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This is a follow-up question to the one I ask here. Gono's answer indicates that the following is true:

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

I had been under the impression that this was only guaranteed to be valid if $\varphi$ is injective, since injectivity is one of the conditions given in a multivariate setting. In fact, I thought that I had a counterexample in my question. Note, however, that with the example I gave, the $f(\varphi(t))\varphi'(t)$ resulting from my substitution failed to coincide with the integrand $t^4$ over $[-1,1]$.

So, here's my question: is the above statement true? Is there a corresponding statement that holds in the multivariate generalization regardless of whether $\varphi$ is injective?

Community
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Ben Grossmann
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1 Answers1

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Let \begin{align*} F(u) &= \int _{{a}}^{{u}}f(\varphi (t))\cdot \varphi '(t)\,{\mathrm {d}}t \\ G(u) &=\int _{{\varphi (a)}}^{{\varphi (u)}}f(x)\,{\mathrm {d}}x\end{align*}

Then $F$ and $G$ are functions with the same derivative everywhere (namely $F'(u)= f(\varphi (u))\cdot \varphi '(u)$, and with the same value (zero) at $u= a$ Thus they have the same value at $b$. In short, this is really just an application of two parts of the FTC, plus the chain rule.

For the multivariate setting, the question of orientation becomes really critical, and I don't know an easy way to say the right multi-variable statement. It's something like "if φφ maps the boundary of the domain in an orientation-preserving way, and [probably some other stuff here], then the integrals are the same." But I've never seen it written out. We have the advantage in 1-dimensional integrals that reversing the limits means something. :)

John Hughes
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  • That seems a bit more obvious now... does something fall apart here in the multivariate case? I suppose the FTC isn't there in the same way... – Ben Grossmann Dec 22 '16 at 22:24
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    Aha! That's interesting. Right, I guess it's the boundary of the region we need to worry about a la Stokes'. I'll accept this as soon as I can. – Ben Grossmann Dec 22 '16 at 22:25
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    Right; if you simply define "integral over a region" as "average value times area of the regions", then by reversing orientation, you change sign in the det of $d\phi$, so you need the usual absolute value signs that we see. Once you do integration on chains, though, you can clean this up a bit, as I recall. – John Hughes Dec 22 '16 at 22:41