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?

  • 1
    $\begingroup$ What is discontinuous about these functions? Only $1/x$ and $12/x$ have a single discontinuity at $x = 0$, but $0 \notin [1,\infty)$. $\endgroup$ – TMM Mar 11 '12 at 15:14
  • $\begingroup$ I want to integrate from $0$ to $\infty$. $\endgroup$ – seporhau Mar 11 '12 at 15:54
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    $\begingroup$ A finite number of points of discontinuity make no difference. In fact, a "small" (technically, measure $0$) infinite set of points of discontinuity makes no difference. The thing that makes your integral diverge is not the point of discontinuity. It is the fact that $1/x$ decreases too slowly, so that informally the area under $y=1/x$, from $x=a$ to "infinity," is always infinite. $\endgroup$ – André Nicolas Mar 11 '12 at 21:21

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:

$$ \int_{(0,1]}f(x)\,dx = \lim_{a\to 0}\int_a^1f(x)\,dx. $$


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