convergence of $\int_a^b \frac{1}{x^2} dx$ Why is it true that $\int_0^a \frac{dx}{x^2} = \infty$ but $\int_a^\infty \frac{dx}{x^2} < \infty$?
Shouldn't it be symmetric? 

By definition:
$$\begin{align}&\int_0^a\frac{dx}{x^2}=\left.\lim_{\epsilon\to 0^+}\left(-\frac1x\right)\right|^a_\epsilon=-\frac1a+\lim_{\epsilon\to0^+}\frac1\epsilon=\infty\\{}\\
&\int_a^\infty\frac{dx}{x^2}=\left.\lim_{b\to\infty}\left(-\frac1x\right)\right|_a^b=-\lim_{b\to\infty}\frac1b+\frac1a=\frac1a\end{align}$$
Thus, the second improper integral converges, whereas the first one diverges.
