TITLE: Can we integrate discontinuous functions
QUESTION [6 upvotes]: Does it make sense to integrate the function $f(x) = 6$ for $x\in [0,1)$ and $f(x) = 1/x$ for $x$ in $(1,\infty)$?
Does it make sense to integrate the function $f(x) = 6$ for $x\in [0,1]$ and $f(x) = 12/x$ for $x$ in $[1,\infty)$?
What's the difference?

REPLY [5 votes]: These are what is called improper integrals.  You integrate them by taking limits of finite integrals.  For your second example we would let
$$
\int_{[1,\infty)}\frac{12}{x}\, dx=\lim_{b\to\infty}\int_1^b\frac{12}{x}\,dx.
$$
If you carryout the integration, you are left with
$$
\lim_{b\to\infty}(\ln(b)-\ln(1).
$$
Since that limit does not exist, we say that it doesn't converge.  If you try the same thing with $f(x)=\frac{12}{x^2}$, you will get a limit that does exist and we would call the integral convergent.  To deal with integrals over open intervals, we would take limits again:
$$
\int_{(0,1]}f(x)\,dx = \lim_{a\to 0}\int_a^1f(x)\,dx.
$$