Potential due to line charge Is it possible to calculate the electric potential at a point due to an infinite line charge? Because potential is defined with respect to infinity.
 A: It is not possible to choose $\infty$ as the reference point to define the electric potential because there are charges at $\infty$.  This is easily seen since the field of an infinite line $\sim 1/r$ so the standard definition of $V(\vec r)$ as the integral
$$
V(r)=-\int_{r}^{\infty}\frac{\lambda}{2\pi\epsilon R}dR
=-\frac{\lambda}{2\pi \epsilon}\left(\log(\infty)-\log(r)\right)
$$
is clearly not well-defined because of the $\log(\infty)$.  Rather, it is often found in this case convenient to define the reference potential so that
$$
V(r)= -\frac{\lambda}{2\pi\epsilon}\int_{r}^{1}\frac{dR}{R}= -\frac{\lambda}{2\pi \epsilon}\left(\log(1)-\log(r)\right)=\log(r) \, .
$$ 
If there is a natural length scale $R_0$ to the problem, one can also define the dimensionless variable $\rho=r/R_0$.  Since $dR/R = d\rho/\rho$, the result is now that the potential at $\rho=1$, i.e. at $r=R_0$, is now set to $0$.  
Of course if you’re only interested in the potential difference between $r_0$ and $r_1$, the limits of the integrals are then $r_0$ and $r_1$ and the integral is perfectly well defined, as is the difference in potential between these two points.
A: Since this an infinite line - not an infinite sphere - there are plenty of points in space infinitely removed from it, which you can use as your zero reference points.
So, once you know how the field of the infinite charged line looks like (you can check here), you can calculate the electric potential due to this field at any point in space.
It is worth noting, that the electric field of an infinite line will be diverging, so, unlike the field of an infinite plane, it will be approaching zero at infinity and, therefore its potential at a random point in space won't be infinitely high.
A: It is possible. Electrosatic potential is just a scalar field whose negative gradient is the electric field. Due to this defintion it is indeterminate to the extent of an additive constant. (if you increase it everywhere equally, its slope remains the same everywhere) Only the potential difference between two points is measurable, which is called voltage. It is a convention that potential in the infinty is often taken zero, which is usefull, but 
you could easily call for example a point 2 meters away zero potential and obtain the same function only offset by a constant, but yielding the exact same forces.
And yes, as V.F. had said, there are infinite number of points being infinitely far from your line, so you could even use infinity as zero point, and easily obtain the potential by integration and symmetry considerations.
