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 know that If $f(x)\geq g(x)\geq 0 \forall x$ on the interval $[a,\infty)$ then, if $f(x)$ is convergent,$g(x)$ also must converge.

In doing a exercise, I have come across a problem where the use of this theorem indicates that an improper integral is convergent, but the solutions indicate otherwise.

The question at hand is determining the convergence of the following integral: $$\int_2^\infty \frac{\sqrt{x^{10} -2}}{x^7}$$

To find a function which is greater than the function, in fulfillment of the theorem's requirements, I came up with: $$\frac{\sqrt{x^{10}}}{x^7}=\frac{x^5}{x^7}=x^{-2}$$

Since we know that $\frac{1}{x^2}$ converges on the infinite interval, and that it is always greater than the original integrand, it can be concluded that the improper integral is convergent.

However, this is not the correct answer, apparently the integral diverges. So, I am left wondering why this could be. The only idea that I have come up with are the following:

  • The fact that both $x^{10}-2$ and $x^{10}$ both grow as $O(x^{10})$, thus making some math I don't know kick in that would render them 'equal' on an infinite integral.

Any help with this would be greatly appreciated.

share|cite|improve this question
Why are you saying "apparently integral diverges "? – Mhenni Benghorbal Jan 11 '14 at 18:52
I am not saying it diverges, but the assignment I am doing does. I am convinced it converges, but I am checking here to make sure I'm not missing something. – Richard P Jan 11 '14 at 18:52
What you have done is correct. The integral does converge. – Mhenni Benghorbal Jan 11 '14 at 18:56
Thank you, time to take it up with my professor. Wish me luck. – Richard P Jan 11 '14 at 19:34

Your Answer


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

Browse other questions tagged or ask your own question.