Finding convergence/divergence of improper integral

[IntegrateMe]

Determine the convergence or divergence of this following improper integral:

\int_2^∞ \frac {1}{(x^3+7)^{\frac{1}{3}}}

So I'm trying to find something easy to compare this to, any help?

 

[Robert1986]

Well, as I'm sure you can tell, it is a lot like 1/x. So, I would try to find something that is close to 1/x to which you can compare this.

 

[IntegrateMe]

Why can't I just compare it to 1/x?

 

[Ivan92]

I don't know if you have gone over this, but you can compare it to

\int\limits_1^\infty {\frac{1}{{{x^n}}}}

If n<2, it is divergent. If n≥2, then it is convergent.

 

[Robert1986]

Well,
\frac{1}{(x^3+7)^{\frac{1}{3}}} &lt; \frac{1}{x}

isn't it? Since [
\int_2^∞ \frac {1}{x}
diverges, this isn't any help.

So, I'd try showing that
(x^3+7)^{1/3} &lt; x + b

where b is a number. (In particular, b is a number that appears in your problem).

Then realize that

\int_2^∞ \frac {1}{x+b}

diverges, and you're done.

 

[IntegrateMe]

How can I show that \int_2^∞ \frac {1}{x+7} is divergent?

 

[Robert1986]

How can I show that \int_2^∞ \frac {1}{x+7} is divergent?

First of all, I was kind of wrong when I said that "b" was in your problem. Use 7^{1/3} in stead of 7. As for showing that

\int_2^∞ \frac {1}{x+7^{1/3}}

is divergent, is there some theorem in your book that would help? I don't know if there is or not. But, you could say that

lim \frac{1}{x+7^{1/3}} = \frac{1}{x}

and argue from there. Or, you could just integrate \int_2^∞ \frac {1}{x+7^{1/3}}

You have done so to show that \int_2^∞ \frac {1}{x}

is divergent, right? Just use a change of variables.