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Consider the integral $\int_{-1}^1 \frac{1}{x^2} dx$ then as a Riemann integral it diverges, also as a Lebesgue integral it is $\infty$. However,

for a moment forget $0 \in [-1,1]$ and "integrate": $$\int_{-1}^1 \frac{1}{x^2} dx = \left. -\frac{1}{x} \right|_{-1}^1 = -2$$

Is there a way that one can make sense of this correspondence? Or it is completely absurd...

Martin Sleziak
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user16015
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5 Answers5

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It makes sense as an integral from -1 to 1 that bypasses the origin in the complex plane, e.g. by moving above it.

Count Iblis
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  • I think a common thing to do is to take the average of two bypasses: a tiny semicircle that bypasses above and a tiny semicircle that bypasses below, and then take the limit as the radius goes to zero. (but since this integrand is analytic, everything works out the same here) –  Sep 13 '15 at 06:49
5

The anti-derivative of $\frac{1}{x^2}$ is $-\frac{1}{x}$ when $x\ne 0$. Inasmuch as the integration limits include the singularity point $x=0$, the anti-derivative is not $-\frac{1}{x}$.

And attempting to split the integral into two improper Riemann integrals leads to the result

$$\begin{align} \int_{-1}^1\frac{1}{x^2}\,dx&=\lim_{\epsilon_1\to 0^{+}}\int_{-1}^{-\epsilon_1}\frac{1}{x^2}\,dx+\lim_{\epsilon_2\to 0^{+}}\int_{\epsilon_2}^{1}\frac{1}{x^2}\,dx\\\\ &=\lim_{\epsilon_1\to 0^{+}}\left(\frac{1}{\epsilon_1}-1\right)+\lim_{\epsilon_2\to 0^{+}}\left(\frac{1}{\epsilon_2}-1\right)\\\\ &=\infty \end{align}$$

even if interpreting the integral as a Cauchy Principal value with $\epsilon_1=\epsilon_2$.

Mark Viola
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    Wait a second, why would you say $\lim_{\epsilon_1 \to 0^{-}}$ when the bounds of the integral are from $-1$ to $-\epsilon_1$. Since $\epsilon_1$ is assumed to be negative, then $-\epsilon_1$ would be positive, so the bound of the integral would include the point $0$ which is what you are trying to avoid doing – ASKASK Sep 12 '15 at 20:23
  • @askask Nice catch. +1 Edited. – Mark Viola Sep 12 '15 at 21:05
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The statement is: if $f \colon \left[a,b\right] \to \Bbb R$ is a continuous function and $F$ is a primitive of $f$ in $\left[a,b\right]$, then $\int_a^b f(x)\,{\rm d}x = F(b)-F(a)$.

Your computation is invalid because $1/x^2$ is not continuous on $\left[-1,1\right]$, in fact, is not defined for $x=0$. And no matter what (real) value you pick for $f(0)$, $f$ won't be continuous and you can't apply the FTC. If you're integrating on $[-1,1]$ then you're not really ignoring $0$.

Ivo Terek
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The hard part is justifying why $-1/x$ is the antiderivative you choose.

The problem is that the "constant of integration" only needs to be locally constant: the domain of your functions here include two disjoint intervals, and each one gets its own constant of integration. The complete set of antiderivatives to $1/x^2$ is all of the piecewise defined functions of the form

$$ \begin{cases} -\frac{1}{x} + C & x < 0 \\ -\frac{1}{x} + D & x > 0 \end{cases} $$

where $C$ and $D$ are any constants, and your "forgetting" method results in $-2 + D - C$.

In order to "make sense of this", you need to find some systematic reason to set $C=D$ here. There are surely classes of examples where you can find such a reason, but not in full generality.

0

You already have explanations that take care of the singularity at zero. Here is another argument:

It is complete nonsense for the simple reason that you have a positive integrand and a negative end result. This goes against (one of) the (main) motivation(s) of integration, namely to represent area.

mickep
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    Right, that was the first thing I thought of... but still wanted to ask this question... who knows :) since people can make sense of the correspondence $1+2+3+... = -\frac{1}{12}$ – user16015 Sep 12 '15 at 18:58
  • Of course, then it depends on what one means by "make sense of"... – mickep Sep 13 '15 at 07:05