1
$\begingroup$

$\int{_0^5\frac{x}{x-2}dx}$

This integral produces a finite value of 5+ln(9/4). However, according to Wolfram Alpha, it diverges (http://www.wolframalpha.com/input/?i=integral+of+x%2F%28x-2%29+from+0+to+5). How can I tell that this integral is divergent without using Wolfram Alpha? Or is it actually convergent from 0 to 5?

$\endgroup$
  • $\begingroup$ How did you get the value $5+\ln(9/4)$? $\endgroup$ – Tim Raczkowski Feb 6 '15 at 3:27
  • 1
    $\begingroup$ Did you scroll down to the bottom of the Wolfram Alpha result? It provides the same principal value for the integral which you claimed. $\endgroup$ – David H Feb 6 '15 at 3:32
  • $\begingroup$ @DavidH Yes, I did. Nevertheless, I was still confused by "integral does not converge". I assumed the Cauchy principle value is similar to the complex value solutions that Wolfram Alpha often gives. $\endgroup$ – Leo Jiang Feb 6 '15 at 3:35
1
$\begingroup$

We must be very careful with what we mean by "converges" when talking about improper integrals. The usual definition of $\int_a^b f(x)dx$ converging is that $\int_a^b f^+(x) dx$ and $\int_a^b f^-(x)dx$ are both finite, where $f^+(x) = \max\{f(x),0\}$ and $f^-(x) = \max\{-f(x),0\}$.

In the case of $\mathbb{R}$ (as opposed to $\mathbb{R}^n$) if $a$ or $b$ is infinite it is common to say an integral converges if, e.g. the case where $b$ is infinite $\lim_{M \to \infty} \int_a^M f(x)dx$ exists in $\mathbb{R}$.

For the case you present we can talk about another kind of convergence, namely existence of the principal value of the integral. (Most would not call this "convergence" though I could see why some might.)

In your case, I would use the first (i.e. the most standard) definition of convergence to conclude that the integral diverges.

$\endgroup$
  • $\begingroup$ By "first definition", you mean when $\int_a^b f^+(x) dx$ and $\int_a^b f^-(x)dx$ are both finite right? Aren't they both finite in my example? If a=0 and b=5, then the integral produces 5+ln(9/4) $\endgroup$ – Leo Jiang Feb 6 '15 at 3:57
  • $\begingroup$ The integral does not produce $5 + \log (9/4)$. The integrand does not satisfy the hypotheses of the fundamental theorem of calculus, and so the calculation that you probably did to get that answer is not valid. $\endgroup$ – nullUser Feb 6 '15 at 4:10
  • $\begingroup$ @MattSamuel see my second paragraph where I address exactly that case. $\endgroup$ – nullUser Feb 6 '15 at 4:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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