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?

  • $\begingroup$ 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) $\endgroup$ – user258700 Jun 9 '16 at 23:23
  • $\begingroup$ 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. en.wikipedia.org/wiki/Line_integral $\endgroup$ – Michael Jun 9 '16 at 23:25
  • $\begingroup$ 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 $\endgroup$ – DWade64 May 7 '18 at 17:47
  • $\begingroup$ 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 $\endgroup$ – DWade64 May 7 '18 at 17:58

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.

  • $\begingroup$ 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? $\endgroup$ – nowhere Jun 10 '16 at 1:08
  • $\begingroup$ @nowhere : Well, I added some minor observations related to this in my answer above. $\endgroup$ – Michael Jun 10 '16 at 15:33

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,

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


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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.