Potential Difference due to a infinite line of charge When a line of charge has a charge density $\lambda$, we know that the electric field points perpendicular to the vector pointing along the line of charge.
When calculating the difference in electric potential due with the following equations.
$$\nabla V=-\vec{E}$$
Therefore
$$\Delta V = -\int_{\vec{r_o}}^\vec{r_f}E\cdot \vec{dr}$$
knowing that
$$\vec{E} = \frac{\lambda}{2\pi\epsilon_or}\hat{r}$$
and that
$$\left\lVert\vec{r_f}\right\lVert < \left\lVert\vec{r_o}\right\lVert $$
Carrying out the integration (Hopefully correctly) I got
$$\Delta V = \frac{\lambda}{2\pi \epsilon_o} \ln(\frac{r_f}{r_o})$$
What confuses me is that the $\ln()$ is negative. I assume that the value should be positive since we move closer towards the line of charge should give us a positive change in electric potential. My best guess for my problem is that I missed a negative somewhere, but looking at online solutions they've got the same answer that I got.
 A: No, it's okay. The pontential difference increases as you go farther. The less you move away, the more similar potential you have (little difference). 
By the way


*

*You can't integrate in three dimensions that way. You're using cylindrical coordinates (because of the symmetry of the problem), and you integrate along $r$, which is $|\vec{r}|$. 

*The limits of integration are thus scalars. However, $\vec{E}$ is a vector, and you do the scalar product inside the integral, but fortunately the angle is 0 degrees.

*You missed the minus sign in front of the integral, so it appears outside the $\ln$. Was that your question? Because now 


$$\Delta V = -\dfrac{\lambda}{2\pi\varepsilon_0} \ln \left(\frac{r_F}{r_o}\right)$$
and the voltage difference increases when you go further, but in a negative sense, which means it becomes "more negative" as you move away.
A: To elaborate a bit on Bill's comment, you might consider a curve defined as follows in some cylindrical $(r,\theta,z)$ coordinate system:
$$\gamma(t) = \big(r(t),\theta(t),z(t)\big) = (t, 0, 0)$$
$$ t \in [r_0,r_f]$$
The tangent vector to this curve is
$$\frac{d\vec r}{dt} = \hat r $$
so
$$\Delta V = -\int_\gamma \vec E \cdot d\vec r = -\int_{r_0}^{r_f} \vec E \cdot \frac{d\vec r}{dt} dt = -\frac{\lambda}{2\pi\epsilon_0}\int_{r_0}^{r_f} \frac{dt}{t}  = -\frac{\lambda}{2\pi\epsilon_0}\ln\left(\frac{r_f}{r_0}\right) $$
$$ =\frac{\lambda}{2\pi\epsilon_0}\ln\left(\frac{r_0}{r_f}\right) $$

Whenever things like this happen, I find it useful to introduce an explicit, unambiguous parameterization of my curve, which usually resolves the issue.
