Classify the following improper integral with the parameter

[Hernaner28]

Homework Statement

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 }}}}

Homework Equations

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]?

Thanks for your help.

 

[Hernaner28]

ANy help?

 

[Mute]

Homework Statement

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 }}}}

Homework Equations

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]?

Thanks for your help.

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.

 

[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

 

[Mute]

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

Perhaps I misunderstood your original post. It appeared that you were trying to bound the function with x^3a rather than x^3a/(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?

 

[SammyS]

Homework Statement

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 }}}}
...

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

 

[Mute]

Actually, your integral diverges for α ≥ ...

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

 

[SammyS]

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

Oops ! I edited that out.

You might want to also.

 

[Mute]

Oops ! I edited that out.

You might want to also.

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.