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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!

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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.

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  • $\begingroup$ Right, from that equality one is able to treat the complex integral (somewhat) as real and 'prove' the formula. However, in that question I mentioned, the proof provided requires that we assume that as long as all the intervals tend to 0 we will get the same result. $\endgroup$ – Leo Apr 19 '17 at 22:40
  • $\begingroup$ What I meant with the last comment is that I´m in need of understanding the formula to prove that equality you just wrote, so using that equality to prove the formula would be circular. $\endgroup$ – Leo Apr 19 '17 at 22:45
  • $\begingroup$ If you want to prove the definition in terms of a real valued integral in terms of the definition which uses a partition of the contour, you have to accept the latter as a definition. $\endgroup$ – Mortified Through Math Apr 19 '17 at 22:58

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