logarithmic integral

**Dinkydoe** · Oct 10th 2010, 08:21 AM

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.

**zzzoak** · Oct 10th 2010, 10:20 AM

Integrating this integral by parts

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

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

**Dinkydoe** · Oct 10th 2010, 10:52 AM

Thanks!

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

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