One way to visualize an antiderivative is that the area under the derivative is added to the initial value of the antiderivative to get the final value of the antiderivative over an interval.

The Riemann Series Theorem essentially says that you can basically get any value you want out of a conditionally convergent series by changing the order you add up the terms.

Now consider the function: $$f(x) = 1/x^3$$The function is unbounded over the interval $(-1,1)$, so it not integrable over this interval.

If you break $f(x)$ into Riemann rectangles over the interval $(-1,1)$ and express the area as a Riemann sum, you essentially get a conditionally convergent series. And because of the Riemann Series Theorem, you can make the sum add up to anything. In other words, you can make the rectangles add up to whatever area you want by changing the order in which you add them up. This is in fact why a function needs to be bounded to be integrable - otherwise the area has infinite values/is undefined.

So my question is, in cases like this, how does the antiderivative "choose" an area? I mean in this case the antiderivative, $\frac{1}{-2x^2}$, choose to increase by $\frac{1}{-2(1)^2} - \frac{1}{-2(-1)^2} = 0.$ In other words, the antiderivative "choose" to assign a area of zero to the area under $1/x^3$from $-1$ to $1$ even though the Riemann Series Theorem says the area can be assigned any value.

How did the antiderivative "choose" an area of $0$ from the infinite possible values?


1 Answer 1


This is a problem with the visualization, not the antiderivative: the antiderivative simply chooses to be a function whose derivative is $1/x^2$ and doesn't care about area.

Also, don't forget that, with the Riemann approach to integration, the integral over $(-1,1)$ isn't "conditionally convergent": it is simply undefined. The Riemann approach makes this an improper integral and splits it into the sum of two integrals $(-1,a)$ and $(b, 1)$, and takes the limit as $a \to 0^-$ and $b \to 0^+$.

It may help noting that the antiderivatives of $1/x^3$ on the set of nonzero real numbers are precisely the functions that can be written in the form

$$ f(x) = \begin{cases} -\frac{1}{2x^2} + C_1 & x < 0 \\ -\frac{1}{2x^2} + C_2 & x > 0 \end{cases} $$

where $C_1$ and $C_2$ are any constants.

  • $\begingroup$ My attempt to explain the same. $\endgroup$
    – ryang
    Commented Jan 28, 2023 at 23:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .