What comparison can I make to tell this improper integral is divergent?

[nombreuso]

It's the integral from 0 to infinity of dx/(x-2)^3. I don't know to what function should I compare it to know if it's divergent or convergent.

 

[pka]

It's the integral from 0 to infinity of dx/(x-2)^3. I don't know to what function should I compare it to know if it's divergent or convergent.

Why are you comparing it to anything? [MATH]\int {\frac{1}{{{{\left( {x - 2} \right)}^3}}}dx}=\dfrac{-1}{2(x-2)^2} [/MATH].

 

[lex]

All you have to do is try to evaluate it.

Do [MATH]\int_0^n \dfrac{1}{(x-2)^3} dx[/MATH]
and take the limit as [MATH]n \to \infty[/MATH]

 

[nombreuso]

Why are you comparing it to anything? [MATH]\int {\frac{1}{{{{\left( {x - 2} \right)}^3}}}dx}=\dfrac{-1}{2(x-2)^2} [/MATH].

Because the result is convergent. However, I know it's divergent at x=2. So the only thing I need to know is if the integral from 2 to 3, for example, or from 0 to 2, is convergent or not. But I need to do it by comparison.

 

[lex]

Because the result is convergent. However, I know it's divergent at x=2. So the only thing I need to know is if the integral from 2 to 3, for example, or from 0 to 2, is convergent or not. But I need to do it by comparison.

Oops - I missed that! (So my post #3 is not correct - that is not all you have to do. You also have to investigate the existence of the integral at the vertical asymptote x=2).

However, to investigate the existence of the integral about 2, 'all you have to do is': [MATH]\int_0^{2-\tfrac{1}{n}} \dfrac{1}{(x-2)^3} dx[/MATH] and take the limit as [MATH]n \to \infty[/MATH]
So, the integral doesn't exist (because of the function's behaviour 'near x=2' rather than its behaviour as x tends to infinity).

 

[Dr.Peterson]

Because the result is convergent. However, I know it's divergent at x=2. So the only thing I need to know is if the integral from 2 to 3, for example, or from 0 to 2, is convergent or not. But I need to do it by comparison.

Are you saying "need" because you were instructed to do so? That is certainly not the only way, and probably not the best way.

I'm also not sure whether you understand the meaning of "convergent" and "divergent" in this context. Are you saying you know ahead of time that the integral is convergent, or are you just referring to the behavior of the integrand?