# Intuitive idea of integral on a smooth curve

I am reading about smooth curves and the book states that the integral of a continuous function $f$ on a smooth curve $\gamma$ is $\int_{\gamma} f(z) dz = \int_{a}^{b} f(z(t))z'(t) dt.$ Why is the integral defined like this or rather how did they derive this?

• You can safely think about this as an old fashioned substitution in single variable (real) calculus. – Doug M Apr 19 '17 at 22:40
• This is essentially the only way to define an integral on curves that is a) parametrisation-invariant and b) agrees with the usual sort of integral when taken on straight lines. – Chappers Apr 19 '17 at 22:41
• This post tackles geometrical interpretation of curve integrals, although they restrict themselves to real numbers. – user432158 Apr 19 '17 at 23:07

The idea is to first choose a smooth parameterization of the curve $\gamma$ of the form $z(t)$, $t \in [a,b]$. Then the right hand side at least makes sense.

But there is still an issue of whether this gives a well-defined integral, independent of the choice of parameterization. If $w(s)$, $s \in [c,d]$ is a different smooth parameterization which travels in the same direction along $\gamma$ then there exists a smooth change-of-variables map $s=h(t)$ which is a smooth map $h : [a,b] \to [c,d]$ taking $a$ to $c$ and $b$ to $d$, and having smooth inverse $h^{-1} : [c,d] \to [a,b]$. The change-of-variables formula then implies that $$\int_a^b f(z(t)) \, z'(t) \, dt = \int_c^d f(w(s)) \, w'(s) \, ds$$ Thus, $\int_\gamma f(z) \, dz$ is well-defined.

To summarize, the real reason for this definition is because it works, i.e. which primarily means that it gives a well-defined value of the integral. Other reasons why this definition "works" will no doubt arise from further theorems that you will study.

This is not an answer, but it´s far too long for a comment.

I myself have been asking questions about complex integrals, and from those questions there are a couple of things I can comment.

First. We define the contour γ as $z(t) = R(t)+iI(t)$, $a<t<b$;

and $∫f(z)dz$ (through $γ$) as the limit as $n$ approaches infinity of

$f(z1)[z1−z0]+f(z2)[z2−z1]+...+f(zn)[zn−z(n−1)]$ (Riemann Sum)

where $zn=z(tn) , a=t0<t1<t2<⋯<tn=b$

then, the latter is equal to

$f(z(t1))[z(t1)−z(t0)]+f(z(t2))[z(t2)-z(t1)]+...+f(z(tn))[z(tn)-z(t(n-1))]$

and as $n$ tends to infinity, $dz$ tends to $0$, so

= $f(z(t1))[z'(t1)dt]+f(z(t2))[z'(t2)dt]+...+f(z(tn))[z'(tn)dt]$

and so

$∫f(z)dz$ (through γ) = $∫f(z(t))z′(t)dt$ (from a to b)

The only issue with the latter is that when we first define $∫f(z)dz$ (through γ) as a Riemann Sum we are assuming that the integral will give the same value as long as all the intervals (zn-z(n-1)) tend to 0, for which I have no proof of.

Hope it helps!