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

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

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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.

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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.

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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.

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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

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