Giving meaning to an invalid integral calculation I know that the following calculation is invalid because the integrand is not continuous over the interval $-1 \leq x \leq 1$, but apparently the final result can still be given meaning.
$$\int_{-1}^{1}\frac{dx}{x^{2}} = \left [ -\frac{1}{x} \right ]_{-1}^{1} = -1 - 1 = -2$$
Supposedly a clue lies in the fact that
$$\int_{-\infty }^{-1}\frac{dx}{x^{2}} = \int_{1}^{\infty }\frac{dx}{x^{2}} = 1$$
but I still have no idea what meaning can be gleaned from the result.
Help me out? Thanks (:
 A: The integral diverges at $x \sim 0$. What you can work out is the principal part:
\begin{eqnarray*}
{\cal P}\int_{-1}^{1}{{\rm d}x \over x^{2}}
& \equiv &
\lim_{\epsilon \to 0^{+}}\left(%
\int_{-1}^{-\epsilon}{{\rm d}x \over x^{2}}
+
\int_{\epsilon}^{1}{{\rm d}x \over x^{2}}\right)
=
\lim_{\epsilon \to 0^{+}}\left\lbrack%
\left({-1 \over -\epsilon} - {-1 \over -1}\right)
+
\left({-1 \over 1} - {-1 \over \epsilon}\right)
\right\rbrack
\\
& = &
\lim_{\epsilon \to 0^{+}}\left(-2 + {2 \over \epsilon}\right) = \infty
\end{eqnarray*}
A: The integral gives the area under a curve produced by a continuous curve.
The fact that function has an asymptote at $x=0$ means it isn't continuous and so the integral can't be used to represent the area. This is demonstrated by the fact that you get a negative area even though the order of integration limits ensures it would be positive, and there is obviously non-negative area under the curve when graphed. The meaning is that integration only works for continuous differentiable functions. 
A: Suppose your result was right. Then,
$$\begin{align*}
\int_{-\infty}^\infty \frac{1}{x^2}\ dx &= \int_{-\infty}^{-1} \frac{1}{x^2}\ dx + \int_{-1}^1 \frac{1}{x^2}\ dx + \int_1^\infty \frac{1}{x^2}\ dx \\
 &= 1 - 2 + 1 \\
 &= 0.
\end{align*}
$$
But $\frac{1}{x^2}$ is an even function, and the only even function that integrates to zero over the whole real line is $f(x) = 0$.
