Thread: logarithmic integral

I have trouble showing that $\displaystyle \int_{2}^x\frac{1}{(\log(x))^k}dx=O(\frac{x}{(\log (x))^k})$ as $\displaystyle x\to\infty$

A hint I've been given, is to split the integral in 2: $\displaystyle \int_{2}^{\sqrt{x}}+\int_{\sqrt{x}}^x$

I don't really see the point to that: I guess that: $\displaystyle \int_{\sqrt{x}}^x\to 0 $ as $\displaystyle x\to \infty$

My guess is that I need some powerseries-expansion. Any suggestion will be greatly appreciated.

Integrating this integral by parts

$\displaystyle
\displaystyle{\int_{2}^x\frac{1}{(\log(t))^k}dt=\f rac{t}{(\log(t))^k}|_2^x -\int_2^x td({(\log(t)^{-k})}=}
$

$\displaystyle
\displaystyle{=\frac{t}{(\log(t))^k}|_2^x +k \int_2^x \frac{dt}{(\log(t))^{k+1}}}
$

Thanks!

I shouldve thought of that. Any idea why that hint could be useful?