Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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 have stacked about expression of the text book. The question was that

Find the value of $p$ for which the integral converges and evaluate the integral for those values of $p$ $$\int_0^\infty \frac{1}{x^p} \; dx$$

Do I need to use $p$ test to determine if it were conversion or diversion? OR Do I still use $\displaystyle\lim\limits_{t \to \infty}\int_0^t \frac{1}{x^p} \; dx$?

If you have any idea, please post it on the wall thank you,

share|cite|improve this question
Split the integral up first as $\int_0^1{1\over x^p}\, dx+\int_1^\infty{1\over x^p}\, dx$. Then consider three cases: $p<1$, $p=1$, and $p>1$. For each case determine if the original integral converges (both $\int_0^1{1\over x^p}\, dx$ and $\int_1^\infty{1\over x^p}\, dx$ have to converge in order for the original integral to converge; if one of the two diverges, so does the original integral). – David Mitra Feb 23 '12 at 1:23
Oh, if you have the $p$-test in hand, you could certainly appeal to it... – David Mitra Feb 23 '12 at 1:32
up vote 4 down vote accepted

The integral $\int_0^\infty {1\over x^p}\, dx$ is improper in two ways: the interval of integration is infinite and the integrand "blows up" at 0. Thus, you need to split it up: $$ I=\int_0^\infty {1\over x^p}\, dx= \underbrace{\int_0^1 {1\over x^p}\, dx}_{I_0}\ +\ \underbrace{\int_1^\infty {1\over x^p}\, dx}_{I_\infty} $$

Then the integral $I$ converges if and only if both of the integrals $I_o$ and $I_\infty$ converge. If $I$ converges, it converges to the value of $I_0+I_\infty$. Note that to show $I$ diverges (if it does) it suffices to show that one of $I_o$ or $I_\infty$ diverges.

If you have the $p$-test in hand, this should be an easy problem. Consider the integral $I_\infty$ for $p\le1$, and consider the integral $I_o$ for $p>1$.

If you don't have the $p$-test in hand, you'd compute: $$\tag{1} I_0=\int_0^1 {1\over x^p}\, dx=\lim_{a\rightarrow0^+} \int_a^1 {1\over x^p}\, dx $$ and $$\tag{2} I_\infty=\int_1^\infty {1\over x^p}\, dx =\lim_{b\rightarrow\infty} \int_1^b {1\over x^p}\, dx $$

The integral $I_o$ or $I_\infty$ converges if and only if the respective limit above converges.

It would be best here to consider three cases: $p>1$, $p<1$, and $p=1$. A hint here (as above) is to consider $I_o$ for $p>1$ and $I_\infty$ for the other cases.

share|cite|improve this answer
I got it !! Thank you !! – Ryu Feb 23 '12 at 4:48

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.