# Antiderivative and contour integration

1. Feb 27, 2016

### Incand

I'm wondering if this could be used to calculate the value of a contour integral directly. If a function has an antiderivative on the entire complex plane, this implies the field is conservative so we should always get the same value no matter which path taken. Shouldn't this also mean integrals could just be computed by the fundamental theorem of calculus as
$\int_\gamma f(z)dz = F(z)\bigg|_{z_1}^{z_2}$
where the curve is from $z=z_1$ to $z=z_2$?

Yet I haven't actually seen a single example of this being done, every integral is computed by parametrization of the contour integral. Just doing a few examples It seems to work but doesn't this always work?

2. Feb 28, 2016

### Ssnow

So we can start with these assumptions:

1) $F'(z)=f(z)$ (there is the antiderivative)
2) We have a contour $\gamma$ that we can assume an arc of the circle ( as example from $\alpha_{1}$ to $\alpha_{2}$).

So $\int_{\gamma}f(z)dz=\int_{\alpha_{1}}^{\alpha_{2}}f(e^{i\theta})ie^{i\theta}d\theta=\int_{\alpha_{1}}^{\alpha_{2}}F'(e^{i\theta})ie^{i\theta}d\theta=$
$=\int_{\alpha_{1}}^{\alpha_{2}}[F(e^{i\theta})]'d\theta=F(e^{i\theta})|^{\alpha_{2}}_{\alpha_{1}}=F(z)|^{z_{2}}_{z_{1}}$

where $z_{1}=e^{i\alpha_{1}},z_{2}=e^{i\alpha_{2}}$. So these cases it seems to works, I think this is the complex version of the fundamental theorem of calculus ...

3. Feb 28, 2016

### Samy_A

Yes, this is correct.
See the Wikipedia article for the Cauchy integral theorem for a precise statement:

4. Feb 28, 2016

### Incand

That's good to know! I found the proof for Cauchy formula actually usually uses this in the proof. It's quite weird never having seen any integrals evaluated this way but I guess the focus been on all of those where $f$ isn't entire.