# Improper integral question

Establish convergence/divergence of the following improper integral:

integral from 0 to infinity of 1 / ( x^(1/3)*(/x-5/^(1/3))*(1 + sqrt(x))^0.7) )

My attempt at a solution was to break it up into 3 intergrals: 0 to 1, 1 to 5, and 5 to infinity...I showed that the first two of these integrals converges by comparing them to 1/x^(1/3) and 1//x-5/^(1/3) respectively. I can't establish convergence/divergence of last intergral from 5 to infinity, as hard as I have tried.

Thanks, any help is much appreciated.

## Answers and Replies

Mark44
Mentor
Establish convergence/divergence of the following improper integral:

integral from 0 to infinity of 1 / ( x^(1/3)*(/x-5/^(1/3))*(1 + sqrt(x))^0.7) )

My attempt at a solution was to break it up into 3 intergrals: 0 to 1, 1 to 5, and 5 to infinity...I showed that the first two of these integrals converges by comparing them to 1/x^(1/3) and 1//x-5/^(1/3) respectively. I can't establish convergence/divergence of last intergral from 5 to infinity, as hard as I have tried.

Thanks, any help is much appreciated.

Is there some reason you wrote /x - 5/ using the slashes?
Why did you choose 1 as a limit of integration in two of your integrals? There is nothing unusual happening at x = 1.
In your integrals, what do you have for the integrands? Please show us the work you did.

BTW, there is no such word as "intergral."

By /x-5/, I mean the absolute value of (x-5).

I chose 1 because don't you have to break up the original improper integral into appropriate intervals to see what's going on? There are unusual things happening at x=0 and x=5, and 1 was chosen to 'link up' these separate intervals (for want of a better phrase).

So I have this:
[0 to 1] 1 / ( x^(1/3)*(/x-5/^(1/3))*(1 + sqrt(x))^0.7) ) + [1 to 5]1 / ( x^(1/3)*(/x-5/^(1/3))*(1 + sqrt(x))^0.7) ) + [5 to infinity]1 / ( x^(1/3)*(/x-5/^(1/3))*(1 + sqrt(x))^0.7) )

The first two integrals I have shown to be convergent (by comparison), but I am stuck on the last one. Mabye I am going about this the wrong way altogether, but I'm sort of following a similar example from my lecture notes.

(Sorry about the messiness, I am trying to learn Latex)

1+sqrt(x)>sqrt(x), and for large x, x-5>0.5x

P.S. You also should break [5,infty) into two intervals.

Mark44
Mentor
OK, I understand now why you are using 1 as a limit of integration. For absolute values, you can use this character: |.

To help you out with your LaTeX, your last integral is
$$\int_5^{\infty}\frac{1}{x^{1/3}|x - 5|^{1/3}(1 + \sqrt{x})^{0.7}}dx$$
You'll need to break this into two integrals, say from a to 6 and from 6 to b, and look at the limit of each as a approaches 5 from above and as b approaches infinity. You can click on the integral and see the LaTeX code, and can copy and paste it to create your own.

Since in both integrals, x >= 5, you can drop the absolute value signs so you have (x^2 - 5x)^(1/3). I doubt that the integral can be found by ordinary means, so you'll probably need to use another comparison.

I'll need to think about this one...

Mark44
Mentor
The denominator is roughly comparable with x61/60, with that exponent obtained by adding 2/3 and 7/20. Since this exponent is > 1, the integral of this function, from 6 to infinity, converges. If you can show that your denominator is > x61/60, and thereby that your integrand is < 1/x61/60, you're home free.

Ahh okay, that makes alot more sense now, thanks alot Mark.

When I finish writing it up I will post my final explanation.