I came across a convergence test for improper integrals referred to as the $\mu$-test while I was looking through a textbook. I'm interested in understanding the idea behind the test since no explanation is given in the textbook.
Let $f(x)$ be unbounded at $a$ and integrable in the interval $[a+ \epsilon, b]$ where $0 < \epsilon < b-a$.
If there is a number $\mu$ between $0$ and $1$ such that $\lim_{x \to a^{+}} (x-a)^{\mu} f(x)$ exists, then $\int_{a}^{b} f(x) \ dx$ converges absolutely.
If there is a number greater than or equal to $1$ such that $\lim_{x \to a^{+}} (x-a)^{\mu} f(x)$ exists and is nonzero, then $\int_{a}^{b} f(x) \ dx$ diverges, and the same is true if $\lim_{x \to a^{+}} (x-a)^{\mu} f(x) = \pm \infty$.
In the case $\int_{a}^{b} f(x) \ dx$ is unbounded at $b$, we should find $\lim_{x \to b^{-}} (x-b)^{\mu} f(x)$, other conditions remaining the same.
If $\mu =1$ (or any positive integer) and the limit exists and is nonzero, then $f(x)$ would have a pole at $a$ if $f(x)$ were a function of a complex variable. So from that perspective I can see why the integral would diverge.
If $\lim_{x\to a^+} (x-a)^\mu f(x) = L$, then $|f(x)| < 2L|x-a|^{-\mu}$ near $x=a$.
Since $$\int_a^b |x-a|^{-\mu}\,dx$$ converges for $0 < \mu < 1$, the integral $$\int_a^b f(x)\,dx$$ converges absolutely. The other assertions are similar.