If this integral is calculated analytically,

$$\int_0^1 \frac1{\sqrt{x}} dx = 2\sqrt{1}-2\sqrt{0}=2.$$

However, the graph of $\dfrac1{\sqrt{x}}\to \infty$ as $x\to 0$, so the area under the graph should approach infinity.

In contrast, the integral:

$$\int_0^1 \frac1{x} dx = \ln1-\ln0 =\infty.$$

Indeed, the graph of $\dfrac1x\to \infty $as $x\to 0$, so the area under the graph should approach infinity.

My question: The red line represents $\dfrac1{\sqrt{x}}$ and the blue line represents $\dfrac1x$. How is it that the area under the red line is finite and the area under the blue line is infinite?

enter image description here

  • $\begingroup$ What is the antiderivative of $x^{-1/2}$ ? $\endgroup$ Nov 22 '14 at 10:25
  • $\begingroup$ I've written that implicitly in my calculations. $2\sqrt{x}$ $\endgroup$
    – t.c
    Nov 22 '14 at 10:26
  • $\begingroup$ Yes I know. But, so, I do not any problem since the antiderivative does not have any problem at $x=0$. This is not the same at all with $\log(x)$ which is undefined at $x=0$. Don't think only about the area under the curve. $\endgroup$ Nov 22 '14 at 10:29
  • $\begingroup$ Added a graph for illustration+edited the question title for clarity $\endgroup$
    – t.c
    Nov 22 '14 at 10:38
  • $\begingroup$ @Integrator I can't see how that leads to the conclusion that area under 1/x is infinite but area under x^-0.5 is finite. $\endgroup$
    – t.c
    Nov 22 '14 at 10:39

There is no contradiction: just because a function goes to $\infty$ at $0$ does not mean its integral near $0$ should be infinite.

If you insist on interpreting the integral as area, here's a way to see that the area under the curve $y = 1/\sqrt{x}$ from $0$ to $1$ is finite. The area under the curve $y = 1/\sqrt{x}$ from $0$ to $1$ is the area of the region bounded by the $x$-axis, the $y$-axis, the line $x=1$, and $y = 1/\sqrt{x}$. But if we think about integrating in the $y$-direction, then this can also be interpreted as the area of the region bounded by $y=0$, $x = 1/y^2$, $x=0$, and $x=1$. This is just given by the integral $$ 1 + \int_1^\infty \frac{1}{y^2}~dy = 1 + 1 = 2. $$ (The $1$ at the beginning comes from the square from $0$ to $1$. You should draw the region I described to see what I mean.)

Notice how I've computed the area of the same region, but I've managed to avoid any singularities.

In general this is a very common thing in calculus. There are other seemingly unintuitive results, such as a curve of infinite perimeter bounding a finite area (Koch snowflake) or a surface of revolution with infinite surface are bounding a region of finite volume (Gabriel's horn).

  • $\begingroup$ +1 Integrating over the $y$ axis. Beat me to it by seconds. $\endgroup$
    – UserX
    Nov 22 '14 at 10:47
  • $\begingroup$ +1 Integrating over the y axis. That's a really clever way of looking at it.. $\endgroup$
    – t.c
    Nov 22 '14 at 10:50

Yes, there is an answer and looking at your calculations I suppose you know how to get it:

$$ \int_{0}^{a} \frac{1}{\sqrt{x}} dx = 2 \sqrt{a} - 2\sqrt{0} = 2 \sqrt{a} $$

The point here is that although the graph of $\frac{1}{\sqrt{x}}$ goes to infinity when $x \to 0$, the area between the x axis and this graph is perfectly finite. This might be hard to understand, but you can try to imagine it in the following way. Suppose you take a number $\epsilon > 0$ and now you calculate the area between the x axis, the graph of $\frac{1}{\sqrt{x}}$ and the vertical lines $x=\epsilon$ and $x=a$. With the same integral we've already calculated, you obtain that this area is:

$$ \int_{\epsilon}^{a} \frac{1}{\sqrt{x}} dx = 2 \left( \sqrt{a} - \sqrt{\epsilon} \right) $$

And now you should't be troubled by the fact that the graph of $\frac{1}{\sqrt{x}}$ goes to infinity, because it is perfectly finite in $x = \epsilon$. Now make $\epsilon$ arbitrarily small, that is, take the limit when $\epsilon \to 0$, the area becomes $2 \sqrt{a}$ and you recover the previous result. The key point here is, from an intuitive point of view, that the interval $(0, \epsilon)$ has length $\epsilon$, which tends to zero. Thus, although the function $\frac{1}{\sqrt{x}}$ diverges near $0$, this zone of divergence is so small that you get a finite area. Of course, this is only a small intuitive discussion trying to shed light on your question, not a rigorous reasoning.

To understand a little bit more this idea, suppose you take $p>0$ a real number different from 1. Then, you have:

$$ \int_{\epsilon}^{a} \frac{1}{x^{p}} dx = \left[ \frac{x^{-p+1}}{1-p} \right]_{\epsilon}^{a} = \frac{1}{1-p} \left( a^{1-p} - \epsilon^{1-p} \right) $$

As you can easily check, when you take $\epsilon \to 0$ the result converges if $p<1$ and diverges if $p>1$. Then your example with $\frac{1}{x}$ (which also diverges because of the logarithm) is some kind of limit: functions like $\frac{1}{\sqrt{x}}$ or $\frac{1}{x^{3/5}}$ diverge slow enough near 0 to have a finite area, but $\frac{1}{x}$ or $\frac{1}{x^4}$ don't.


By writing both integrals in form of limit-of-sum we get $$\int_0^1 \frac1{\sqrt{x}} dx =\lim_{n\to\infty}\frac1n\sum_{k=1}^n {\frac {\sqrt n}{\sqrt{ k}} }=\lim_{n\to\infty}\sum_{k=1}^n {\frac {1}{\sqrt{ nk}} }=\lim_{n\to\infty} \frac{H_n^{1/2}}{\sqrt{n}}=2 $$ But following Integral when expressed as limit-of-sum Diverges! $$\int_0^1 \frac1{x} dx =\lim_{n\to\infty}\frac1n\sum_{k=1}^n {\frac{n}{k} } =\lim_{n\to\infty}\sum_{k=1}^n {\frac{1}{k} }$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.