Finding convergence/divergence of improper integral

• IntegrateMe
In summary, the conversation discusses determining the convergence or divergence of the improper integral \int_2^∞ \frac {1}{(x^3+7)^{\frac{1}{3}}}. The participants consider comparing it to \int\limits_1^\infty {\frac{1}{{{x^n}}}} and \int_2^∞ \frac {1}{x+b} to determine its convergence. Ultimately, they suggest using 7^{1/3} instead of 7 and using a change of variables to show that \int_2^∞ \frac {1}{x+7^{1/3}} is divergent, thus proving the original integral is also divergent.
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?

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.

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

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.

Last edited:
Well,
$$\frac{1}{(x^3+7)^{\frac{1}{3}}} < \frac{1}{x}$$

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

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

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.

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

IntegrateMe said:
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.

What is an improper integral?

An improper integral is an integral where one or both of the limits of integration are infinite, or the integrand approaches infinity at some point within the interval of integration.

How do you determine convergence or divergence of an improper integral?

To determine convergence or divergence of an improper integral, you need to evaluate the integral using a limit as one or both of the limits of integration approaches infinity. If the limit exists and is a finite number, the integral is said to converge. If the limit does not exist or is infinite, the integral is said to diverge.

What are some common techniques used to evaluate improper integrals?

Some common techniques used to evaluate improper integrals include using the comparison test, the limit comparison test, and the p-test. These methods involve finding a similar integral that is easier to evaluate and using it to determine the convergence or divergence of the given integral.

Can an improper integral converge even if the integrand is undefined at some point?

Yes, an improper integral can still converge even if the integrand is undefined at some point within the interval of integration. This is because the integral is evaluated using a limit, and the limit may still exist and be a finite number even if the function is undefined at a certain point.

What is the importance of determining convergence or divergence of an improper integral?

Determining convergence or divergence of an improper integral is important because it allows us to determine whether the integral has a finite value or not. This information is crucial in many applications, such as in physics and engineering, where improper integrals are commonly used to model real-world problems.

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