Simple integration: $\int_{-1}^1 \frac{1}{x^2} dx$ Consider the integral $\int_{-1}^1 \frac{1}{x^2} dx$ then as a Riemann integral it diverges, also as a Lebesgue integral it is $\infty$. However,
for a moment forget $0 \in [-1,1]$ and "integrate":
$$\int_{-1}^1 \frac{1}{x^2} dx = \left. -\frac{1}{x} \right|_{-1}^1 = -2$$
Is there a way that one can make sense of this correspondence? Or it is completely absurd...    
 A: It makes sense as an integral from -1 to 1 that bypasses the origin in the complex plane, e.g. by moving above it. 
A: The anti-derivative of $\frac{1}{x^2}$ is $-\frac{1}{x}$ when $x\ne 0$.  Inasmuch as the integration limits include the singularity point $x=0$, the anti-derivative is not $-\frac{1}{x}$.  
And attempting to split the integral into two improper Riemann integrals leads to the result
$$\begin{align}
\int_{-1}^1\frac{1}{x^2}\,dx&=\lim_{\epsilon_1\to 0^{+}}\int_{-1}^{-\epsilon_1}\frac{1}{x^2}\,dx+\lim_{\epsilon_2\to 0^{+}}\int_{\epsilon_2}^{1}\frac{1}{x^2}\,dx\\\\
&=\lim_{\epsilon_1\to 0^{+}}\left(\frac{1}{\epsilon_1}-1\right)+\lim_{\epsilon_2\to 0^{+}}\left(\frac{1}{\epsilon_2}-1\right)\\\\
&=\infty
\end{align}$$
even if interpreting the integral as a Cauchy Principal value with $\epsilon_1=\epsilon_2$.
A: The statement is: if $f \colon \left[a,b\right] \to \Bbb R$ is a continuous function and $F$ is a primitive of $f$ in $\left[a,b\right]$, then $\int_a^b f(x)\,{\rm d}x = F(b)-F(a)$.
Your computation is invalid because $1/x^2$ is not continuous on $\left[-1,1\right]$, in fact, is not defined for $x=0$. And no matter what (real) value you pick for $f(0)$, $f$ won't be continuous and you can't apply the FTC. If you're integrating on $[-1,1]$ then you're not really ignoring $0$.
A: The hard part is justifying why $-1/x$ is the antiderivative you choose.
The problem is that the "constant of integration" only needs to be locally constant: the domain of your functions here include two disjoint intervals, and each one gets its own constant of integration. The complete set of antiderivatives to $1/x^2$ is all of the piecewise defined functions of the form
$$ \begin{cases} -\frac{1}{x} + C & x < 0 \\ -\frac{1}{x} + D & x > 0 \end{cases} $$
where $C$ and $D$ are any constants, and your "forgetting" method results in $-2 + D - C$.
In order to "make sense of this", you need to find some systematic reason to set $C=D$ here. There are surely classes of examples where you can find such a reason, but not in full generality.
A: You already have explanations that take care of the singularity at zero. Here is another argument:
It is complete nonsense for the simple reason that you have a positive integrand and a negative end result. This goes against (one of) the (main) motivation(s) of integration, namely to represent area.
