When dealing with infinitely long line charges (basically a cylindrical geometry) calculating the potential relative to infinity becomes a problem. You have to establish a reference (ground/earth) at a finite location. So, your result of an infinite potential difference is not incorrect, although it is confusing the first time you see it.
This site provides a description of why this happens. You are effectively grounding one end of your distribution, then stacking an infinite amount of charge down the line.
Yes it is a complicated generalization.
The electric field at a point $\mathbf r$ is
$$
\mathbf E(\mathbf r) = k\int \frac{\mathbf r - \mathbf r'}{|\mathbf r - \mathbf r'|^3} dq'.
$$
For the problem you're attempting to solve, let $R$ be the radius of the ring to avoid notational confusion with other "r" variables, then
$$
\mathbf r = (x,0,0), \qquad
\mathbf r'= (R\cos\theta', R\sin\theta', 0).
$$
It follows that
\begin{align}
\frac{\mathbf r - \mathbf r'}{|\mathbf r - \mathbf r'|^3}
= \frac{(x - R\cos\theta', -R\sin\theta', 0)}{((x-R\cos\theta')^2 + R^2\sin^2\theta')^{3/2}}.
\end{align}
For the problem at hand, the charge measure $dq'$ is
$$
dq' = \frac{Q}{\pi R}(Rd\theta') = \frac{Q}{\pi}d\theta'.
$$
Plugging these in reveals that to compute the field at a given $x\neq 0$, we'd need to compute the following integral:
$$
k\int_{-\pi}^0 \frac{(x - R\cos\theta', -R\sin\theta', 0)}{((x-R\cos\theta')^2 + R^2\sin^2\theta')^{3/2}} \frac{Q}{\pi}d\theta'.
$$
This is a hard integral compared to the case in which $x=0$ because in that case it collapses to
$$
-\frac{kQ}{\pi R^2}\int_{-\pi}^0 (\cos\theta', \sin\theta', 0)d\theta' = \left(0, -\frac{2kQ}{\pi R^2}, 0\right).
$$
Best Answer
Let us say you have a charged unit semi-circle ($\rho=1,-\frac{\pi}{2}\leq\varphi\leq\frac{\pi}{2}$) and are trying to calculate the potential in a point with coordinates $(x,y,z)$. You get (I neglect all kinds of constant factors): $V(x,y,z)=\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}d\varphi\frac{1}{\sqrt{(x-\cos(\varphi))^2+(y-\sin(\varphi))^2+z^2}}=\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}d\varphi\frac{1}{\sqrt{x^2+y^2+z^2+1-2 x\cos(\varphi)-2 y\sin(\varphi)}}$. I guess this may be an elliptic integral. To get the field, you need to take a gradient of this integral.
EDIT: To answer MaxMackie's question in the comment: a point charge in a point $\overrightarrow{r'}=(x',y',z')$ creates a field described by potential $V(\overrightarrow{r'},\overrightarrow{r})=\frac{1}{|\overrightarrow{r}-\overrightarrow{r'}|}$, where $\overrightarrow{r}=(x,y,z)$ is the field point (again, constant factors are neglected). If you have a charge distribution $\rho(\overrightarrow{r'})$, the new potential $V_\rho(\overrightarrow{r})=\int d\overrightarrow{r'}\rho(\overrightarrow{r'})V(\overrightarrow{r'},\overrightarrow{r})$. To get the electric field, take a gradient of the potential with respect to $\overrightarrow{r}$.