I am very confused on what path and contour integrals actually represent.

I tried looking for an answer on Google but the only examples were related to Quantum Mechanics.

For instance, real Riemann integrals can be thought as "the area under the curve". What about path/contour integrals?

Also, is there a difference between path and contour integrals, or are they the same thing? I've been thinking of contour integrals as path integrals where the path is a "closed loop", but I'm not sure if that's the case.

It's all so confusing, and not being able to visualise these concepts makes it feel like I'm just memorising formulas and theorems without any reason.

I have been thinking about path integrals as integrating over a curve in three dimensions. For example, in an $xyz$ coordinate system (with $y$ pointing “up”), we have a path on the $x$-"$z$" plane and integrate over that path. That is, instead of integrating over and along "$y=0$", we integrate over a certain path. But I don't know if this is a correct way to think of path integrals.

This issue has been frustrating me a lot because I'm now studying theorems that involve path/contour integrals and I just can't understand what they really do or mean.

By path integral I mean $$\int_{\gamma\vert_{[a,b]}}f(z)\hspace{0.2em}dz = \int_{a}^{b}f(\gamma(t))\cdot\gamma'(t)\hspace{0.2em}dt$$ where $f:U\subseteq\mathbb{C}\rightarrow\mathbb{C}$ is a function that satisfies all conditions for complex integration and $\gamma:[a,b]\rightarrow U\subseteq\mathbb{C}$ is at least a piece-wise regular path.

Any help would be appreciated.

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    $\begingroup$ The way you're thinking about path integrals is how I think about them, and it works for me. I don't know much about contour integrals though. $\endgroup$ Apr 23 '21 at 20:14
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    $\begingroup$ I can't think of any particular nice intuitive explanation of what contour integration represents, but you can see the formula is almost the same as the line integral formula, which does have a nice interpretation: en.wikipedia.org/wiki/Contour_integration $\endgroup$ Apr 23 '21 at 20:14
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    $\begingroup$ Broadly speaking, a line integral can be thought of as "the area under the fence built along this path with height given by the function $f$". This is clearer in the real case --- the notion of complex "areas" is perhaps a bit confusing, but I think of that as a separate thing. $\endgroup$
    – davidlowryduda
    Apr 23 '21 at 20:19
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    $\begingroup$ 'real Riemann integrals can be thought as "the area under the curve"' It can, but it shouldn't. Because then you get stuck when generalising, just like you are now. Integration is the art of adding together very many, very small things. Some times that's the areas of very many, very narrow rectangles, but there are a plethora of other contexts where you can add small things. $\endgroup$
    – Arthur
    Apr 23 '21 at 20:19
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    $\begingroup$ "Contour integral" is a name for the integral along a path in the complex plane. $\endgroup$
    – user
    Apr 23 '21 at 20:41

Integrals are colloquially known as the area under a curve, and contour integrals are no different! But for contour integrals there are now two "curves": 1) the integrand as seen moving along a contour, and 2) the contour.

An ideal scenario

Let's consider the total amount of money you earned from the time you were born to when you turned 16, and then the total amount of money you will earn from the time you turned 16 and the time you retire.

Since your parents were desperately trying to teach you something about money, they gave you money on occasion from age 0 to age 16 for basically sitting around playing video games. Let $f(t)$ be the rate at which they gave you money. This rate changed with time. E.g. the tooth fairy was stingy when you first started losing teeth and became more generous the older you got. Then you started getting a regular allowance and by the time you were 16 you were making bank for just sitting around! The total amount you earned from 0 to 16 is given by the integral,

$$\$ = \int_0^{16} f(t) \, dt$$

Then you got your first job, and your life got complicated....

You've now entered a phase of life where you don't get paid to sit around. You have to navigate a complex plane, where you will have to look for work to earn money. Now the rate at which you can earn money, $f(z)$, depends on where you work, $z$. The path you take in life, $\gamma(t)$, and how quickly you move, $\gamma^\prime(t)$, now matters too! Since $f$ changed from being time-dependent to location-dependent, the total amount of money you will earn from 16 to retirement is given by a contour integral, $$\$\$\$ = \int_{16}^{\text{retirement}} f(\gamma(t)) \gamma^\prime (t) \, dt$$

An imperfect model for life, I know, but also a funny mnemonic device for remembering the form of the contour integral.


Meaning of line- path- and contour integrals

A curve-, line-, path- or contour integral extends the usual definition of an integral to the integration in the complex plane or in a multidimensional space. The term contour integral is typically reserved for line integrals in the complex plane but does not imply integration over a closed contour.

Also note that "Path integral" is a homonym, which on one hand describes integration along a curve and on the other hand is a description in quantum mechanics that generalizes the action principle of classical mechanics.

Thinking about and defining line integrals

Let $\mathcal{C}$ be a smove curve in the complex plane and let $f\colon\mathbb{C}\to\mathbb{C}$ be a continuous function. We now subdivide the interval $[a,b]$ into $a=t_0<t_1<t_2<\dots<t_n=b$. So we are left with $n$ intervals of length $\Delta t = (b − a)/n$.

enter image description here

From the figure we can easily see that the displacement between two points on $\mathcal{C}$ is $\gamma(t_{k})-\gamma(t_{k-1})=\gamma(t_{k-1}+\Delta t)-\gamma(t_{k-1})$.

Now we can form the Riemann sum $$\lim_{n \to \infty} \sum_{k=1}^n f(\gamma(t_k))[\gamma(t_{k-1}+\Delta t)-\gamma(t_{k-1})]=\lim_{n \to \infty} \sum_{k=1}^n f(\gamma(t_k))\left[\frac{\gamma(t_{k-1}+\Delta t)-\gamma(t_{k-1})}{\Delta t}\right]\Delta t.$$ In the square brackets we have a differential quotient. Recall that $\Delta t=(b-a)/n$, so if we let $n\to\infty$, the displacement $\Delta t$ tends to zero. This means that if we take the limit, the expression in the square brackets becomes $\gamma'(t)$.

The integral $$\int_{a}^{b}f(\gamma(t))\cdot\gamma'(t)\,\mathrm{d}t$$ is then the limit of this Riemann sum as the lengths of the subdivision intervals approach zero.


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