To integrate a function $f:\Bbb C \to \Bbb C$, do I need the domain of integration to be a curve? To integrate a function $f:\Bbb C \to \Bbb C$,  do I need the domain of integration to be a curve? Isn't it possible to do something like $$\int_\Bbb C f(z)dz?$$
Consider the function $e^{-z^{10}}$, can we calculate $\int_\Bbb C e^{-z^{10}}dz$?
To be extremely clear, I'm not interested in how to calculate the integral for that especific function, but on how it would be done. I would also like to get a bit of clarification on what things like $dz\wedge d\overline z$ mean.
 A: First of all $f(z)\,dz$ is a "one-form", which means that you can integrate it over one-dimensional objects (i.e. curves). So,
$$
\int_{\mathbb{C}} f(z)\,dz
$$
is nonsense.
But, you can integrate a "two-form", $f(z)\,dz \wedge d\bar z$ or $f(z)\, dx\wedge dy$ (which is almost the same thing), over the complex plane. You may be used to writing the integrand in double integrals without wedges $\wedge$, and that's alright.
On the other hand, your example, $f(z) = \exp(-z^{10})$ does not have a convergent double integral over the complex plane: If $z = re^{it}$, then
$$
|\exp(-z^{10})| = |\exp(-r^{10}(\cos 10t + i\sin 10t))| = \exp(-r^{10}\cos 10 t)
$$
which grows very fast as $r\to\infty$ along rays where $\cos 10t = -1$.
A: It is possible to compute an integral for a function $f : \mathbb C \to \mathbb C$. To do so, you can separate the real and the imaginary parts. Then, you have to compute real two double integrals.
As your map $z \mapsto e^{-z^5}$ is continuous, it is integrable on $\mathbb C$. The value of the integral being potentially infinite.
If we work in polar coordinates: $z=re^{i\theta}$, you get $$f(z)=e^{-z^5}=e^{-r^5e^{5i\theta}}=e^{-r^5(\cos 5\theta +i\sin 5 \theta)}=e^{-r^5 \cos 5 \theta}(\cos(r^5 \sin 5 \theta)- i \sin (r^5 \sin 5 \theta)).$$
