# logarithmic integral

• October 10th 2010, 09:21 AM
Dinkydoe
logarithmic integral
I have trouble showing that $\int_{2}^x\frac{1}{(\log(x))^k}dx=O(\frac{x}{(\log (x))^k})$ as $x\to\infty$

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

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

My guess is that I need some powerseries-expansion. Any suggestion will be greatly appreciated.
• October 10th 2010, 11:20 AM
zzzoak
Integrating this integral by parts

$
\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{=\frac{t}{(\log(t))^k}|_2^x +k \int_2^x \frac{dt}{(\log(t))^{k+1}}}
$
• October 10th 2010, 11:52 AM
Dinkydoe
Thanks!

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