I have $\int_{0}^{\infty}\frac{1}{x^{2} - 1}dx$. I did not initially recognize the vertical asymptote and just compared with $\frac{1}{x^{2}}$. Can I do this, or do I need to split the integral?
2026-04-13 09:45:24.1776073524
Using direct comparison on $\int_0^\infty \frac{1}{x^2 - 1} dx.$
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It is necessary to pay some attention to what's really going on. We define the improper integral $$ \int_0^\infty \frac{1}{x^2 - 1} dx = \lim_{a \to 1} \int_0^a \frac{1}{x^2 - 1} dx + \lim_{\substack{b \to 1 \\ c \to \infty}} \int_b^c \frac{1}{x^2 - 1} dx.$$ From only first principles, it would be inappropriate to compare to the integral of $1/x^2$ on the infinite half-line $[0, \infty)$.
Further, it is not always true that $\frac{1}{x^2} < \frac{1}{x^2 - 1}$ (incorrect at, for example, $1/2$), so it would not be possible to directly compare the two functions over the whole interval anyway.