Computing Multi-Step Improper Integrals

A general method to approach improper integrals:

1. For how many reasons is it improper?

2. Break it up into several improper integrals until

a) each improper integral is only improper for ONE reason

b) AND all asymptotes occur at the endpoints of intervals, not in the interior.

1. Evaluate each improper integral. Set the limit of integration with the issue (asymptote or infinite limit of integration) as a variable, then take the limit as that variable approaches the correct value (from the correct side, if it’s an asymptote).

2. If any of the improper integrals diverge, the original improper integral also diverges. In other words, all the split-up improper integrals must converge in order for the original integral to converge. (Even if by symmetry the area should be 0, each separate integral must converge.)

Example 1: Let's look at $\int_{-1}^1\frac{1}{x}$. It's improper for one reason -- it has a singularity at $x=0$. So we write it as $$\int_{-1}^{0-}\frac{1}{x}\,dx+\int_{0+}^1\frac{1}{x}\,dx,$$ where the $0-$ and $0+$ indicate we take a limit as $x\to0^-$ and $x\to0^+$, respectively. Evaluating the right expression, we get $0-\log(0^+)=+\infty$, so the integral diverges. We can also evaluate the left expression if we want, and that one is $-\infty$. Note that if we forgot about the singularity, we would just get $\log(1)-\log(|-1|)=0$, but this is incorrect!

Example 2: How about $\int_{-\infty}^\infty\frac{1}{x^p}$, for $p>0$? First, for how many reasons is it improper? It has a singularity at $x=0$, and also has limit of integration $+\infty$ and $-\infty$. So we need to break it up into lots of pieces. It's tempting to use $$\int_{-\infty}^0\frac{1}{x^p}\,dx+\int_0^\infty\frac{1}{x^p}\,dx,$$ but each of those integrals are still improper for two reasons, the singularity at $x=0$ and the infinite limit of integration! So we need to break it up even more: $$\int_{-\infty}^{-1}\frac{1}{x^p}\,dx+\int_{-1}^{0^-}\frac{1}{x^p}\,dx+\int_{0^+}^1\frac{1}{x^p}\,dx+\int_1^\infty\frac{1}{x^p}\,dx.$$ Now some of these integrals converge and some diverge. If $p>1$, then the integrals over $[0,1]$ and $[-1,0]$ diverge (though the others converge). If $0<p<1$, then the opposite happens. And if $p=1$, then everyone diverges. In any case, not all four of the integrals converge, so the original improper integral diverges regardless of the value of $p$.

Overview of "multi-step" improper integrals

We now understand two-step improper integrals: those that can be broken into two one-step improper integrals. In general, an improper integral $\int_a^b f(x)\,dx$ should be broken into $N$ one-step improper integrals (each individually evaluated as the limit of a definite integral), where $N$ is the sum of the number of infinite endpoints, the number of infinite singularities at finite endpoints, and twice the number of infinite singularities at (finite) interior points.