Need Help with Improper Integral: Convergent or Divergent?

[ttiger2k7]

This was a problem on one of my previous tests that I got wrong entirely. In preparing for my final, I'm attempting to redo it. I was wondering if someone could check my work.

Determine whether the following improper integral is convergent or divergent. If the integeral is convergent, find its value.

I = \int^{\infty}_{0}\frac{2x}{(x^{2}+1)^{1.2}}

which, after carrying out the 1.2, becomes

\int^{\infty}_{0}\frac{2x}{(x^{2.4}+1)}

That's where I get stuck. I know that the function diverges, but I don't know how to prove it.

I thought about trying to manipulate it somehow so that the function I'm comparing it to is in the form

\frac{1}{x^{p}}

I started off by trying a direct comparison test and compared it to \frac{2x}{x^{2.4}} which can simplify to

\frac{2}{x^{.4}}

So, since the above function has p=.4<1, it diverges, by the Direct Comparison Test, I diverges.

 

[Mute]

Be careful! (a + b)^p \neq (a^{p} + b^p).

Furthermore, the function 2x/x^{2.4} is actually slightly larger than your original integrand, so the fact that the integral over this diverges doesn't actually tell you anything about the convergence/divergence of your original integrand.

The fact that there is a 2x in the integrand suggests you should make the substitution u = x^2 + 1. Then your integral becomes

I = \int^{\infty}_{1}\frac{du}{u^{1.2}} = \int^{\infty}_{1}\frac{du}{u^{6/5}}

This is an integral you can do, and you should be able to tell if it diverges or converges.

 

[HallsofIvy]

This was a problem on one of my previous tests that I got wrong entirely. In preparing for my final, I'm attempting to redo it. I was wondering if someone could check my work.

Determine whether the following improper integral is convergent or divergent. If the integeral is convergent, find its value.

I = \int^{\infty}_{0}\frac{2x}{(x^{2}+1)^{1.2}}

which, after carrying out the 1.2, becomes

\int^{\infty}_{0}\frac{2x}{(x^{2.4}+1)}

No, it doesn't! In general, (x+ y)ⁿ is NOT xⁿ+ yⁿ!

That's where I get stuck. I know that the function diverges, but I don't know how to prove it.

I thought about trying to manipulate it somehow so that the function I'm comparing it to is in the form

\frac{1}{x^{p}}

I started off by trying a direct comparison test and compared it to \frac{2x}{x^{2.4}} which can simplify to

\frac{2}{x^{.4}}

So, since the above function has p=.4<1, it diverges, by the Direct Comparison Test, I diverges.

Let u= x²+ 1, so that du= 2x dx, in your original integral.