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

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?

• Hi. You forgot the exponent in your title. – Nicolas FRANCOIS Jun 17 '20 at 16:05
• Thanks, I corrected it. – 3nondatur Jun 17 '20 at 16:16
• Will this help you? math.stackexchange.com/questions/1282024/… – Oliver Diaz Jun 17 '20 at 16:21
• Yeah, this also helps me. Thanks a lot to both of you. – 3nondatur Jun 17 '20 at 16:22

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.

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

• (+1) Great answer, but FTLOG it’s l’Hôpital. A hospital is where sick people get better. – cansomeonehelpmeout Jun 18 '20 at 12:57
• @cansomeonehelpmeout Forgive me for my laziness. I actually knew that l’Hôpital is a precise name, but was too lazy to search and paste the accented character... – Sangchul Lee Jun 18 '20 at 13:02

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 $$x\geq 2$$.