Why doesn't $$\int_{-1}^1 \frac{1}{x}~\mathrm{d}x$$ converge? I mean you would think that because of symmetry the area from the negative side and positive side cancel out, resulting in the integral equaling zero.

By the way, I'm familiar with the way to test if an integral converges by splitting the integral into $0$ to $1$ and $0$ to $-1$ and adding up the integral sums. I don't really understand why this is valid though in the case of symmetry.

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    $\begingroup$ See Cauchy principal value $\endgroup$ – Petite Etincelle Aug 18 '14 at 19:57
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    $\begingroup$ You don't integrate with symmetry in mind. $\endgroup$ – user157227 Aug 18 '14 at 20:00
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    $\begingroup$ @user157227 <<puts on physicist cap>> On the contrary! Symmetry is best way to integrate. <<puts mathecist cap back on>> Then again, we should probably prove that. $\endgroup$ – David H Aug 18 '14 at 20:08
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    $\begingroup$ @DavidH It's different because everything is absolutely convergent in physics. $\endgroup$ – user157227 Aug 18 '14 at 20:09
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    $\begingroup$ First provide definition of "converge" that you will use. Then check whether this satisfies it. As pointed out below, with the definition from elementary calculus, this does not converge. Perhaps you can define "asd-convergence" and show that this does converge. Then we need to know if there are any useful properties of asd-convergence. $\endgroup$ – GEdgar Aug 18 '14 at 20:55

The integral as you've written isn't well-defined - what you've written is:

$\qquad$Integrate $\frac1x$ along the line running from $-1$ to $+1$

However, since $\frac1x$ isn't defined at 0, this doesn't make sense.

As such, the integral as you've written it is slightly ambiguous. The answer to whether or not the integral converges will depend on how you define it. There are two ways of doing this:

In general, when seeing an integral like this, we are referring to:


$$\lim_{\epsilon, \delta \to 0}\left(\int_{-1}^{-\delta}\frac1x\mathrm dx+\int_\epsilon^1\frac1x\mathrm dx\right)\\=\left(\lim_{\delta \to 0}\int_{-1}^{-\delta}\frac1x\mathrm dx\right)+\left(\lim_{\epsilon \to 0}\int_\epsilon^1\frac1x\mathrm dx\right)$$

In this case, because we are allowing $\epsilon$ and $\delta$ to converge to $0$ at different rates, the integral will not generally converge.

The only way to make this integral to converge will be to impose a relation on $\epsilon$ and $\delta$ such as $\epsilon = \delta$. However, when writing the above integral, we usually mean that we are considering the convergence of $\epsilon$ and $\delta$ without these restrictions, and hence this integral will diverge.

However, we could also consider:


$$\lim_{\delta \to 0}\left(\int_{-1}^{-\delta}\frac1x\mathrm dx+\int_\delta^1\frac1x\mathrm dx\right)$$ Here, the two integrals are converging at the same rate, and in this case, the integral is $0$.

I'm guessing this is what you intuitively had in mind when you said that the positive and negative sides should cancel each other out. However, this integral is not the same as (1), and when writing $\displaystyle\int_{-1}^1\frac1x\mathrm dx$, a writer will normally be referring to ($1$) and not ($2$).

Contrast this to the integral $$\int_2^4\frac1x$$

As in ($1$), we can write this integral as

$$\lim_{\epsilon, \delta \to 0}\left(\int_{2}^{3-\delta}\frac1x+\int_{3+\epsilon}^4\frac1x\right)$$

However, in this case, both integrals converge separately and hence the total integral will converge. This is different to ($1$) where both integrals diverge separately, and hence their sum will diverge.

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    $\begingroup$ I don't understand why $\frac{1}{x}$ not being defined at $0$ would affect integrability. I mean using Riemann sums you can split the function into any number of finite rectangles and the sum of the area of the rectangles would be defined. So why wouldn't this extend to infinite rectangles? $\endgroup$ – asd Aug 18 '14 at 20:03
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    $\begingroup$ The point is that you will have rectangles on the negative side that will be of different widths to the ones on the positive side. The only way for the integral to be zero would be to impose the condition that the rectangles have the same width - but this is not a natural assumption to make. $\endgroup$ – Mathmo123 Aug 18 '14 at 20:07
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    $\begingroup$ @asd There are two issues at play here. You are not allowed to re-arrange your infinite sum because it is not absolutly convergent. By the Riemann Series Theorem you would be able to make your integral equal to whatever you would like. $\endgroup$ – user157227 Aug 18 '14 at 20:07
  • $\begingroup$ A function being undefined on a set of measure zero doesn't seem to be much of a barrier to integrating. As an example, integrate $1/(sgn(x) * |x^{(1/2)}|)$ from -1 to 1 -- it is undefined at 0 as well. $\endgroup$ – Yakk Aug 18 '14 at 20:27
  • $\begingroup$ @Yakk I haven't said that it makes a difference. Just that $\int_{-1}^1\frac1x$ is essentially an abuse of notation that needs to be explained. Also - for the record, if there is an essential singularity at a single point, it makes a huge difference when trying to integrate.... so a function not being defined on a measure $0$ set can and does make a difference. The point here is that there is a pole at $0$, and that the integral is unbounded near $0$, not that the function is undefined at $0$ $\endgroup$ – Mathmo123 Aug 18 '14 at 20:29

The main issue here is that, despite the odd-ness of the integrand function, $\frac{1}{x}$ is not a Riemann integrable function over $(0,1)$, hence the value of such integral, if existing, would be something like $\infty-\infty$.

Also notice that: $$\int_{(-1,-a/n)\cup(b/n,1)}\frac{dx}{x}=-\log\frac{n}{a}+\log\frac{n}{b}=\log\frac{a}{b}.$$ By choosing different values for $a$ and $b$ and taking the limit as $n\to +\infty$, you get different hypothetical values for your integral.

  • $\begingroup$ Why is the function not Riemann integrable? I mean visually speaking you can split the function into any number of (finite) rectangles with the sum of the rectangles being defined. So why can't you extend this to infinite rectangles? $\endgroup$ – asd Aug 18 '14 at 20:01
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    $\begingroup$ In order that a function is Riemann-integrable over $[-1,1]$, it must be Riemann integrable over any sub-interval of $[-1,1]$. Now, what $\int_{0}^{1}\frac{dx}{x}$ should be? Since the antiderivative of $\frac{1}{x}$ is $\log x$, the value of the integral should be $-\log(0)$, i.e. $+\infty$. This gives that the Riemann sums over $[0,1]$ do not converge, hence $f(x)$ is not Riemann-integrable over a superset of a neighbourhood of zero. $\endgroup$ – Jack D'Aurizio Aug 18 '14 at 20:07
  • $\begingroup$ Why must a function be Riemann integrable over any subinterval for it to be integrable over an interval? This seems like an arbitrary condition. $\endgroup$ – asd Aug 18 '14 at 20:11
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    $\begingroup$ It just follows from the definition of Riemann integrability. The limit values of the Riemann sums must be finite. If this does not happen on a sub-interval, it cannot happen on the big interval, too. $\endgroup$ – Jack D'Aurizio Aug 18 '14 at 20:13
  • $\begingroup$ Sorry Jack, I don't see how that follows. I looked over the definition of Riemann sums and don't see how it implies that if it doesn't happen on a subinterval it cannot happen on the big one too. $\endgroup$ – asd Aug 18 '14 at 20:34

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