Why $2n \int ^\infty _n \frac{c}{x^2 \log(x)} \sim n \frac{C}{n \log(n)}$? I want to understand why we have
$$
2n \int ^\infty _n \frac{c}{x^2 \log(x)} \operatorname*{\sim}_{n\to\infty} n \frac{C}{n \log(n)}
$$
where $c$ is a normalizing constant. 
I am unable to understand how the integral is removed. 
 A: As usual in this business, L'Hospital's rule is quite useful:
$$ \lim_{n\to\infty} \frac{\int_{n}^{\infty} \frac{dx}{x^2 \log x}}{\frac{1}{n \log n}}
= \lim_{n\to\infty} \frac{-\frac{1}{n^2 \log n}}{-\frac{1}{n^2 \log n} - \frac{1}{n^2 \log^2 n}} = 1.$$
A: Given the $n$ on both sides and you uses of constants, your question is equivalent to asking why, for some absolute constant $\alpha >0$, we have
$$
\int_n^\infty \frac{dx}{x^2\ln x} \operatorname*{\sim}_{n\to\infty} \frac{\alpha}{n\ln n}
$$
To see why, let's notice that, defining $f$ on $[2,\infty)$ by $f(x)=-\frac{1}{x\ln x}$, we have that $f$ is differentiable and
$$
f'(x) = \frac{\ln x+1}{x^2\ln^2x}.
$$
Now, since $$
\frac{1}{x^2\ln x} \operatorname*{\sim}_{x\to\infty} f'(x)
$$
and that the integral $\int_2^\infty \frac{dx}{x^2\ln x}$ converges, then theorems of comparisons of integrals (for positive integrands) guarantee that the remainders are equivalent, i.e.
$$
\int_t^\infty \frac{dx}{x^2\ln x} \operatorname*{\sim}_{t\to\infty} \int_t^\infty f'(x) = -f(t) =\frac{1}{t\ln t}.
$$
This implies your result.
A: Integration by parts gives
$$
\begin{align}
\int_n^\infty\frac{\mathrm{d}x}{x^2\log(x)}
&=\frac1{n\log(n)}-\int_n^\infty\frac{\mathrm{d}x}{x^2\log(x)^2}\\
&=\frac1{n\log(n)}+O\!\left(\frac1{\log(n)^2}\int_n^\infty\frac{\mathrm{d}x}{x^2}\right)\\
&=\frac1{n\log(n)}+O\!\left(\frac1{n\log(n)^2}\right)\tag{1}
\end{align}
$$

Asymptotic Series
Integration by parts gives
$$
\int_n^\infty\frac{(k-1)!\,\mathrm{d}x}{x^2\log(x)^k}
=\frac{(k-1)!}{n\log(n)^k}-\int_n^\infty\frac{k!\,\mathrm{d}x}{x^2\log(x)^{k+1}}\tag{2}
$$
Iterating $(2)$ yields
$$
\int_n^\infty\frac{\,\mathrm{d}x}{x^2\log(x)}
=\frac1{n\log(n)}\sum_{k=0}^{m-1}\frac{k!}{\log(n)^k}+O\!\left(\frac1{n\log(n)^{m+1}}\right)\tag{3}
$$
