Improper Integrals: Divergence at x=0?

  • Thread starter Thread starter aaaa202
  • Start date Start date
  • Tags Tags
    Integrals
aaaa202
Messages
1,144
Reaction score
2

Homework Statement


Check if the following integrals diverge:
Int(1/(x^3-2x^2+x), from 0 to 1) and Int(1/(x^3+x^2-2x)

Homework Equations


Ratio-test(not sure if that's the name)

The Attempt at a Solution


I have solved the problem and found that both integrals diverge at x=0. I just want to check if you guys agree with me.
 
Physics news on Phys.org
Ratio test is for infinite series, not a improper integrals. It would be helpful to see your work.
 
aaaa202 said:

Homework Statement


Check if the following integrals diverge:
Int(1/(x^3-2x^2+x), from 0 to 1) and Int(1/(x^3+x^2-2x)

Homework Equations


Ratio-test(not sure if that's the name)


The Attempt at a Solution


I have solved the problem and found that both integrals diverge at x=0.
Both integrands are undefined at x = 0, which is why the integrals are improper. Both integrands are also undefined at the other endpoint, x = 1.

You need to use limits to evaluate these integrals - then you can decide whether either one diverges.
aaaa202 said:
I just want to check if you guys agree with me.
 
Okay ratio test was then not the correct name. What I did was use the test that if limx->0[g/f>0] then g also diverges if f diverges. So I took f as 1/x^3. Isn't this correct, and don't you guys get that both integrals diverge?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top