## Line Integral of a Space-Filling Curve

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### Line Integral of a Space-Filling Curve

Would it be possible to take the line integral of a function along a space-filling curve? I assume that would require an explicit equation for the curve, and I haven't found one in my (brief) wikipedia-ing. Assuming it is possible, would the answer be finite, assuming a well-behaved scalar function? Would the result be the same as taking a double integral of the function over the same domain?

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Last updated 6/29/108

Last updated 6/29/108

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### Re: Line Integral of a Space-Filling Curve

Generally speaking, a line integral is defined as

[math]\int_C f\, ds = \int_a^b f(\mathbf{r}(t)) |\mathbf{r}'(t)|\, dt[/math]

(for a scalar field) or

[math]\int_C \mathbf{F}(\mathbf{r})\cdot\,d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt[/math]

(for a vector field) so that it doesn't depend on how you parametrize your curve. Since a space-filling curve is not differentiable, this definition doesn't work, so my first answer would be no. My second answer would be that an appropriate generalization of the first definition could probably be found (the second looks hopeless), in which case the line integral would almost certainly diverge for any continuous function which is not identically zero, and probably for any function not identically zero.

You could, of course, find [imath]\int_a^b f(\mathbf{r}(t)) \, dt[/imath] for some parametrization r of your curve, but that would depend on the parametrization.

[math]\int_C f\, ds = \int_a^b f(\mathbf{r}(t)) |\mathbf{r}'(t)|\, dt[/math]

(for a scalar field) or

[math]\int_C \mathbf{F}(\mathbf{r})\cdot\,d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt[/math]

(for a vector field) so that it doesn't depend on how you parametrize your curve. Since a space-filling curve is not differentiable, this definition doesn't work, so my first answer would be no. My second answer would be that an appropriate generalization of the first definition could probably be found (the second looks hopeless), in which case the line integral would almost certainly diverge for any continuous function which is not identically zero, and probably for any function not identically zero.

You could, of course, find [imath]\int_a^b f(\mathbf{r}(t)) \, dt[/imath] for some parametrization r of your curve, but that would depend on the parametrization.

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"With math, all things are possible." —Rebecca Watson

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### Re: Line Integral of a Space-Filling Curve

You could try evaluating the integral along the curves approaching the space filling curve, for many of the constructions the curve is piecewise, so these will be possible to integrate. Then see if the limit exists. I would imagine that you'd get something sensible for integrating around the Koch curve if the function is nice enough, so I don't see why you necessarily wouldn't get a limit for something space filling.

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### Re: Line Integral of a Space-Filling Curve

HenryS wrote:I don't see why you necessarily wouldn't get a limit for something space filling.

If your function is bigger than epsilon in an open set, then the integral along the portion of a curve inside that open set is epsilon times the length of that part of the curve. For a curve with infinite length, this is definitely going to be bad. It's true that integrating over the whole curve might give you positive and negative parts which cancel out, and so the limit could exist, but it's not really the "true" value of the integral for the same reason that [imath]\int_\mathbb{R} \frac{1}{x}[/imath] is undefined, and not zero.

I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

"With math, all things are possible." —Rebecca Watson

### Re: Line Integral of a Space-Filling Curve

Well, I was thinking along the lines of contour integrals of holomorphic or meromorphic functions, when you should get a sensible answer of integrating around the Koch snowflake, as long as there aren't any poles too near the curve.

I see what you mean though: even in this case the limit only exists as a result of lucky cancellation.

I see what you mean though: even in this case the limit only exists as a result of lucky cancellation.

- skeptical scientist
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### Re: Line Integral of a Space-Filling Curve

HenryS wrote:I see what you mean though: even in this case the limit only exists as a result of lucky cancellation.

Exactly. And when that happens, the limit depends not just on the curve, but on the sequence of approximations used, which makes it hard to call it a path integral.

I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

"With math, all things are possible." —Rebecca Watson

### Re: Line Integral of a Space-Filling Curve

Although, if the function is holomorphic, then the integral is 0 no matter the sequence of approximations.

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