Principal value of $\frac{1}{x^2}$ over $\mathbb{R}$ If I look at the the integral
$$
PV\int_{-\infty}^{\infty}\frac{1}{ax^2+bx+c}\;\mathrm{d}x
$$
under the condition $b^2>4ac$, I can conclude this equals zero. Graphically this makes sense, the divergent positive and negative areas "cancel" with each other. If instead $b^2=4ac$ there is no negative area to cancel the divergence of the positive area, so I would think it should be infinite.
However, if in my work I take a limit as $b\rightarrow0$ with $a=1$, $c=0$, the integral continues to be zero. Wolfram also gives the result 
$$
PV\int_{-\infty}^{\infty}\frac{1}{x^2+bx}\;\mathrm{d}x = 0 \;\text{for}\ b\in \mathbb{R}
$$
Notably this includes $b=0$. Symbolically I keep coming to the conclusion that
$$
PV\int_{-\infty}^{\infty}\frac{1}{x^2}\;\mathrm{d}x = 0
$$
but geometrically I'm missing something. Other principal values I've seen I can look at a graph and reason some form of area cancelation, among other things, but $\displaystyle\frac{1}{x^2}$ is positive everywhere. I also only get to this result by introducing other constants in the function then having them tend to, or equal $0$, which is different than any other principal values I've done.
Is this principal value actually correct? If so, why can $0$ be associated with an integral over a function positive everywhere?
 A: Let us start with the case $b^2> 4 a c$. Then there are two real solutions $x_1, x_2$ (let us say $x_1<x_2$) and we can write $ax^2+b x+c = a (x-x_1)(x-x_2)$. Your integral then assumes the form (PV is implied)
$$ \int_{-\infty}^{\infty}\!dx\, \frac{1}{a (x-x_1)(x-x_2)} = \lim_{\epsilon_1 ,\epsilon_2 \to 0^+} \int_{(-\infty,x_1-\epsilon_1) \cup (x_1 +\epsilon_1,x_2-\epsilon_2\cup (x_2+\epsilon_2,\infty)} \!dx\,\frac{1}{a (x-x_1)(x-x_2)}.$$ Note that the antiderivative is given by $$\int\!dx\frac{1}{a (x-x_1)(x-x_2)} = \frac{\ln[(x-x_2)/(x-x_1)]}{a(x_2-x_1)}.$$
We obtain the result
$$ \int_{-\infty}^{\infty}\!dx\, \frac{1}{a (x-x_1)(x-x_2)} = \lim_{\epsilon_1 ,\epsilon_2 \to 0^+} \frac{1}{a(x_2- x_1)} \left[ - \ln\left(\frac{\epsilon_2}{x_2-x_1}\right) + \ln\left(\frac{\epsilon_2}{x_2-x_1}\right) -  \ln\left(\frac{x_2-x_1}{\epsilon_1}\right) +  \ln\left(\frac{x_2-x_1}{\epsilon_1}\right) \right] = 0.$$
For the case $b^2 < 4 ac$, there is no real solution. The solutions are $x_1$ and $\bar{x}_1$ (with $\mathop{\rm Im}x_1 >0$).  Thus, we can forget about the principal value and obtain the result (e.g., via residue theorem)
$$ \int_{-\infty}^{\infty}\!dx\,\frac{1}{ax^2+b x+c} = \frac{\pi}{a \mathop{\rm Im}x_1} = \frac{2\pi}{ \sqrt{4 a c -b^2}}.$$
The case $b^2 = 4 ac$, is special. The limit starting from the first case is 0. The limit starting from the second case diverges. As you have noted the integral of $1/(x-x_1)^2$ is positive throughout the integration interval, so the correct result clearly diverges.
We have thus the general solution
$$ \mathop{\rm P.V.}\int_{-\infty}^{\infty}\!dx\,\frac{1}{ax^2+b x+c} = \begin{cases} \frac{2\pi}{ \sqrt{4 a c -b^2}}, &4 a c > b^2,\\
0 ,& 4 a c < b^2,\\
\infty,&4 a c = b^2.\end{cases}  $$
