# A convergence test for improper integrals ($\mu$-test)

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.

• It seems that the second case covers all the conditions possible for $\mu\ge 1$ ? Oct 8, 2013 at 13:19

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.