Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
What is discontinuous about these functions? Only $1/x$ and $12/x$ have a single discontinuity at $x = 0$, but $0 \notin [1,\infty)$. –  TMM Mar 11 '12 at 15:14
    
I want to integrate from $0$ to $\infty$. –  seporhau Mar 11 '12 at 15:54
    
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. –  André Nicolas Mar 11 '12 at 21:21

1 Answer 1

up vote 4 down vote accepted

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:

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.