This is my first year teaching calculus.

On a recent quiz, I asked my students to give an argument that $\int^0_1(1/x)dx$ does not exist. I was looking for arguments that appealed to Riemann sum definition of the integral.

since $1/x$ is unbounded as $x \to \infty$, the area under the curve given by $y = 1/x$ must also be unbounded.

Which makes sense to the intuition, but is actually incorrect. Using the integral evaluation formula for example, we know that $\lim_{x \to 0}(1/\sqrt{x})$ does not exist, but $\int^0_1(1/\sqrt{x})dx = 2.$ While constructing this counterexample, I considered the fact that $\int_a^\infty (1/x^2)dx$ exists, and $(1/\sqrt{x})$ is the inverse of $(1/x^2)$ on $(0,\infty),$ so $\int^0_1(1/\sqrt{x})dx$ must be a finite area.

After introducing the counterexample, I then began to show the class that $\int^0_1(1/x)dx$ does not exist using the Riemann sum definition. So, we construct a Riemann sum for the function on $[0,1]$, $$S = \sum_{k=1}^{n}(1/c_k)(1/n) $$

Since the left endpoint of the interval of integration is $0$, the first term of any left-handed Riemann sum is unbounded, and since the rest are positive since $1/x$ is positive on $(0,1]$, the value of the left-handed sums must be unbounded for every value of $n$, and so $\lim_{n \to \infty} S$ does not exist. Since the limit of any Riemann sum of a function on an interval converges to the value of the integral of the function over the interval whenever the integral exists, either all Riemann sums of $1/x$ on $[0,1]$ must have the same limit as $n \to \infty$, namely they are all unbounded, or they converge to different values and the integral does not exist.

However on closer inspection, this argument doesn't seem much better than the one the students gave above, since the first term of the left hand Reimann sums of $1/\sqrt{x}$ on $[0,1]$ are also unbounded.

(1) Is there an error in my argument above that $\int^0_1(1/x)dx$ does not exist?

(2) Appealing to the Riemann sum definition of the integral, what is a good way to demonstrate that $\int^0_1(1/x)dx$ does not exist, but $\int^0_1(1/\sqrt{x})dx$ does?

My next inclination is to consider limits of Riemann sums for these functions over $[a,1]$ as the left endpoint approaches $0$, but I want to keep the discussion as intuitive as possible to support the definition of the integral and to avoid moving forward into a discussion of improper integrals just yet.

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    $\begingroup$ The Riemann integral is strictly defined for functions that are bounded on an interval [a,b]. So trying to prove that the improper integral over (0,1] does not exist is not a "fair" approach. If you want to approach this improper integral using Riemann sums, then you really do have to look at Riemann sums over [a,1]. $\endgroup$
    – user52817
    Jan 19, 2016 at 18:31
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    $\begingroup$ @user52817 It might be more honest to say that it is a theorem that unbounded functions are never Riemann integrable. $\endgroup$ Jan 20, 2016 at 6:17
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    $\begingroup$ @StevenGubkin In fact that depends on the approach. Many books in Germany use upper and lower Riemann integrals and for upper integrals to exist, one usually assume that the function is bounded right from the start. $\endgroup$
    – Dirk
    Jan 20, 2016 at 9:59
  • $\begingroup$ @Dirk that is a good point. $\endgroup$ Jan 20, 2016 at 14:26
  • $\begingroup$ @StevenGubkin Yes, my undergrad analysis text (that I kept for a reference) uses the limit of upper and lower sums to define the integral. Unfortunately, this class is an AP class, and I have to teach their curriculum. Thank you kindly for your input. $\endgroup$
    – Andrew
    Jan 20, 2016 at 14:43

2 Answers 2


When we write $\int_0^1 \frac{1}{\sqrt{x}} d x = 2$, we do not mean that the Riemann integral of $\frac{1}{\sqrt{x}}$ on $[0,1]$ exists and is equal to $2$, because, as you note, this is false. One can construct sequences of riemann sums for this function which converge to any number $>2$ than you wish, or sequences which diverge to infinity, or sequences which alternate between $2$ and $3$.

It is unfortunate, but we actually use the same symbol for a different mathematical object. In this case the integral sign is an "improper integral" which has a different definition. The definition of this improper integral is

$$ \lim_{a \to 0} \int_a^1 \frac{1}{\sqrt{x}} d x $$

which you can prove converges.

$\int_0^1 \frac{1}{x} dx$ is also an improper integral, but this one diverges because the corresponding limit diverges.

So your argument via riemann sums is indeed flawed: these integrals are not riemann integrals, so the riemann sum argument you gave is irrelevant.

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    $\begingroup$ Thank you Steven. I'll be sure to address this with the class and reconsider how I approach this discussion in the future. $\endgroup$
    – Andrew
    Jan 20, 2016 at 14:46

If there is a singularity in the integration range, the integral can only be defined by the appropriate limit(s). And limits do or do not exist, for various reasons.

This is by far the cleanest way to explain this, reduce the new phenomenon to already known pieces, and use what is known about them.


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