Calculate $\int\limits_{-2}^2 \frac{1}{x^2-4}\mathrm dx$ unless it diverges.

My attempt

I calculated the indefinite integral $$\int\frac{1}{x^2-4}\mathrm dx = \frac14\left( \ln(2-x) - \ln(2+x) \right) + C$$

and notice from there that the limits on the integral will cause $\ln(0)$ problems.


From this I decided that the integral must be divergent, but does that hold? Or is there some simple rule that would have allowed me to conclude this before calculating the indefinite integral?

Any tips appreciated!

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    $\begingroup$ You cannot conclude that yet. The two issues with ln might "cancel". You can try combining the natural logarithms by ln(a/b)=lna-lnb. Note the fact that the function trails off to infinity at the two endpoints is no guaruntee that the integral is divergent. $\endgroup$ – Jacob Wakem Nov 18 '16 at 12:43
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    $\begingroup$ However the function is symmetric about x=0 so you can get away with only doing one natural logarithm and multiplying by 2. If it diverges, you are done! $\endgroup$ – Jacob Wakem Nov 18 '16 at 12:44
  • $\begingroup$ @Alephnull - But I get the same-ish problem with $x=2$ causing $ln0$-difficulties, even if I just look at the integral from 0 to 2. Either it will do it at the $\ln(2-x)$ part, or after using the rule you mentioned, I get $\ln(\frac{0}{4})$. $\endgroup$ – Alec Nov 18 '16 at 12:54
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    $\begingroup$ @Alephnull: Since the integrand is negative everywhere between $-2$ and $2$, there can't be any cancelling anyway. And even so, such "cancellation" is not completely well-behaved and should be called out by saying that all we find is a Cauchy principal value of the improper integral. $\endgroup$ – Henning Makholm Nov 18 '16 at 12:56
  • $\begingroup$ Another idea is to compare it to a polynomial function. It is almost a polynomial function. $\endgroup$ – Jacob Wakem Nov 18 '16 at 13:06

Heuristically, you could have noted that $x^2-4$ has nonzero, finite slope near $x=2$, so it behaves approximately like $c(x-2)$ for some $c$ there.

Therefore its reciprocal will diverge mostly in the same way as $\frac{1}{c(x-2)}$, and we know that the integral of that kind of diverging functions will itself diverge. (This would be in contrast to, for example, $c(2-x)^{-1/2}$, where the function itself diverges but the area below it is finite).

This could be made into a more rigorous proof by comparing it the integral of $\frac{1}{x^2-4}$ near $x=2$ with that of $\frac{1}{k(x-2)}$ for a sufficiently large $k$ that $|k(x-2)|>|x^2-4|$ in a punctured neighborhood of $2$. In this particular case that is probably not much of an improvement over finding the full indefinite integral, but it is valuable to be able to use the heuristic reasoning.


$\displaystyle \int\limits_{-a}^a\frac{dx}{x^2-4} = 2\int\limits_0^a\frac{dx}{x^2-4}=\frac{1}{4}\ln\frac{2-a}{2+a}$ with $0<a<2$ .

Then we see what happens for $a\to 2^-$ .

  • $\begingroup$ I woild caution that this will give you incorrect number answers for integrals that diverge because of a discontinuity at zero. $\endgroup$ – Mark S. Nov 18 '16 at 12:47
  • $\begingroup$ @Mark S. : At $\pm 2$, not $0$. $\endgroup$ – user90369 Nov 18 '16 at 12:51
  • $\begingroup$ The trouble with this strategy is that, for example, $$\lim_{a\to 2^-} \int_{-a}^a \frac{x}{x^2-4}\,dx \ne \lim_{a\to 2^-} \int_{-a}^{a/2+1} \frac{x}{x^2-4}\,dx$$ and it's troublesome to argue that one is a better way to approach $\int_{-2}^2$ than the other. $\endgroup$ – Henning Makholm Nov 18 '16 at 13:01
  • $\begingroup$ @Henning Makholm : Thanks, yes, your argument is of course correct and needs to be considered. $\endgroup$ – user90369 Nov 18 '16 at 13:09
  • $\begingroup$ @HenningMakholm Isn't that the feature? If they both get arbitrarily close to the point and disagree then we have divergence. $\endgroup$ – Jacob Wakem Nov 18 '16 at 13:12

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