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Let $\alpha: [a,b] \rightarrow \mathbb{R}^2$ be a smooth path (i.e. $\alpha'$ is continuous on $[a,b]$), and let $f$ be a continuous vector field. The line integral of $f$ along $\alpha$ is defined as $\int_a^b f[\alpha(t)]\cdot \alpha'(t) dt$. In a typical calculus textbook, it is written as $\int_C f_1 dx + f_2 dy$, suggesting that the line integral depends only on $C$ - the image of the path, not the actual parameterization $\alpha$.

In general however, this is not true. For example, $\oint_C -ydx + x dy =2\pi$ when integrated along the path $\alpha: [0,2\pi]\rightarrow \mathbb{R}^2, \alpha(t) = (\cos(t), \sin(t))$. The value becomes $4\pi$ when integrated along $\beta: [0,4\pi]\rightarrow \mathbb{R}^2, \beta(t)= (\cos(t),\sin(t))$, even though $\alpha$ and $\beta$ have the same image.

My question: what additional assumption is required to make the notation $\int_C f_1dx +f_2 dy$ unambiguous?

nowhere
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  • None. It is always independent of the chosen parameterization. In the example you give, $\beta$ is not actually a parameterization of the image of $\alpha$ (i.e. $\alpha$ and $\beta$ have different images) –  Jun 09 '16 at 23:23
  • Looking at the wikipedia link here, it requires the curve to be a bijection. However, the result can depend on the orientation (clockwise/counter-clockwise) in the case of a loop. https://en.wikipedia.org/wiki/Line_integral – Michael Jun 09 '16 at 23:25
  • This comment is really late. I don't know if this is correct or not, but if the curve $\mathcal{C}$ is not closed, then you have no issue. Different parameterizations of open curves give the same value for the line integral [the only way parameterizations can vary is by 1) speed and 2) backtracking/forward tracking on the same curve $\mathcal{C}$. Any back/forward tracking comes in equal/opposite contributions and sum to 0. Different speeds also don't affect the integral ]. With closed curves however, then you have issues. So as long as the curve isn't closed, then the value is unambiguous – DWade64 May 07 '18 at 17:47
  • This only works if both parameterizations start and stop and the same point. If you have some curve $\mathcal{C}$ with 2 different parameterizations (each with say a different starting but the same endpoint) then my above comment doesn't hold. You'll get two different values for the line integral. So as long as the curve $\mathcal{C}$ is open and the start/endpoints are given, the value is unambiguous – DWade64 May 07 '18 at 17:58

2 Answers2

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More generally, you can say this: Let $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ be a continuous function. Let $\alpha:[0,1]\rightarrow\mathbb{R}^n$ and $\beta:[0,1]\rightarrow\mathbb{R}^n$ be two smooth curves. Suppose there is a smooth curve $\theta:[0,1]\rightarrow[0,1]$ such that $\theta(0)=0, \theta(1)=1$, and $\beta(t) = \alpha(\theta(t))$ for all $t \in [0,1]$. Then: $$ \int_0^1 f(\alpha(t))\cdot \alpha'(t)dt = \int_0^1 f(\beta(t))\cdot \beta'(t)dt $$

Proof: By the chain rule we have $\beta'(t) = \alpha'(\theta(t))\theta'(t)$. Thus: \begin{align} \int_0^1 f(\beta(t))\cdot \beta'(t) dt &= \int_0^1 f(\alpha(\theta(t)))\cdot \alpha'(\theta(t))\theta'(t)dt \\ &= \int_0^1 f(\alpha(u))\cdot \alpha'(u)du \end{align} where the second equality is by the change of variables rule for integration of real-valued functions. $\Box$

In your example of a loop traversed once by a curve $\alpha$ and then twice by a curve $\beta$, you cannot write $\beta(t) = \alpha(\theta(t))$ for all $t \in [0,1]$. Defining $\theta(t) = 2t$ does not work because that would only be defined for $t \in [0,1/2]$. In another example of a clockwise loop versus a counter-clockwise loop, you again cannot find such a $\theta(t)$ function.

Edit: Some additional observations:

Allowing for non-injectivity:

The condition on $\theta$ above does not require the functions $\alpha, \beta$, or $\theta$ to be injective. For example, it works for the following: Let $\alpha(t)$ be any smooth curve defined over $t \in [0,1]$ (possibly having repeated values over that interval) and define $\beta(t) = \alpha(\theta(t))$, where $$ \theta(t) = \sin^2\left(\frac{5\pi t}{2}\right) $$ The $\beta$ curve retraces the $\alpha$ curve 5 times, and the first four times cancel each other out.

Path independence when $f$ is a gradient

Suppose there is a smooth function $H:\mathbb{R}^n\rightarrow\mathbb{R}$ such that $f(x) = \nabla H(x)$ for all $x \in \mathbb{R}^n$. Then path integrals of $f$ depend only on the endpoints of the path, not on the intermediate values of the path. Indeed: $$ \int_0^1 f(\alpha(t))\cdot \alpha'(t) dt = \int_0^1\frac{d}{dt}\left[H(\alpha(t))\right] dt = H(\alpha(1))-H(\alpha(0)) $$

Cauchy integral formula

There are also connections to complex numbers and the Cauchy integral formula. For example, in some cases the path integral of a loopy curve depends only on the number of times the curve winds clockwise around the origin.

