Bounding the integral $\int_{2}^{x} \frac{\mathrm dt}{\log^{n}{t}}$ If $x \geq 2$, then how do we prove that $$\int_{2}^{x} \frac{\mathrm dt}{\log^{n}{t}} = O\Bigl(\frac{x}{\log^{n}{x}}\Bigr)?$$
 A: Put
$$
f(x)=\int_2^x {\frac{{dt}}{{\log ^n t}}} \;\; {\rm and} \;\; g(x)=\frac{x}{{\log ^n x}}.
$$
Then,
$$
f'(x)=\frac{1}{{\log ^n x}} \;\; {\rm and} \;\; g'(x)=\frac{{\log ^n x - n\log ^{n - 1} x}}{{\log ^{2n} x}}.
$$
Hence
$$
\mathop {\lim }\limits_{x \to \infty } \frac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to \infty } \frac{{f'(x)}}{{g'(x)}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\log ^n x}}{{\log ^n x - n\log ^{n - 1} x}} = \mathop {\lim }\limits_{x \to \infty } \frac{1}{{1 - n/\log x}} = 1.
$$
That is, $f$ is asymptotically equal to $g$ as $x \to \infty$ (written as $f \sim g$); in particular,
$f = O(g)$ as $x \to \infty$.
EDIT: The asymptotic equivalence
$$
\int_2^x {\frac{{dt}}{{\log ^n t}}} \sim \frac{x}{{\log ^n x}} \;\; {\rm as} \; x \to \infty,
$$
derived elementarily above (and which is much more than the OP asked for), also follows as a very special case from the theory of Regular variation. The function $L(x)=1/(\log ^n x)$, $x \geq 2$, is slowly varying (at infinity), for any $n \in \mathbb{R}$, and is bounded on every compact subset of $[2,\infty)$. Thus, by Karamata's integral theorem (Theorem A.9 here), 
(a) for $\alpha > -1$, 
$$
\int_2^x {t^\alpha  L(t)dt} \sim \frac{{x^{\alpha  + 1} }}{{\alpha  + 1}}L(x) \;\; {\rm as} \; x \to \infty;
$$
(b) for $\alpha < -1$, 
$$
\int_2^x {t^\alpha  L(t)dt} \sim - \frac{{x^{\alpha  + 1} }}{{\alpha  + 1}}L(x) \;\; {\rm as} \; x \to \infty.
$$
In particular (letting $\alpha = 0$ in (a)), 
$$
\int_2^x {L(t)dt} \sim xL(x) \;\; {\rm as} \; x \to \infty,
$$
that is,
$$
\int_2^x {\frac{{dt}}{{\log ^n t}}} \sim \frac{x}{{\log ^n x}} \;\; {\rm as} \; x \to \infty.
$$
A: Here is yet another approach:
Let $2\leq f(x)\leq x$  be some function which we will soon chose depending on $x$. Then splitting the integral we have $$\int_{2}^{x}\frac{1}{\log^{n}t}dt\leq\int_{2}^{f(x)}\frac{1}{\log^{n}t}dt+\int_{f(x)}^{x}\frac{1}{\log^{n}t}dt\leq f(x)+\frac{x}{\log^{n}\left(f(x)\right)}.$$ Hence, taking $f(x)=\frac{x}{\log^{n+1}x}$  (or even $f(x)=xe^{-c\sqrt{\log x}}$ ) we conclude $$\int_{2}^{x}\frac{1}{\log^{n}t}dt\ll\frac{x}{\log^{n+1}x}. $$
A: 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
$$\int_c^{\ln x}{e^u \over u^n}\,du = {2x \over (\ln{x})^n} - {2e^c \over c^n}$$
$$< {2x \over (\ln{x})^n}$$
A: Here is a way to prove it without knowing the answer before hand.
Let $c > 2$ be a number such that $c > e^{n+1}$.
It is enough to consider
$$\int_{c}^{x} \frac{dt}{\log^n t}$$
Integrating by parts, we get
$$\int_{c}^{x} \frac{dt}{\log^n t} =  \frac{x}{\log^n x} - K + n \int_{c}^{x} \frac{dt}{\log^{n+1} t}$$
Now $$n\int_{c}^{x} \frac{dt}{\log^{n+1} t} \le \frac{n}{n+1} \int_{c}^{x} \frac{dt}{\log^n t} $$
as $\log t \ge n+1$ for $t \ge c$
Thus $$\frac{1}{n+1}\int_{c}^{x} \frac{dt}{\log^n t} < \frac{x}{\log^n x}  $$
In fact, we have that
$$\int_{2}^{x} \frac{dt}{\log^n t} = \theta(\frac{x}{\log^n x})$$
A: Consider
$$ \frac{d}{dx} \frac{x}{\log^n(x)} = \frac{1}{\log^n(x)} - \frac{n}{\log^{n+1}(x)} = \frac{1}{\log^n(x)}\left( 1 - \frac{n}{\log x}\right)$$
For sufficiently large $x$, the RHS is bigger than $\frac12 \frac{1}{\log^n(x)}$, and hence we get the asymptotic bound we need. The usual arguments then gives you a sufficiently large constant for the Big O. 
