# Integration: Convergence/Divergence Tests

1. ### WaterPoloGoat

14
Does anyone know of helpful tests that can help me determine whether an improper integral converges or diverges? Specifically ones where you don't have to solve the integral?

For example, the problem I'm solving has a very complicated solution to the integral:

the problem: integrate dx/(sqrt(x+x^3)) from 0 to $$\infty$$.

I have tried integrating, and ended up using Maple, only to get a very complicated answer.

But I only need to prove whether it converges or diverges. That's it.

Any ideas?

2. ### lanedance

3,307
how about an integral comparison test, you'll probably nee dto split it into two integrals though...

3. ### WaterPoloGoat

14
I've started doing a comparison test, but I got a little stuck. I need to play around with it more.

I noted that 1/sqrt(x^3+x) is < 1/sqrt(x^3) = x^(-3/2).

If the integral x^(-3/2) converges, then so does the original problem. However, I ran into a little snag:
The integral of x^(-3/2) is 1/sqrt(x), and evaluating the integral from 0 to b (where b is approaching infinity) yields:

1/sqrt(b)-1/sqrt(0), I have a fraction of form 1/0. So...

Then I considered 1/sqrt(x^3+x) < 1/sqrt(x^2) , but ran into a similar snag.

What do you mean by splitting it into two integrals?

4. ### lanedance

3,307
i meant use the property of integrals
$$\int_a^b dx \frac{1}{\sqrt{x^3+x}} = \int_a^c dx \frac{1}{\sqrt{x^3+x}} + \int_c^b dx \frac{1}{\sqrt{x^3+x}}$$

then if you pick c correctly, your approach should work for the 2nd part, but you'll have to think about the 1st... maybe consider a variable change

5. ### WaterPoloGoat

14
Ah. I see. Thank you. I'll work on that.

6. ### sagardip

38
Whether an integral is convergent or divergent can be determined by a no. of ways. For example you can apply the trapezoidal rule or the Simpsons rule to approximate a definite integral.