"Can move it an infinite distance" is no way to measure energy. Give an object any amount of kinetic energy, and in the presence of no other forces, It will move an infinite distance.
$\int_{r}^{\infty} \vec{E} \cdot \vec{dl}$
Represents the amount of work done by the electric field in moving an object from r to $\infty$
For a point charge, For all r ≠ 0, this qauntity is finite.
Probably not what your asking :
The formula for electric field energy
$\iiint \frac{1}{2}\epsilon_0 |\vec{E}|^2 dv $
Is derived by finding the amount of work to assemble a charge distribution
This is derived from the discrete version of this formula, finds the work to build up a distribution of point charges. Deriving this formula needs you to disregard the potential of the charge you're building up, at that particular moment in time, ( since it does not repell itself)
In deriving the continuous version, the generalisation to a distribution ignores this condition since each element $\rho dv$'s potential is zero for a finite $\rho$. ( as discussed in griffiths) so it yields the same result either way.
However, For a point charge $\rho = Q\delta^3(r)$
This is infinite, so our formula somewhat breaks down for point charges
This leads to an incorrect reading that the energy of a single point charge is infinity.
This result of infinities when dealing with problematic point charges is often manually subtracted, as the formula is only meant to find the potential energy of the distribution, and not the added infinities.
Modeling charges a spherical balls of charge with a finite radius however fixes this problem and allows the formula to work without ignoring certain assumptions