Principal value with non-analytic function? Can the principal value integral $$PV\int_{-1}^{2} \frac{\log{x}}{x} \ dx$$be defined? The definition involving contour integration clearly doesn't work, but what about the 'recipe' involving taking the limit as you approach the singularity?
Presumably you have to somehow define the logarithm on the negative real axis. I imagine we'll want to make sure that our definition is consistent with the real logarithm on the positive side, but I'm not well-versed enough to know if a full definition can be consistently given.
Considering a second case, is $$\int_{0}^{3} \frac{\log{x}}{x-2} \ dx $$ any better? Here we avoid questions about the log on the negative real axis, but it still has a branch point at $0$.
 A: Up to the definition of CPV,
$$PV\int_{-1}^{2} \frac{\log{x}}{x} \ dx = \lim_{\epsilon \to 0+}\left(\int_{-1}^{-\epsilon} \frac{\log{x}}{x} \ dx +\int_{\epsilon}^{2} \frac{\log{x}}{x} \ dx \right)$$ as the principal value of $\log(x)$ has a singularity at $x=0$. Because for $x<0$ the principal value of $\log(x)=\log|x|+\pi i$, the  limit under consideration  does not exist: consider the integral of  $i\pi/x$ over $(-1,-\epsilon)$. The one also may be treated as $\infty$. Next, up to Maple, $$ int(ln(z)/(z-2), z = 0 .. 3, CauchyPrincipalValue) $$ produces $$1/6\,{\pi }^{2}- \left( \ln  \left( 2 \right)  \right) ^{2}-{\it dilog
} \left( 3/2 \right)
  .$$ See dilog for info. 
A: By the definition,
$$
\begin{align}
&\mathrm{PV}\int_{-1}^2\frac{\log(x)}{x}\,\mathrm{d}x\\
&=\lim_{\epsilon\to0^+}\left(\int_{-1}^{-\epsilon}\frac{\log(x)}{x}\,\mathrm{d}x
+\int_\epsilon^2\frac{\log(x)}{x}\,\mathrm{d}x\right)\\
&=\lim_{\epsilon\to0^+}\left(\int_{-1}^{-\epsilon}\frac{i\pi}{x}\,\mathrm{d}x
+\color{#C00000}{\int_{-1}^{-\epsilon}\frac{\log|x|}{x}\,\mathrm{d}x
+\int_\epsilon^1\frac{\log|x|}{x}\,\mathrm{d}x}
+\int_1^2\frac{\log|x|}{x}\,\mathrm{d}x\right)\\
&=\left(\lim_{\epsilon\to0^+}\int_{-1}^{-\epsilon}\frac{i\pi}{x}\,\mathrm{d}x\right)
+\int_1^2\frac{\log(x)}{x}\,\mathrm{d}x\\
&=i\pi\lim_{\epsilon\to0^+}\log(\epsilon)
+\int_1^2\frac{\log(x)}{x}\,\mathrm{d}x\\
&=-i\infty+\frac12\log(2)^2
\end{align}
$$
So the Principal Value would exist if only the $i\pi$ were not in the $\log(x)$ for $x\lt0$. With this in mind, and mimicking the proof above, we get that
$$
\begin{align}
\mathrm{PV}\int_{-1}^2\frac{\log|x|}{x}\,\mathrm{d}x
&=\int_1^2\frac{\log(x)}{x}\,\mathrm{d}x\\
&=\frac12\log(2)^2
\end{align}
$$

For the second case
$$
\begin{align}
&\mathrm{PV}\int_0^3\frac{\log(x)}{x-2}\,\mathrm{d}x\\
&=\mathrm{PV}\int_0^3\left(\frac{\log(x)}{x-2}+\frac{\log(x)}{2}\right)\,\mathrm{d}x
-\int_0^3\frac{\log(x)}{2}\,\mathrm{d}x\tag{1}\\
&=\mathrm{PV}\int_0^3\frac{x\log(x)}{2(x-2)}\,\mathrm{d}x-\frac32\Big(\log(3)-1\Big)\tag{2}\\
&=\mathrm{PV}\int_0^3\frac{2\log(2)}{2(x-2)}\,\mathrm{d}x
+\int_0^3\frac{x\log(x)-2\log(2)}{2(x-2)}\,\mathrm{d}x-\frac32\Big(\log(3)-1\Big)\tag{3}\\
&=\small\mathrm{PV}\int_1^3\frac{\log(2)}{x-2}\,\mathrm{d}x
+\int_0^1\frac{\log(2)}{x-2}\,\mathrm{d}x+\int_0^3\frac{x\log(x)-2\log(2)}{2(x-2)}\,\mathrm{d}x-\frac32\Big(\log(3)-1\Big)\tag{4}\\
&=\small\mathrm{PV}\int_{-1}^1\frac{\log(2)}{x}\,\mathrm{d}x
+\int_0^1\frac{\log(2)}{x-2}\,\mathrm{d}x+\int_0^3\frac{x\log(x)-2\log(2)}{2(x-2)}\,\mathrm{d}x-\frac32\Big(\log(3)-1\Big)\tag{5}\\
&=\int_0^1\frac{\log(2)}{x-2}\,\mathrm{d}x+\int_0^3\frac{x\log(x)-2\log(2)}{2(x-2)}\,\mathrm{d}x-\frac32\Big(\log(3)-1\Big)\tag{6}\\
&=-\log(2)^2+\int_0^3\frac{x\log(x)-2\log(2)}{2(x-2)}\,\mathrm{d}x-\frac32\Big(\log(3)-1\Big)\tag{7}\\
\end{align}
$$
where the integrand is a bounded function over a finite interval.
Note that
$$
\begin{align}
\mathrm{PV}\int_{-1}^1\frac{2\log(2)}{2x}\,\mathrm{d}x
&=\lim_{\epsilon\to0^+}\left(\int_{-1}^{-\epsilon}\frac{2\log(2)}{2x}\,\mathrm{d}x
+\int_\epsilon^1\frac{2\log(2)}{2x}\,\mathrm{d}x\right)\\
&=\lim_{\epsilon\to0^+}0\\
&=0\tag{8}
\end{align}
$$
Motivation:
$(1)$: remove the possible problem of integrating $\log(x)$ near $0$
$(2)$: algebra and integration
$(3)$: move the singularity to an easier to handle form, away from the logs
$(4)$: break out the symmetric part around the singularity
$(5)$: change variables $x\mapsto x+2$
$(6)$: apply $(8)$
$(7)$: integration
A: The complex logarithm changes its value when you evaluate the function at the negative real line. Consider the familiar expression $$z=re^{i\theta}$$
Taking the logarithm on both sides give you $\log[z]=\log[r]+i(\theta+2n\pi)$. Therefore if $z$ is negative, $\theta\ge \frac{\pi}{2}$, and the imaginary part is not really well defined unless you want the principal value to be $$\log[r]+\pi i$$instead. If you defined it this way, then integrating $\frac{\log[r]}{r}$ from $-2$ to $3$ is no different from integrating $\int^{2}_{0}\frac{\log[r]}{r}+\int^{3}_{0}\frac{\log[r]}{r}+2\pi i$, where the first two are real valued integrals. But I think $\int^{2}_{0}\frac{\log[r]}{r}$ cannot be well defined, as one has to define it by 
$$\frac{1}{2}\log[r]^{2}|^{2}_{\epsilon}$$
which clearly diverges. Since the two integrals add up instead of cancelling each other at the singularity, we cannot compute the "principal value" as you desired. 
The computation is at here. 
