The Pauli Exclusion principle states that the quantum states of all fermions must be orthogonal. This follows from the assertion that the wavefunction of the universe is antisymmetric under the exchange of any two identical fermions.

Another way of saying the same thing is, "If you were to take two identical fermions in the universe and switch them instantaneously, then the sign of the wavefunction of the universe would be reversed but otherwise the wavefunction would remain the same."

So, let's see what this implies. Suppose two identical fermions are in the same quantum state. Since the particles are in the same exact state, swapping them should have no effect upon the wavefunction of the universe. However, swapping always reverses the sign of the wavefunction of the universe! That's a change!

The solution is that the wavefunction of this situation is zero. If you reverse the sign of zero, you get zero back. Thus you can swap them, and get the same thing back, at the same time as swapping them and reversing sign. And the wavefunction which is zero is the wavefunction for a situation which is not occurring. Therefore, it cannot be that two identical fermions are in the same state.... because if it was true, it wouldn't be!

Now, we just considered the case where the fermions were in identical states. The thing about quantum mechanics is that real states are composed of an infinite number of components. Fortunately, we are free to choose how to divvy up these components, and consider them separately, so long as we make them all orthogonal. So, for any two fermions' wavefunctions, we split off their 'shared' component, and make it zero. This renders the full wavefunctions orthogonal (actually, if you want to be super-correct, you antisymmetrize them so that swapping back and forth actually looks like a multiplication by -1; but the only observable effects of this are sufficiently complicated that I won't do that here).

There is one last gap here: to connect this orthogonality to the whole idea of not overlapping. Orthogonality implies that two fermions' wavefunctions can't have the same sign of value in the same places without cancelling exactly elsewhere. If we try to put two fermions into the same place, we will have to put some of both of them into a second location where they cancel. But now we have two fermions and two locations. That's OK. Let's add a third fermion. Well, the math gets trickier, but it ends up being that we have to add a third location... hopefully, you see where this is going. In reality, you won't use discrete locations, but integrate over regions - but the same idea applies: in order for the wavefunctions to be orthogonal, they must not overlap, except in ways that do not make the heuristic statement that they are 'not in the same place' obscenely inaccurate.

This heuristic only holds if you understand (or at least accept... :) ) that two electrons are considered in different 'places' even if the only difference in their state is that they have opposite quantum spin. That is why you can fit two electrons into one S orbital, for example.
To tie this in: the way one expresses 'opposite spin' is via the spinor of a particle... and spinors that refer to opposite directions of spin will be orthogonal to each other. No, spinors are not the axial ω vectors like in classical mechanics. ω vectors contain direction and magnitude of ω. Spinors contain direction, phase, and magnitude... of the wavefunction, not the frequency.


Now, it may seem that all of this is saying that a mathematical principle is capable of exerting a force upon an object (i.e. degeneracy forces that hold up atoms and the degenerate matter of white dwarves and neutron stars). It's not a force like the other forces of nature, though. The fermion does not exert forces on the particles it's excluding with to maintain that exclusion; it just can't go there. The other particle will end up pushed on, though, because antisymmetrization means that it's also getting pushed on by whever force is pushing on the first electron. So what's holding up white dwarves and neutron stars? The kinetic part of the energy of a particle is proportional to the curvature of its wavefunction. If you try to pack a fermion into a tight space, it needs to go from being low density to very very high density back down to low density in a very short distance. This requires high curvature, and thus high energy. Looked at from a k-space perspective, you always have unlimited room at shorter wavelengths, but pushing particles into those short wavelength (and thus high energy) states requires a lot of energy.

siren: Fermions' wavefunctions need not by antisymmetric (odd) functions. Indeed, the very concept of fermions' wavefunctions being "antisymmetric" makes no sense as a universal principle, since it implies that there is a preferred center for the universe for fermions (and thus, pretty much, the entire universe) to be antisymmetric around. Also, this would require all fermions to exist equally on one half of the universe as the other. While this is a conceivable rule, it certainly does not lead to the results we are talking about.

The flaw is that when you calculate the probability associated with a wavefunction, you first multiply it by its complex conjugate (essentially, square it). This removes all negative portions, so there cannot be any cancellation of left versus right, because it's all positive. All cancellation must occur as addition of wavefunctions to each other, before this step. What you said about integrating odd functions is not relevant (though it is true).