Comparison of Improper Integrals

[UMich1344]

Homework Statement

For what values of p does the following integral converge or diverge?

$$\int^{\infty}_{2}\frac{dx}{x[ln(x)]^{p}}$$

The Attempt at a Solution

I have graphed y=$$\frac{1}{x[ln(x)]^{p}}$$ for different values of p (negative, zero, positive), and it looks as though in all three cases the integral converges.

However, when I found the antiderivatives of $$\frac{1}{x[ln(x)]^{p}}$$ for

p=-p

$$\int^{\infty}_{2}\frac{dx}{x[ln(x)]^{p}}=\frac{[ln(x)]^{p+1}}{p+1}\right|^{\infty}_{2}$$

p=+p

$$\int^{\infty}_{2}\frac{dx}{x[ln(x)]^{p}}=\frac{[ln(x)]^{1-p}}{1-p}\right|^{\infty}_{2}$$

p=0

$$\int^{\infty}_{2}\frac{dx}{x[ln(x)]^{p}}=ln(|x|)\right|^{\infty}_{2}$$

When taking the limits of the antiderivatives as x approaches infinity, it appears that for any value of p the integral will diverge.

I'm doing something wrong here, so any feedback will be greatly appreciated.

 

[sutupidmath]

$$\int^{\infty}_{2}\frac{dx}{x[ln(x)]^{p}}$$ sub $$u=lnx=>du=\frac{dx}{x}$$

$$\int^{\infty}_2\frac{du}{u^p}=\frac{u^{1-p}}{1-p}|^{\infty}_{2}=\frac{(lnx)^{1-p}}{1-p}}|^{\infty}_{2}$$ Now what u need to do is:


$$\lim_{b\rightarrow \infty}\frac{(lnx)^{1-p}}{1-p}}|^{b}_{2}=\frac{1}{1-p}\lim_{b\rightarrow \infty}[(ln(b))^{1-p}-(ln2)^{1-p}]$$

and see for what values of p this limit gives u a nr.