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This is probably a silly (and maybe duplicate) question, but I haven't been able to find a standard answer so far.

Let $f$ be a real-valued function on $\mathbb{R}^n$, and let $F$ be a function from $\mathbb{R}^n$ to $\mathbb{R}^n$. Let $\phi:[a,b] \to \mathbb{R}^n$ be a continuously differentiable path.

Define an Integral of Type 1 to be $$\int_{\phi} f = \int_a^b f(\phi(t))|\phi'(t)|dt$$ and and Integral of type 2 to be $$\int_{\phi} F = \int_a^b F(\phi(t)) \cdot \phi'(t)dt$$

It seems that by using the unit tangent vector to the image of the curve $\phi$ one should be able to convert back and forth between these two kinds of integrals: But it doesn't seem like the unit tangent vector is always well-defined.

Is this allowed? If not is it allowed under some restrictions, for example that $\phi$ is injective?

Thank you!

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  • $\begingroup$ Ordinarily one assumes a path is (piecewise-) smooth with a regular parametrization (nowhere-vanishing derivative) except at the "corners." $\endgroup$ – Ted Shifrin Jan 2 '14 at 16:06
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You can't always convert between these, even if the parametrization has nonvanishing derivative everywhere. Consider this curve, in which the middle part is double-traced in opposite directions:

double-traced

An integral of type 2 does not depend on the values of $F$ on the double-traced part, since it cancels out. But an integral of type 1 does.

A sufficient condition for being able to convert between two types is: $\phi $ is a diffeomorphism onto its image. In this case conversion is straightforward since the velocity vector is nonzero.

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  • $\begingroup$ @bryanj Yes, this is what constitutes being a diffeomorphism. $\endgroup$ – Post No Bulls Jan 3 '14 at 3:11

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