Michael
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  • Thank you. I want to follow up on your answer: we now know that given a curve $C$, there can be two different parameterizations $\alpha$ and $\beta$, such that there is no $\theta$ satisfying $\beta = \alpha \circ \theta$. I find this problematic. Requiring that parameterizations be bijective, as you mentioned earlier, is one way to resolve this issue. But this assumption might be too restrictive, as it forces you to only consider simple curves. Is there a milder assumption? – nowhere Jun 10 '16 at 01:08
  • @nowhere : Well, I added some minor observations related to this in my answer above. – Michael Jun 10 '16 at 15:33
  • @Michael Can every curve be parameterized? I am asking this because in order to prove Gradient Theorem we parameterize the path. But if some path can't be parameterized how do we know that the integral would depend only on the endpoints of that path? – Antonios Sarikas Mar 22 '20 at 20:56
  • @user599310 : What do you mean by "curve" or "path" and what do you mean by "parameterized"? – Michael Mar 22 '20 at 22:33
  • @Michael Suppose I want to integrate a vector field along a path. I can do this by parameterizing the path. Gradient theorem says that the value of this integral if the vector field is conservative is indepedent of path. But in order to prove Gradient theorem Wiki uses a parameterized path (composite function). What I am asking is what if it exists some path that it can't be parameterized. Then we can't say for sure that the integral must be indepedent of path because in the proof we started stating that the path is parameterized. https://en.wikipedia.org/wiki/Gradient_theorem (Section proof) – Antonios Sarikas Mar 23 '20 at 18:10
  • @user599310 : What do you mean by an "unparameterized" path? Usually a path is defined as some curve $\vec{f}(t) \in\mathbb{R}^2$ for $t \in [0,1]$. What other format do you have that specifies your path? – Michael Mar 23 '20 at 18:56
  • @Michael Can every path have a function $\vec{f}(t)$ representing it? – Antonios Sarikas Mar 23 '20 at 22:43
  • @user599310 : Please define what you mean by a "path." If the definition of "path" is a function of the type $\vec{f}(t)$ then your question is like asking "do all squares have four sides?" I have asked you to specify what other format you have in mind for "path" but you have refused to answer. – Michael Mar 24 '20 at 05:29
  • @Michael I define a path as a trajectory (e.g. moving in 3D space). I know that my definition isn't mathematical rigorous but that how I have the term path in my mind. Integrating along a path=Integrating along a trajectory. What I am asking can we have a function that describes every path you can follow to get from a point A to a point B (e.g. straight line, curve)? – Antonios Sarikas Mar 24 '20 at 16:15
  • I would say a "path" is the same as a "trajectory" which is a continuous function $\vec{f}(t)$ defined for $t \in [0,1]$, so that $\vec{f}(0)$ and $\vec{f}(1)$ define the start and end points. No, we cannot have a function that describes every path you can follow from point A to point B. – Michael Mar 24 '20 at 19:12
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In general, a line integral of type

$$ I = \int_{C} \sum_{i} F_{x_i} \mathrm{d} x_i, $$

is independent of parameterization when integrand is an exact differential. This means that exists $F$ with total derivative:

$$ \mathrm{d} F = \dfrac{\partial F}{\partial x_1} \mathrm{d} x_1 + \dfrac{\partial F}{\partial x_2} \mathrm{d} x_2 + \dotsb, $$

where, for all variable $x_i$,

$$ F_{x_i} = \dfrac{\partial F}{\partial x_i}. $$

Note that $dF$ must be integrable on $C$. In fact, let $F \,\colon D\subseteq\mathbb{R^n}\to\mathbb{R}$ be a continuous and differentiable funtion and let $\gamma \,\colon \left[ 0,1 \right] \to D$ be a smooth path or piecewise smooth path (integrable path), then

$$ \int_{\gamma} \mathrm{d}F = \int_{0}^{1} \dfrac{\mathrm{d}F(\gamma(t))}{\mathrm{d}t} \mathrm{d}t = F(\gamma(1)) - F(\gamma(0)). $$

Example of formal proof

Let's assume that $-y dx + x dy$ is an exact differential, then exists $F$ such that

$$ \begin{aligned} \mathrm{d} F &= \dfrac{\partial F}{\partial x} \mathrm{d} x + \dfrac{\partial F}{\partial y} \mathrm{d} y \\ & = -y dx + x dy, \end{aligned} $$

which implies

$$ \begin{aligned} \dfrac{\partial F}{\partial x} &= -y, & \dfrac{\partial F}{\partial y} &= +x \end{aligned}. $$

Calculating its partial cross-derivatives,

$$ \begin{aligned} \dfrac{\partial^2 F}{\partial x \partial y} &= -1, & \dfrac{\partial^2 F}{\partial y \partial x} &= +1 \end{aligned}, $$

where

$$ \dfrac{\partial^2 F}{\partial x \partial y} \neq \dfrac{\partial^2 F}{\partial y \partial x} $$

is a contradiction by Schwarz's or Clairaut's theorem. Therefore, $-x dx + y dy$ is not an exact differential, and its line integral is not independent of the chosen parameterization.

Noir
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