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

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?

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  • $\begingroup$ How did you get the value $5+\ln(9/4)$? $\endgroup$ – Tim Raczkowski Feb 6 '15 at 3:27
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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