma (t)) d\gamma(t)$ where the last integral is a Riemann Stieltjes one. That is, $I$ is the only value such that for every $\varepsilon>0$ there exist $\delta>0$ such that if $P={a=t_0<t_1< \cdots< t_m=b}$ is any partition with $\max_{1\leq k \leq m}{t_k-t_{k-1}}<\delta$ then $$ \left| I- \sum_{k=1}^m f(\gamma(s_k))[\gamma(t_k)-\gamma(t_{k-1})]\right|<\varepsilon $$ where $s_k \in [t_{k-1},t_k]$ – Alonso Delfín Jul 29 '15 at 22:34
  • Ah, I see. Why should $\varphi$ be continuous? – user251257 Jul 29 '15 at 22:40
  • @user251257 If $\sigma$ is a reparametrization of $\gamma$ then there exist $\varphi$ continuous such that $\sigma=\gamma\circ \varphi$. If you mean why $\varphi$ need to be continuos to prove that $\int_\gamma f=\int_{\gamma \circ \varphi} f$, that is because in the proof you will use the uniform continuity of $\varphi$ on $[c,d]$ to see that $\left|\int_\gamma f-\int_{\gamma \circ \varphi}f \right|< \varepsilon$ – Alonso Delfín Jul 29 '15 at 22:45
  • I meant, does such continuous $\varphi$ exists for any pair of parameterizations of $C$ of same orientation, or do you define that two parameterizations are equivalent if such continuous transformation exists? – user251257 Jul 29 '15 at 22:48
  • @user251257 The definition I know of a reparameterization of a curve is exactly when such a continuos bijection $\varphi$ exist ! – Alonso Delfín Jul 29 '15 at 22:54
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    Ah. Thank you very much. – user251257 Jul 29 '15 at 23:01