TITLE: How can we find $\lim_{n \rightarrow +\infty}\int_{1}^{n} \frac{\cos(nx)(x-1)}{\sqrt{\ln^3(x)(1+x^4)}}dx $?
QUESTION [8 upvotes]: I have a question about improper integrals:
How can we find $\lim_{n \rightarrow +\infty} \int_{1}^{n} \frac{\cos(nx)(x-1)}{\sqrt{\ln^3(x)(1+x^4)}}dx$?
$\textbf{Some effort:}$
We know that $-1 \leq \cos(nx) \leq 1$, so we will try to sandwich it:
$$-\lim_{n \rightarrow +\infty} \int_{1}^{n} \frac{(x-1)}{\sqrt{\ln^3(x)(1+x^4)}}dx \leq \lim_{n \rightarrow +\infty} \int_{1}^{n} \frac{\cos(nx)(x-1)}{\sqrt{\ln^3(x)(1+x^4)}}dx \leq \mbox{$\displaystyle\lim_{n \rightarrow +\infty} \int_{1}^{n} \frac{(x-1)}{\sqrt{\ln^3(x)(1+x^4)}}dx$} $$
The next step is to get rid of the natural logarithm, which will be by using the change of variables and letting $\ln(x)=u$. But this makes the situation worse. I am interested of using again sandwich rule for natural logarithm as well. But I do not know that $\ln(x) < ?$ and $\ln(x) > ?$ for $x \in [0,+\infty]$?
Can you please give me an idea?
Thanks!

REPLY [3 votes]: Note that the integral
$$\int_1^{+\infty}  {\frac{{x - 1}}{{\sqrt {(1 + {x^4}){{\ln }^3}x} }}dx} $$ converges absolutely, the only problematic point is $1,+\infty$. At $x=+\infty$, it converges because $$\int_{}^{+\infty} \frac{1}{x\ln^{3/2} x} dx$$ converges. You can easily show that it is convergent at $x=1$ too.

From now on, I will denote $(x-1)/\sqrt{(1+x^4)\ln^3 x}$ by $f(x)$. Then
$$\lim_{n \to +\infty} \int_{1}^{n} f(x)\cos(nx) dx = \lim_{n \to +\infty} \int_{1}^{+\infty} f(x)\cos(nx) dx - \lim_{n \to +\infty} \int_{n}^{+\infty} f(x)\cos(nx) dx$$
The first limit tends to $0$ by Riemann-Lebesgue lemma since $f$ is absolute convergent, for the second, simply note that
$$\left| \int_{n}^{+\infty} f(x)\cos(nx) dx \right| \leq \int_{n}^{+\infty} |f(x)| dx \to 0$$
so the overall limit is $0$.