$\mu$ test for convergence of improper integral of first kind

When $L \neq 0$, this limit comparison can be use to prove both absolute convergence when $\mu > 1$ and absolute divergence when $\mu \leqslant 1$.

Suppose $\lim_{x \to \infty} x^\mu f(x) = L$. Then for $\epsilon = |L|/2$ there exists $x_0 > 0$ such that when $x \geqslant x_0> 0$ we have

$$| |x^\mu f(x)| - |L|| \leqslant |x^\mu f(x) - L| \leqslant |L|/2.$$

Whence,

$$-|L|/2 \leqslant |x^\mu f(x)| - |L| \leqslant |L|/2\\ \implies |L|/2 \leqslant |x^\mu f(x)| \leqslant 3|L|/2 \\ \implies (|L|/2)x^{-\mu} \leqslant | f(x)| \leqslant (3|L|/2)x^{-\mu} $$

and the integrals of $|f(x)|$ and $x^{-\mu}$ must converge or diverge together.

The limit comparison test is often presented only for the case where $f(x) \geqslant 0$ for all $x > a$. In that case, if $\lim_{x \to \infty}x^\mu f(x) = L > 0$, then there exists $x_0$ such that when $x \geqslant x_0$ we have

$$0 < \frac{L}{2}x^{-\mu} < f(x) < \frac{3L}{2}x^{-\mu},$$

and $\displaystyle \int_a^\infty f(x) \, dx$ diverges if $\mu \leqslant 1$ (using the left inequality) and converges if if $\mu > 1$ (using the right inequality).

An example of the former case is $f(x) = \sin(1/x)$ - where $f$ is eventually positive. Since $\lim_{x \to \infty}x \sin(1/x) = 1$, then $\displaystyle \int_1^\infty \sin(1/x) \, dx$ diverges.

If $L = 0$, then for any $\epsilon > 0$ there exists $x_0$ such that when $x \geqslant x_0$ we have

$$-\epsilon x^{-\mu} < f(x) < \epsilon x^{-\mu},$$

and we can only conclude that $\displaystyle \int_a^\infty f(x) \, dx$ converges if $\mu > 1$ (using the right inequality). An example is the gamma function

$$\displaystyle \Gamma(s) = \int_0^\infty x^{s-1}e^{-x} \, dx,$$

where proof of convergence is facilitated using the test function $x^{-2}$.

If $L = 0$, then we cannot conclude divergence if $\mu \leqslant 1$ since the left lower bound is negative and the integral could converge to a positive value.