Determine whether the limit exists and evaluate the integral if it does

[resolvent1]

Homework Statement

Determine whether the limit

\lim_{n \rightarrow \infty} \int_{-1}^1 \frac{1}{\sqrt[3]{1-x^{2n}}} dx

exists and evaluate the integral if it does

Homework Equations

Dominated convergence theorem (I think) and a power series representation.

The Attempt at a Solution

I've been attempting to find a dominating function for \frac{1}{\sqrt[3]{1-x^{2n}}} by looking at the power series representation for \frac{1}{1-x^{2n}}, but I'm not seeing it. I'd appreciate any help.

 

[SammyS]

Homework Statement

Determine whether the limit

\lim_{k \rightarrow \infty} \int_{-1}^1 \frac{1}{\sqrt[3]{1-x^{2n}}} dx

exists and evaluate the integral if it does

Homework Equations

Dominated convergence theorem (I think) and a power series representation.

The Attempt at a Solution

I've been attempting to find a dominating function for \frac{1}{\sqrt[3]{1-x^{2n}}} by looking at the power series representation for \frac{1}{1-x^{2n}}, but I'm not seeing it. I'd appreciate any help.

What does k have to do with this?

Why take the limit as k → ∞ ?

 

[jackmell]

Much more interesting if we consider:

\lim_{n\to\infty}\int_{-1}^1 \frac{dx}{\sqrt[3]{1-x^{2n}}}

and in that case isn't:

x^{2n}\leq x^2

in that interval?

 

[resolvent1]

Sorry, that should be an n, not a k.

It's true that x^{2n} \leq x^2 on [-1,1], so \frac{1}{\sqrt[3]{1-x^{2n}}} \leq \frac{1}{1-x^2} on [-1,1], but \frac{1}{1-x^2} isn't in L^1(-1,1)

So I'm not sure what to do with it.

 

[I like Serena]

What about $1 \over \sqrt{1-x^2}$?

 

[resolvent1]

Thanks, I think that works perfectly. (I can't believe I didn't see that before - sorry - and that problem was really starting to piss me off.)

 

[I like Serena]

Good! :)

Does that mean that you also evaluated the integral in its limit?

 

[resolvent1]

Yeah, I knew that several hours ago, just not the dominating function. THank you for your help.