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
<br /> \lim_{\epsilon \to 0^{+}}\int_\epsilon^1 x^a(1 + x^b)^{-1}\,dx<br />
and
<br /> \lim_{R \to \infty} \int_1^R x^a(1 + x^b)^{-1}\,dx<br />
converge. Since b - 1 &gt; a &gt; -1 we have b &gt; 0, so 0 &lt; x^b &lt; 1 if 0 &lt; x &lt; 1 and 0 &lt; x^{-b} &lt; 1 if x &gt; 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!