# Classify the following improper integral with the parameter

1. Jul 3, 2012

### Hernaner28

1. The problem statement, all variables and given/known data
Discuss for alpha the convergence of the following improper integral:
$$\displaystyle \int\limits_{0}^{3}{\frac{{{x}^{3\alpha }}}{{{\left( 9-{{x}^{2}} \right)}^{\alpha }}}}$$

2. Relevant equations

3. The attempt at a solution
Well, my attempt was to simplify the integral to:

$$\displaystyle \int\limits_{0}^{3}{{{\left( \frac{{{x}^{3}}}{9-{{x}^{2}}} \right)}^{\alpha }}}$$

After that my idea was to bound the function above by $$\displaystyle {{x}^{3a}}$$ but I don't know if that's correct, because $$\displaystyle {{x}^{3a}}$$ is greater than the original function but it could happen that it's not greater for all x, so how can I find a function I can use to bound the function in the interval of integration [0,3]?

2. Jul 4, 2012

ANy help?

3. Jul 4, 2012

### Mute

For positive $\alpha$ your integrand diverges at x = 3, where the denominator is zero. How are you going to find a function which bounds that? I think you need a revised strategy to figure out the convergence properties of this integral. You might try choosing a function (or functions) which is greater than your integrand but has the same divergence, and which you can evaluate the integral of.

Similarly, for negative $\alpha$ your integrand diverges at x = 0.

4. Jul 4, 2012

### Hernaner28

Well, I did find two functions which bounds that:

$$\dfrac{4x}{9-x^2}\leq \dfrac{x^3}{9-x^2}\leq \dfrac{9x}{9-x^2}$$

Why you say I cannot bound it?

Now what can I do with that? It's still tough for me

5. Jul 4, 2012

### Mute

Perhaps I misunderstood your original post. It appeared that you were trying to bound the function with x3a rather than x3a/(9-x2).

In any event, I'm not sure if just using bounds alone are going to be the easiest way to show the result you want, at least for positive alpha. You might want to try changing variables first, and then finding a bound.

Another piece of knowledge that will be useful is the answer to this question: for what values of $\alpha$ does

$$\int_0^c du~\frac{1}{u^\alpha}$$

converge, where the upper limit c is a finite constant?

6. Jul 4, 2012

### SammyS

Staff Emeritus
BTW:

I have this pet peeve about leaving the differential symbol out of integrals, dx, in this case.

$\displaystyle \int\limits_{0}^{3}{\frac{{{x}^{3\alpha }}}{{{\left( 9-{{x}^{2}} \right)}^{\alpha }}}}\,dx$

Last edited: Jul 4, 2012
7. Jul 4, 2012

### Mute

I think the entire point of the OP's problem is to find that out! You probably shouldn't just give it away! =P

8. Jul 4, 2012

### SammyS

Staff Emeritus
Oops ! I edited that out.

You might want to also.

9. Jul 5, 2012

### Mute

Too late to completely edit it out the fact that it diverges for some values of alpha; fortunately I had the foresight to remove the value from the quoted post when I posted.