Explain to me why I can't use $u$-substitution with $u=e^{ix}$? To be clear, I know you can't do this, since then any integral of the form
$$
\int_{-\pi}^\pi f(t)dt
$$
would have to equal zero by the fact that $e^{i\pi}=e^{-i\pi}=-1$, so the bounds of integration would be from -1 to -1, which is nonsense.
But is there a clear explanation as to why I can't introduce a complex function substitution in this fashion? As opposed to saying "well you can't do it unless you prove you can", which is, of course, true.
 A: You absolutely can use this substitution, but you have to bear in mind that the endpoints are not the only thing that matters. The entire interval of $[-\pi,\pi]$ is changed: that is, your new range of integration should be the image of $[-\pi,\pi]$ under the function $u(x) = e^{ix}$.
It just so happens that in ordinary calculus, with appropriate substitutions and integrands, only the endpoints matter - and hence we forget the importance of everything in-between. (This is also true in somewhat limited circumstances in complex analysis.) But all $u$-substitution is doing is using the chain rule, to transform an integral into an equivalent one. We may think of the area under the curve being transformed by the substitution, even -- into a different shape with the same area.
The key realization: The substitution $u = e^{ix}$ turns $[-\pi,\pi]$ into the unit circle in the complex plane: it may "start" and "end" at $-1$, but every point in-between is mapped onto the unit circle too. (This can be justified with Euler's formula.) This is especially obvious if you consider breaking up $[-\pi,\pi]$ into very many smaller subintervals, and consider the impact of the $u$-substitution on each of them, and then adding up the corresponding integrals.
