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If $x \geq 2$, then how do we prove that $$\int_{2}^{x} \frac{\mathrm dt}{\log^{n}{t}} = O\Bigl(\frac{x}{\log^{n}{x}}\Bigr)?$$

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5 Answers 5

up vote 15 down vote accepted

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

$$\int_{2}^{x} \frac{dt}{\log^n t} = \theta(\frac{x}{\log^n x})$$

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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.

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Just for the fun of dropping oil on a dead fire : I like this proof better, more clean. =) Haha! –  Patrick Da Silva Jun 15 '11 at 6:01

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 $$\int_{2}^{x}\frac{1}{\log^{n}t}dt\ll\frac{x}{\log^{n+1}x}. $$

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Thank you for this answer, Eric! I was wondering, though, how you obtained the second inequality? I can't really see how the $f(x)$ pops up without being within some integral. –  rwols Sep 25 at 11:35

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, $$ \int_2^x {\frac{{dt}}{{\log ^n t}}} \sim \frac{x}{{\log ^n x}} \;\; {\rm as} \; x \to \infty. $$

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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 $$\int_c^{\ln x}{e^u \over u^n}\,du = {2x \over (\ln{x})^n} - {2e^c \over c^n}$$ $$< {2x \over (\ln{x})^n}$$

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