Riemann Integral over functions with poles

Question

Im wondering when (and if so why?) it is allowed to integrate a function which has poles on the interval $[a,b]$ where $-\infty<a<0<b<\infty$. For example take $\frac{1}{x^2}$ or $\frac{1}{x}$. I would argue that this is possible because I know the antiderivative of both functions: $-\frac{1}{x}+C$ and $\ln|x|+C$ and now I can argue with the fundamental theorem of calculus that $\int_{a}^b f(x)=F(b)-F(a)$ and obviously I can evaluate the antiderivatives above on the points $a ,b$. Or am I not allowed to use the fundamental theorem since the functions are not bounded on the interval $[a,b]$?

Answer 1

The area under the curve $1/x^2$ between $x=-1$ and $x=1$ is obviously positive. But your method gives $-2$. So clearly something doesn't work. The Riemann integral is defined with certain limits, and if the function is unbounded on the interval, those limits may not exist. So each case has to be handled separately.

In general, you can't just switch limiting processes. For example

$$\lim_{x\to \infty} \lim_{y\to \infty} \frac{x}{x+y} = 0$$

but

$$\lim_{y\to \infty} \lim_{x\to \infty} \frac{x}{x+y} = 1.$$

But this is what you do when you integrate across a pole. You're taking a limit outside the integral and pushing it inside the limit in the definition of the integral. So you get silly results like $0=1$ or $\infty=-2.$