# Proof of the Contour Integral Formula

A couple days ago I asked a question concerning complex integrals

The person that answered explained to me that the integral of the contour $C$ defined by $γ(t),a≤t≤b$  can be written as follows

$\int_Cf(z)dz$ (about $C$) is equal to the limit as $n$ tends to infinity of

$f(z_1)[z_1-z_0] + f(z_2)[z_2-z_1] + ... + f(z_n)[z_n-z_{n-1}]$

where $z_n = γ(t_n)$, $a = t_0<t_1<t_2<\cdots<t_n=b$

The issue with the latter is that it implies that if all the $z_j-z_{j-1}$ tend to $0$, we will get the same result. I suppose this assumption is correct, but I haven´t been able to find any proof on the internet

If someone could direct me to or show me the most clear proof you know I would truly appreciate it!

We first need to assume the contour is rectifiable, which is to say it can be assigned an arc length. If the parameterization $t \mapsto \gamma(t)$ is given then it is possible to write $$\int_\gamma f(z)dz = \int_a^b f(\gamma(t))\gamma'(t)dt.$$ The definition of the complex integral thus follows from that of the real integral. The rectifiability of $\gamma$ is needed to ensure good behavior for the measure $dz = \gamma'(t)dt$.
To see how this is equivalent to your definition, recall how the real integral (of Riemann) can be defined as a limit of Riemann sums calculated over successive refinements of the partitions of $[a,b]$. Clearly each partition of $[a,b]$ induces a partition of the contour by setting $\gamma(p)$ as a member of the contour partition whenever $p$ is a member of the partition of $[a,b]$. As the Riemann sum approaches the integral, each $\gamma(p)$ asymptotically approaches a straight line which connects its endpoints $z_j$ and $z_{j-1}$. In the limit of an infinitely refined partition these two will be identical (assuming again that $\gamma$ is rectifiable), and thus the definitions coincide.