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!
Here are some informal, notional thoughts. For a rigorous treatment you can see the text of Alfhors.
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.