Indefinite integral for $p \neq 1$:

\begin{equation} \begin{aligned} \int \frac{1}{x^p} dx = \frac{1}{1-p} x^{1-p} + C \end{aligned} \end{equation}

Evaluated on $[0,1]$:

\begin{equation} \begin{aligned} \int_0^1 \frac{1}{x^p} dx = \lim_{c \to 0} \frac{1}{1-p} (1-c^{1-p}) \end{aligned} \end{equation}

For $0 < p < 1$:

\begin{equation} \begin{aligned} \lim_{c \to 0} \frac{1}{1-p} (1-c^{1-p}) = \frac{1}{1-p} \end{aligned} \end{equation}

So the integral converges. The area under the graph is finite. But

\begin{equation} \begin{aligned} \lim_{x \to 0^+} \frac{1}{x^p} = \infty \end{aligned} \end{equation}

for $0 < p < 1$. I am looking for intuition for why the limit above diverges, but the area under the graph is finite. My intuition right now tells me that the part of the x-axis under the infinitely high spike, must be infinitely thin. I would like to better my understanding, and make it more rigorous. Any help is appreciated.

  • 1
    $\begingroup$ The thing is that it's true the function goes up to infinity the closer you get to zero (from the right) but the speed at which it does is sub-logarithmic. It goes to infinity very very slowly. So the integral converges. $\endgroup$ Commented Oct 1, 2017 at 8:07
  • 2
    $\begingroup$ The Reimman integral is defined as the areas of rectangles where each base is in the x-plane, imagine you're looking at $\int_1^\infty 1/x^q$, but you're looking at it where each area is defined as the sum of small bases rectangle where each base in the $y$ lane, now flip your graph, what do you get? $\endgroup$
    – Rab
    Commented Oct 1, 2017 at 8:09

1 Answer 1


Indeed, it gets thin as it approaches the y-axis, but that's not enough. It need to get thin fast enough.

To illustrate, let's look at another function in another limit $y=\frac{1}{x^2}$ as $x \longrightarrow \infty$. I've also plotted $y=\frac{1}{x}$ (red line) for comparison.

As we integrate the area going to the right, we do add more area with increasing x, but so little that the total converges.

Note that as x values increase, the $y=\frac{1}{x^2}$ curve get much thinner than the other curve. This other $1/x$ curve's tail is not thin enough to have a finite integral. What we said about $y=\frac{1}{x^2}$ is true for any $y=\frac{1}{x^q}$ where $q>1$. The tail is too thick in the case of $q \leq 1$

enter image description here

Now, coming to your question, consider the curve $y=\frac{1}{x^p}$ where $p<1$. Here's a plot of this for $p=\frac{1}{2}$. It's basically the same as the first graph with axes exchanged. enter image description here

From exactly the same reasoning as before, the graph of $\frac{1}{x^p}$ gets close to the y axis very quickly (or in other words, y goes to $\infty$ very slowly) so that the area under the curve is finite. Again, in the case of $p \geq 1$, the tail is too thick for that to happen.


You must log in to answer this question.

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