# Complex antiderivative

I am confused on a couple things:

1.) Why is it that an integral of a complex valued function of a complex variable exists if f(z(t)) is piecewise continuous (and/or piecewise continuous on $\mathbb{C}$) and not continuous, like a real?

2.) Why is it that one cannot make use of an antiderivative to evaluate an integral of a function like $1/z$, on a contour of something like $z=2e^{i\theta}$ positively oriented with $-\pi \le \theta \le \pi$? That is, because $F'(z)=1/z$ is undefined at $0$, it is disqualified (I think). But why?

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I don't understand the first part. The integral can be piecewise continuous in order for it to exist? If the integral has properties, then it exists, regardless of the specific properties, no? – joriki Apr 24 '12 at 3:26
Yes, I think I put it poorly! A complex integral of a function exists if the function is piecewise continuous in the interval of that integral. The function is also piecewise continuous on the contour C. I'm sure for real valued integrals of real variables, the function must be continuous - not piecewise continuous. – user5262 Apr 24 '12 at 3:32
We just define it as the sum of the integrals of the continuous pieces of the curve – Pedro Apr 24 '12 at 3:35
@Joshua: Real-valued functions that are integrable definitely do not have to be continuous. Integrability is a very weak condition. In fact, it's not too much of an exaggeration to say that every nonnegative, bounded (real-valued) function on a closed interval of $\mathbb{R}$ is integrable; there are counterexamples, but you cannot write one down "explicitly"; you need some form of the axiom of choice to find one. – Ted Apr 24 '12 at 6:07
@Perdo: We just define it as, but it actually isn't? I'm confused. What we define something as should cohere with what it 'is' in the most humanly way possible; or we are defining it, in order to make it what it is. – user5262 May 9 '12 at 3:37

If the function $f(z)$ has an antiderivative $F(z)$ on a domain that includes a contour $\Gamma$ that goes from point $a$ to point $b$, then $\int_\Gamma f(z)\ dz = F(b) - F(a)$. The problem is that in many cases such an antiderivative can't exist. A case in point is your example, $\int_\Gamma \frac{dz}{z}$, where $\Gamma$ is the counterclockwise circle of radius $2$ centred at $0$. You'd like to say that an antiderivative of $f(z) = 1/z$ is $\log(z)$, but there is no version of $\log(z)$ that is continuous on that whole circle. Indeed that must be so, because since $b=a$ (this contour ends in the same place where it starts), you'd have $\int_\Gamma \frac{dz}{z} = F(a) - F(a)=0$. But, parametrizing the contour as $z(t) = 2 e^{it}$, $$\int_\Gamma \frac{dz}{z} = \int_{-\pi}^{\pi} \frac{2 i e^{it}\ dt}{2e^{it}} = \int_{-\pi}^\pi i\ dt = 2 \pi i$$