Physics-Electrostatics: Find electric potential on a center of charged rod
OK, as now we have another answer, I'll add my comment as answer, too:

The math is correct, if the integral is calculated, one gets
$$V_{1/2}=kQ'\int_0^a\frac{1}{x}\,dx=kQ'(\ln(a)\underbrace{-\ln(0)}_{=+\infty})=+\infty$$

If this looks odd, keep in mind the well known potential of a point charge:

$$V(r)=\frac{Q}{4\pi\varepsilon_0}\frac{1}{r}$$

For \$r\to0\$, this also rises to \$+\infty\$ . 



About @Andy aka's answer: 


  Capacitance is zero because gap is infinite therefore V is infinite.


While it sounds elegant, it's in general wrong.

I guess this arises from the formula 
$$C=\varepsilon_0\frac{A}{d}$$ 
for the capacitance of a parallel-plate capacitor. Indeed, when d becomes infinite, C becomes zero. But this formula is only valid for an ideal capacitor, where d is not too large, compared to the size of the plates.

If the distance is large, the plates do not behave like an ideal parallel plate capacitor any more, the field looks more like that of two charges spheres. And the capacitance of two spheres of radius R and distance D is:

$$C=2\pi\varepsilon_0\left(\frac{1}{\frac{1}{R}-\frac{1}{D-R}}\right)\quad\xrightarrow{D\ \to\ \infty}\quad C=2\pi\varepsilon_0 R$$

If the spheres have the same origin but different radii \$R_{inner}\$ and \$R_{outer}\$:

$$C=4\pi\epsilon_0 \frac{1}{\frac{1}{R_{outer}}-\frac{1}{R_{inner}} } \quad\xrightarrow{R_{outer}\ \to\ \infty}\quad C=4\pi\epsilon_0 R_{inner}$$

So, just having an infinite distance does not lead to a capacitance of zero. (This effect is calles self capacitance, by the way)

For the math, see below.



It is interesting that a point charge has an infinite potential, and a one-dimensional charge distribution as in this question also has an infinite potential. Thinking about this, I came to the conclusion, that a charge distribution without volume has an infinite potential. 

The reason for this can be understood from Maxwell's first equation (or the Gauß-equation):

$$\int_A\vec{E}\, d\vec{A}=\varepsilon_0\int_V\rho\, d\vec{V}$$

If we have a look at a charged point / line / plane, we can always put a closed surface around it, such that the charge is fully enclosed  and the electric field has the same strength everywhere (implies that E is perpendicular on surface, as it's also conservative). Then, the equation becomes as simple as:

$$E\cdot A=\varepsilon_0\cdot Q$$

Now, we can choose a smaller and smaller surface, like the shells of an onion. A goes against zero, and because Q is constant, E has to raise against infinity. And since potential is the integral of the E-field (along a path), the potential also raises to infinity.

If the charge distribution has a volume, the enclosed charge is not constant. For example, a solid, homogeneous charged sphere of radius R has the potential

$$V(r)=\begin{cases}4\pi\varepsilon_0\cdot\frac{r}{R^2}&amp;\text{for}&amp; r\le R \\ 4\pi\varepsilon_0\cdot r&amp;\text{for} &amp;r \ge R\end{cases}$$



About the math:

The potential field of a charges sphere is that of a point charge in its center, i.e. \$V(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}\$

Same sized spheres:

For the two same-sized spheres with opposite charge, the potential field is
$$V_{tot}(r)=V(r)-V(|D-r|)=\frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{r}-\frac{Q}{|D-r|}\right)$$

Now, the spheres have radius R, so the potential difference is
$$U=\Delta V_{tot}=V_{tot}(D-R)-V_{tot}(R)=...=\frac{2}{4\pi\varepsilon_0}\left(\frac{Q}{R}-\frac{Q}{D-R}\right)$$

Now, it's just
$$C=\frac{Q}{U}=2\pi\varepsilon_0\left(\frac{1}{\frac{1}{R}-\frac{1}{D-R}}\right)$$

Spheres with same center:

$$V_{inner}(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r} \quad\text{for}\quad r\ge R_{inner}$$

$$V_{outer}(r)=\frac{1}{4\pi\varepsilon_0}\frac{-Q}{r} \quad\text{for}\quad r\ge R_{outer}$$

Outside the outer sphere, the field cancel each other. The potential at the outer sphere is generated by the inner sphere only. So, the potential difference is 

$$U=V_{inner}(R_{inner})-V_{inner}(R_{outer})$$

and

$$C=\frac{Q}{U}=4\pi\varepsilon_0\left(\frac{1}{\frac{1}{R_{outer}}-\frac{1}R_{inner}}\right)$$