TITLE: Bounding the integral $\int_{2}^{x} \frac{\mathrm dt}{\log^{n}{t}}$
QUESTION [11 upvotes]: If $x \geq 2$, then how do we prove that $$\int_{2}^{x} \frac{\mathrm dt}{\log^{n}{t}} = O\Bigl(\frac{x}{\log^{n}{x}}\Bigr)?$$

REPLY [2 votes]: Another way of dealing with this integral is to let $t = e^u$, so you are trying to bound
$$\int_{\ln 2}^{\ln x}{e^u \over u^n}\,du$$
Clearly it suffices to replace $\ln 2$ with any fixed constant $c$ since the difference is a constant that doesn't affect asymptotics.
Integrating by parts gives
$$\int_c^{\ln x}{e^u \over u^n}\,du = {x \over (\ln{x})^n} - {e^c \over c^n}+\int_c^{\ln x}{ne^u \over u^{n+1}}\,du$$
If $c$ is large enough the integrand on the right is less than half that on the left, so the integral is less than half of the left integral. Subtracting gives
$$\int_c^{\ln x}{e^u \over u^n}\,du = {2x \over (\ln{x})^n} - {2e^c \over c^n}$$
$$< {2x \over (\ln{x})^n}$$