Convergence/Divergence of Integrals

[amcavoy]

Determine whether the following integral converges or diverges without evaluating it.

$$\int_{0}^{1-}\frac{dx}{\sqrt{1-x^4}}$$

Thanks for your help.

 

[dextercioby]

That function behaves ~1/x^{2},so it should be integrable in that domain.U have to find a function which is greater than your function,however,it's integral on [0,1] needs to be finite...

$$\int_{0}^{1} \frac{dx}{\sqrt{1-x^{4}}} =\frac{1}{4}\left(\sqrt{\pi}\right)^{3}\frac{\sqrt{2}}{\Gamma^{2}\left(\frac{3}{4}\right)}$$

Daniel.

 

[Hurkyl]

The problem point is x=1, not x=0. √(1 - x⁴) behaves like 2√(1-x) there.

P.S. 1/x² isn't integrable over [0, 1].

 

[dextercioby]

I didn't say "integrable".And i didn't say =1/x^{2}...

Daniel.

 

[mathwonk]

i believe it certainly converges. now let me think why i say that.

if y^2 = 1-x^4 then 2ydy = -4x^3 dx, so dx/sqrt(1-x^4) = dx/y = -2dy/4x^3. which makes perfectly good sense at x = 1.


huh?

formulas have always seemed like magic to me.

 

[dextercioby]

Yeah,but it's still infinite,because your limits of integration will still involve 0...

Daniel.

 

[saltydog]

That function behaves ~1/x^{2},so it should be integrable in that domain.U have to find a function which is greater than your function,however,it's integral on [0,1] needs to be finite...

$$\int_{0}^{1} \frac{dx}{\sqrt{1-x^{4}}} =\frac{1}{4}\left(\sqrt{\pi}\right)^{3}\frac{\sqrt{2}}{\Gamma^{2}\left(\frac{3}{4}\right)}$$

Daniel.

Can you explain how this integral is evaluated in terms of the Gamma function?

 

[dextercioby]

I dunno.Basically,it returned my a value for the Legendre elliptic integral and then expressed this one in terms of Gamma factor...

Daniel.

 

[dextercioby]

If u make a substitution,u can transform the integral to

$$\int_{1}^{+\infty} \frac{dx}{\sqrt{x^{4}-1}}$$

whose antiderivative,Mathematica finds

Daniel.

P.S.My Maple gave me the results...