Prove that $\int_2^x \frac{dt}{\log(t)^n} = \mathcal{O}\bigg(\frac{x}{\log(x)^n}\bigg)$

Question

I am stuck at the following exercise:

Show that for $n \in \mathbb{N}$ holds

$$\int_2^x \frac{dt}{\log(t)^n} = \mathcal{O}\bigg(\frac{x}{\log(x)^n}\bigg).$$

I do not see how I could prove this. I know that the following identity holds:

$$\int_2^x \frac{dt}{\log(t)^n} = \frac{t}{\log(t)^n} \bigg\vert^x_2 + n \int_2^x\frac{dt}{\log(t)^{n+1}},$$

but I do not see how this could help here. Could you give me a hint?

Answer 1

You could use the following theorem : if f and g are positive, $\int_a^x f$ and $\int_a^x g$ both have infinite limit and $f=o(g)$, then $\int_a^x f = o\left(\int_a^x g\right)$.

So $\int_2^x \frac{{\rm d}t}{(\log(t))^{n+1}} = o\left(\int_2^x \frac{{\rm d}t}{(\log(t))^n}\right)$, and your proof gives you the result.

Answer 2

I am surprised that no one mentioned L'Hospital's Rule. By the L'Hospital's Rule,

$$ \lim_{x\to\infty} \frac{\int_{2}^{x} \frac{\mathrm{d}t}{\log^n t}}{\frac{x}{\log^n x}} = \lim_{x\to\infty} \frac{\frac{\mathrm{d}}{\mathrm{d}x}\int_{2}^{x} \frac{\mathrm{d}t}{\log^n t}}{\frac{\mathrm{d}}{\mathrm{d}x}\frac{x}{\log^n x}} = \lim_{x\to\infty} \frac{\frac{1}{\log^n x}}{\frac{1}{\log^n x}-\frac{n}{\log^{n+1} x}} = 1.$$

This immediately proves that $\int_{2}^{x} \frac{\mathrm{d}t}{\log^n t} = \mathcal{O}\left(\frac{x}{\log^n x}\right)$.

Answer 3

By L'Hospital's rule $$ \mathop {\lim }\limits_{x \to + \infty } \frac{{\int_2^x {\frac{{dt}}{{\log ^n t}}} }}{{\frac{x}{{\log ^n x}}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{1}{{\log ^n x}}}}{{\frac{1}{{\log ^n x}} - \frac{n}{{\log ^{n + 1} x}}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{1}{{1 - \frac{n}{{\log x}}}} = 1. $$ So $$ \int_2^x {\frac{{dt}}{{\log ^n t}}} \sim \frac{x}{{\log ^n x}} $$ as $x\to +\infty$. In particular, there is a $C>0$ such that $$ \int_2^x {\frac{{dt}}{{\log ^n t}}} \le C\frac{x}{{\log ^n x}} $$ for all $x\geq 2$.