Why can the equivariant volume of a non-compact space be finite? I am very confused with equivariance (equivariant cohomology etc). In specific when one tries to evaluate the equivariant volume of, say, $\mathbb{R}^2$ (with coordinates $x,y$) one finds that it is $2\pi/\epsilon$ where $\epsilon$ is the infinitesimal parameter that multiplies the Hamiltonian field,
$$ \text{Vol}[\mathbb{R}^2] =  \int e^{\omega - \epsilon H} $$
where $\omega = dx \wedge dy$ and $H=\frac{1}{2}(x^2+y^2)$.
How can I get some intuition on what the equivariant volume represents? And what is the difference with the normal volume? 
A reference is this one (mainly targeted to physicists though) page 18 for example.
 A: Your question is rather psychological. But nevertheless I will write some remarks.
Remark 1.      Maybe the formulation of your question is not the best. A non compact space may have finite volume (I mean just classical volume, not equivariant). For instance you may consider $\mathbb{C} = \mathbb{CP}^1   \ \backslash \ \{ \infty \}$. And consider symplectic volume of $\mathbb{C}$ with respect to restriction of Fubini-Study form. It is finite $\DeclareMathOperator{\Vol}{Vol} \Vol[\mathbb{C}] = \Vol[\mathbb{CP}^1]$. So the right question is as follows: let $M$ be symplectic, with infinite volume. Why the equivariant volume of $M$ can be finite?
Remark 2. Limit $\epsilon \rightarrow 0$.
$$\Vol[X] = \int e^{\omega - \epsilon H}.$$
If you consider case $\epsilon = 0$ then you get just the usual volume (divided by $n!$). In your example we see $$\infty = \lim_{\epsilon \rightarrow 0} \frac{2 \pi}{\epsilon}.$$
Remark 3. So some intuition comes from the words “regularization” and “deformation”. By “regularization” I mean that we have a procedure which gives a finite answer for infinite quantity (volume) expressed via some regulator $\epsilon$. “Deformation” means the following: even if the answer is finite, it still depends on $\epsilon$ (see the example of the sphere).
Remark 4. One should be careful about this procedure for non compact space. You may consider anticlockwise rotation instead of clockwise. This changes the sign of Hamiltonian. Then for positive $\epsilon$ your integral just diverges.
$$\int e^{\epsilon(x^2+y^2)} dx \wedge dy.$$
I would give the following interpretation of this result. Maybe it is more convenient to get intuition not just about equivariant volume, but about integrals of equivariantly closed forms. It is safe to integrate closed form over a compact manifold. One can check that for $\epsilon > 0$ form $e^{\omega - \epsilon(x^2+y^2)}$ extends to regular form on sphere $S^2 = \mathbb{R}^2 \cup \infty$. Also one can apply localization formula for this form and see that contribution of infinity is 0. That is why the naive localization formula works in this case (“naive formula” means sum over fixed points of $\mathbb{R}^2$).
For $\epsilon > 0$ form $e^{\omega + \epsilon(x^2+y^2)}$ does not extend to $S^2$.
Remark 5. In general, it is not possible to solve such a problem of divergence by changing the sign of the Hamiltonian. For instance, you can consider $\mathbb{R}^4$ with symplectic form $dx_1 \wedge dx_2 + dx_3 \wedge dx_4$ and Hamiltonian $$H = \frac{1}{2}(x_1^2 + x_2^2 - x_3^2 - x_4^2).$$
So, it seems that there is no good theory of equivariant volume in general. It is not a problem for physicists. They just apply the localization formula, pretending that the contribution of infinity is 0.
