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

$$\int_0^\infty \frac{7x^7}{1+x^7}$$

Im really not sure how to even start this. Does anyone care to explain how this can be done?

share|cite|improve this question
What is the limit of the integrand as $x$ tends to infinity? – Mark Bennet Mar 14 '12 at 15:35

The answer by Davide Giraudo has all the right elements, but I think the OP may appreciate to see all of the details spelled out. First of all, an improper integral is defined as a limit, as follows: $$\int_0^\infty \frac{7x^7}{1+x^7} dx = \lim_{N\to \infty} \int_0^N \frac{7x^7}{1+x^7} dx.$$ Next, we use the fact that $1+x^7\leq 2x^7$ when $x\geq 1$, and the properties of the definite integral to bound the integral inside the limit: $$\begin{align*} \int_0^N \frac{7x^7}{1+x^7} dx &=\int_0^1 \frac{7x^7}{1+x^7} dx + \int_1^N \frac{7x^7}{1+x^7} dx \\ &\geq \int_0^1 \frac{7x^7}{1+x^7} dx+ \int_1^N \frac{7x^7}{2x^7} dx \\ &= \int_0^1 \frac{7x^7}{1+x^7} dx+ \int_1^N \frac{7}{2} dx \\ & = \int_0^1 \frac{7x^7}{1+x^7} dx+ \frac{7(N-1)}{2}.\end{align*}$$ Hence: $$\int_0^\infty \frac{7x^7}{1+x^7} dx = \lim_{N\to \infty} \int_0^N \frac{7x^7}{1+x^7} dx\geq \lim_{N\to \infty} \left(\int_0^1 \frac{7x^7}{1+x^7} dx+ \frac{7(N-1)}{2} \right)= \infty.$$ Therefore, the improper integral diverges.

share|cite|improve this answer

The only problem is in $+\infty$. We have for $x\geq 1$ that $1+x^7\leq 2x^7$ so $\frac{7x^7}{1+x^7}\geq \frac 72\geq 0$ and $\int_1^{+\infty}\frac 72dt$ is divergent, so $\int_1^{+\infty}\frac{7x^7}{1+x^7}dx$ is divergent. Finally, $\int_0^{+\infty}\frac{7x^7}{1+x^7}dx$ is divergent.

share|cite|improve this answer

An inproper integral will diverge if the limit of the function at infinity is not zero (as Chris pointed out, it's a different business if the limit doesn't exist). Here, $$ \lim_{x\to\infty}\frac{7x^7}{1+x^7}=7, $$ so the integral diverges.

share|cite|improve this answer
Just as a comment, this is only a necessary condition if the limit exists. Take a triangle with very small base and very large height and area $\frac{1}{2^n}$ and place it at the nth integer. The integral of this function is bounded above by one but no pointwise limit exists as $x\to \infty$. This is essentially the reason that not all continuous $L^1$ functions must lie in $C_0$. – Chris Janjigian Mar 14 '12 at 15:59
Good point! I'll rephrase my answer. To my shame, I actually contributed the very same example you mention a couple weeks ago on another question... – Martin Argerami Mar 14 '12 at 16:00

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.