Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to get bound for the following integral $$ \int_0^{\infty}\frac{1}{|x|^r}dx, \mbox{for } 1\leq r< \infty $$ In particular, the bound of the form $\frac{constant}{r}$.

Sorry, we can omit the absolute value. And probably consider interval $(a,\infty)$ for some $a$ very close to 0. So, the integral is $$ \int_a^{\infty}\frac{1}{x^r}dx, \mbox{for } 1\leq r< \infty $$

share|cite|improve this question
Do you want $\int_a^\infty {1\over |x|^r}\,dx$ for some $a>0$? As it stands, the integral never converges. – David Mitra Jan 21 '12 at 21:05
I think that the absolute value signs can be omitted, or should we interpret it as a complex integral? – wnvl Jan 21 '12 at 21:15
I would assume that you can find the exact value of the integral. – Fabian Jan 21 '12 at 23:28

Let us first do some inital work. Here we first safely assume $r>1$. Then our integral turns into (We exclude the limits for now)

$$ \int_{0}^{\infty} \frac{1}{x^r} \, \text{d}x = \left[ \frac{1}{1-r} x^{1-r} \right]_{0}^{\infty} $$

Quite easily we can see that since $r>0$

$$ \lim_{x \to \infty} \frac{1}{1-r}x^{1-r} = 0 $$

This is quite good! Now we know that

$$ \int_{0}^{\infty} \frac{1}{x^r} \, \text{d}x = \left[ \frac{1}{1-r} x^{1-r} \right]_{0}^{\infty} = 0 - \lim_{x \to 0} \frac{1}{1-r}x^{1-r}$$

The problem is that this last limit does not exist for any values of r. It simply tends to $- \infty$ for all values of $r$.

Let us now plot the problem for a few values of r, to really strike home the point


As we can see the problem is how these functions behave at values very close to zero and very close to infinity. In a nuttshell, the problem with evaluating this integral is that

$ \frac{1}{x^r} $

Approaches $\infty$ at a faster rate when r> increases but, approaches 0 at a slower rate when r increases. Similarly

Approaches $\infty$ at a slower rate when r decreases but, approaches 0 at a faster rate when r increases. As can be seen from the graph.

Although a similar integral that does converge is

$$ \int_{0}^{\infty} \frac{1}{k^r} \, \text{d}x = \frac{1}{r - 1} $$

So to sum up

$$ \int_{0}^{\infty} \frac{1}{x^r} \, \text{d}x = -\infty $$

when $1<r<\infty$

share|cite|improve this answer

The derivative of $x\mapsto \dfrac1{x^{r-1}}$ is $x\mapsto-\dfrac{r-1}{x^r}$, hence, for every $r\gt1$ and $a\gt0$, $$ \int_a^{+\infty}\frac{\mathrm dx}{x^r}=\left[-\frac1{(r-1)x^{r-1}}\right]_{x=a}^{x=+\infty}=\frac1{(r-1)a^{r-1}}.$$ There is no upper bound independent on $a\gt0$. If $a=0$ or if $r=1$, the integral diverges.

share|cite|improve this answer

The integral is not bounded.

For r=1 the primitive function of the integrand is log(x) which diverges at 0 and infinity.

For r>1 the primitive function of the integrand is $\frac{1-r}{x^{r-1}}$ which diverges at 0.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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