Using Antiderivative to compute complex integral I have been practicing solving complex integrals. When am I allowed to simply use the antiderivative of the integral to apply the Fundamental Theorem of Calculus to solve the integral? Is it when it is easy to find the antiderivative?
 A: Let $\gamma: [a, b] \to \Bbb{C}$ be continuously differentiable. The path-integral of a complex-valued function $f: \Bbb{C} \to \Bbb{C}$ over the path $\gamma$ is defined to be
\begin{align}
\int_\gamma f(z) \, dz := \int_a^b f( \gamma(t)) \gamma'(t) \,dt\,.
\end{align}
Note that on the right we have a regular Riemann integral. The integrated function can be written in the form
$$
f(\gamma(t)) \gamma'(t) = u(t) + i v(t)\,.
$$
Now 
$$
\int_\gamma f(z) \, dz = \int_a^b u(t) \, dt + i \int_a^b v(t) \, dt\,.
$$
The integrals $\int_a^b u(t) \, dt$ and $\int_a^b v(t) \, dt$ are regular real Riemann integrals, so you can apply the fundamental theorem of calculus to them whenever you can find the antiderivative. So assume we have functions $V$, $U$ s.t. $V'(t) = v(t)$ and $U'(t) = u(t)$ for all $t$. Now
$$
\int_\gamma f(z) \, dz = U(b) - U(a) + i \big( V(b) - V(a) \big)\,.
$$
But in practice, we often don't bother to write $f(\gamma(t))\gamma'(t)$ as $u(t) + i v(t)$. We may use shortcuts like the fact that $\frac{d}{dt} e^{it} = i e^{it}$: consider the example $f(z) = 1$ and $\gamma(t) = e^{it}$, $t \in [0, 2\pi]$. Now
$$
\int_\gamma f(z) \, dz = \int_0^{2\pi} 1 \cdot i e^{it} \, dt \,.
$$
Now we could write $i e^{it}$ as $i( \cos t + i \sin t) = - \sin t + i \cos t$ and find the antiderivatives of the imaginary and real parts, but it's more convenient to use the "complex antiderivative" $e^{it}$. So we have
$$
\int_\gamma f(z) \, dz = e^{i 2 \pi} - e^{i \cdot 0}\,.
$$
So you are pretty much always using the fundamental theorem of calculus, unless you work with Riemann sums, which I doubt you do too often. Just remember that the path integral is defined as a certain "regular" Riemann integral over an interval. Sometimes the integrated function is complex valued though, but it doesn't really change anything.
