Proving convergence given inequalities of powers

[anniecvc]

Homework Statement

Show that if a>-1 and b>a+1 then the following integral is convergent:

∫(x^a)/(1+x^b) from 0 to ∞

The Attempt at a Solution

x^-1 < x^a < x^a+1 < x^b

x^-1/(1+x^b) < x^a/(1+x^b) < x^a+1/(1+x^b) < x^b/(1+x^b)

I also know any integral of the form ∫1/x^p when p>1 is convergent (from any number t to ∞)

Honestly not sure how to attack this problem. I'm trying to bound it but not sure how to show the parameters.

 

[pasmith]

Homework Statement

Show that if a>-1 and b>a+1 then the following integral is convergent:

∫(x^a)/(1+x^b) from 0 to ∞

The Attempt at a Solution

x^-1 < x^a < x^a+1 < x^b

x^-1/(1+x^b) < x^a/(1+x^b) < x^a+1/(1+x^b) < x^b/(1+x^b)

I also know any integral of the form ∫1/x^p when p>1 is convergent (from any number t to ∞)

Honestly not sure how to attack this problem. I'm trying to bound it but not sure how to show the parameters.

You need to show that both
$$\lim_{\epsilon \to 0^{+}}\int_\epsilon^1 x^a(1 + x^b)^{-1}\,dx$$
and
$$\lim_{R \to \infty} \int_1^R x^a(1 + x^b)^{-1}\,dx$$
converge. Since $b - 1 > a > -1$ we have $b > 0$, so $0 < x^b < 1$ if $0 < x < 1$ and $0 < x^{-b} < 1$ if $x > 1$ so you can use the binomial theorem to expand $(1 + x^b)^{-1}$ in powers of $x^b$ or $x^{-b}$ as appropriate.

 

[anniecvc]

Thank you!