What does the comparison test not determine about an improper integral?

For the benchmark p-integral with $p>0$, $\int_1^\infty x^{-p}\,dx$ converges when $p>1$ and diverges when $0<p\le1$. The case $p=2$ provides a convenient convergent upper bound for comparison tests.

Conditions: The exponent $p$ is a positive real number ($p>0$).; The integral is evaluated from $1$ to $\infty$.

The comparison test decides convergence or divergence by bounding the target integrand with a simpler nonnegative function. If $0 \le g(x) \le f(x)$ on the tail, convergence of the larger integral $\int_a^\infty f(x)\,dx$ forces convergence of the smaller integral $\int_a^\infty g(x)\,dx$.

Conditions: The functions $f(x)$ and $g(x)$ are nonnegative on the tail interval $[a, \infty)$.; The inequality $0 \le g(x) \le f(x)$ holds for all $x \ge a$.; Finite-interval integrability is assumed for the functions involved.

To prove convergence, split the integral into $[1,2]$ and $[2,\infty)$. On $[1,2]$, the integrand is continuous, so the finite piece is well-defined.

Conditions: The integral is $\int_1^\infty \frac{x-2}{x^3+1}\,dx$.; The comparison function is $\frac{1}{x^2}$.; The inequality holds for $x \ge 2$.