# Integrate Improper Int: 0 to ∞: Does it Converge or Diverge?

• dtl42
In summary, the homework statement asks if the integral from 0 to infinity of \int\frac{1}{\sqrt x \sqrt{x+1}\sqrt{x+2}}dx converges or diverges. I tried to integrate it, but I haven't even been able to do that so I couldn't then evaluate the limit of the integral. So, I can't answer the question.
dtl42

## Homework Statement

Does the integral from 0 to Infinity of $$\int\frac{1}{\sqrt x \sqrt{x+1}\sqrt{x+2}}dx$$ converge or diverge?

None.

## The Attempt at a Solution

I tried to integrate it, but I haven't even been able to do that so I couldn't then evaluate the limit of the integral.

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Well, since you can't use an anti-derivative of the integrand to answer the question, can you use some other technique?

To give you a hint:
Were you asked to actually CALCULATE the value of the integral, or just whether or not the integral converged or diverged?

Wow, I'm getting dizzy looking at your problem. Is this correct ...

$$\int\frac{1}{\sqrt x \sqrt{x+1}\sqrt{x+2}}dx$$

Yea that is correct, and the question does just ask about its convergence.

dtl42 said:
Yea that is correct, and the question does just ask about its convergence.
Now:

Note that this integral is improper in two radically different ways:
1. At x=0 (lower limit), the integrand is infinitely large.
2. The upper limit is not a real number.

Now, let us first try to analyze two distinct cases, where each of these has only one of these pathological traits:

a) Suppose that we investigate the case where only 1 holds, say that the integral goes from x=0 to x=1.
Does this integral converge or diverge?

b) Suppose we investigate the other special case, say looking at the integral from say..x=1 to infinity. Does this integral converge or diverge?

What can you conclude from your analysis?

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I am fairly that both of those cases converge, although I had to use a Ti-89 to evaluate them. Since they both converge, the entire integral should converge as well. Though, I am struggling to reach the convergence of each case without an 89.

dtl42 said:
I am fairly that both of those cases converge, although I had to use a Ti-89 to evaluate them. Since they both converge, the entire integral should converge as well. Though, I am struggling to reach the convergence of each case without an 89.
Well, can you compare each case with two simpler integrals that you CAN calculate, and utilize those results to reach your (correct) conclusion?

In particular, in terms of inequalities, what must be your guiding principle to determine whether your original (part-)integrals must converge as well?

Use integral test for series; hopefully I am not making forced connection between the two things.

So, your integral would converge if the series converges

nvm.. this does not work

or.. does works ...
1/n^1.5 is similar to your series which converges

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I am stuck with which functions that are larger for all x, that will converge. Help?

dtl42 said:
I am stuck with which functions that are larger for all x, that will converge. Help?

A good idea!

Now, instead of finding a single, continuous function that is larger than the integrand for ALL non-negative x, try first to find a nice function that is larger than the integrand for x between 0 and 1, and whose integral you know will converge.

## 1. What is an improper integral?

An improper integral is an integral with one or both limits of integration being infinite or the integrand having an infinite discontinuity within the interval of integration.

## 2. How do you determine if an improper integral converges or diverges?

To determine if an improper integral converges or diverges, you need to evaluate the limit of the integral as the upper limit approaches infinity. If the limit exists and is a finite number, the integral converges. If the limit does not exist or is infinite, the integral diverges.

## 3. What is the difference between a convergent and divergent integral?

A convergent integral is one that evaluates to a finite number, meaning the area under the curve is finite. A divergent integral is one that does not have a finite solution, meaning the area under the curve is infinite.

## 4. How do you determine the convergence or divergence of an improper integral using the comparison test?

The comparison test states that if 0 ≤ f(x) ≤ g(x) for all x ≥ a, and if ∫g(x)dx converges, then ∫f(x)dx also converges. If ∫g(x)dx diverges, then ∫f(x)dx also diverges. This means that if the integrand of the improper integral is smaller than or equal to a known convergent integral, the improper integral also converges. If the integrand is larger than or equal to a known divergent integral, the improper integral also diverges.

## 5. Can an improper integral have both convergent and divergent parts?

Yes, an improper integral can have both convergent and divergent parts. This is known as a semi-convergent integral. In this case, the integral will converge in some parts and diverge in others. The overall convergence or divergence of the integral will depend on the behavior of the integrand at the points of discontinuity.

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