# Calculate $\lim_{n\rightarrow \infty}\int_{1}^{\infty}{\frac{\sqrt{x}\log{nx}\sin{nx}}{1+nx^{3}}}$

I have to calculate (if it exists) $\displaystyle\lim_{n\rightarrow \infty}\displaystyle\int_{1}^{\infty}{\dfrac{\sqrt{x}\log{nx}\sin{nx}}{1+nx^{3}}}$.

I think I have to use Lebesgue dominated convergence theorem. The problem is that I cannot find an integrable function $g$ such that: $$\forall\text{ } n\in\mathbb{N}, \left|\dfrac{\sqrt{x}\log{nx}\sin{nx}}{1+nx^{3}}\right|\leq g(x) \text{ a.e.}$$ If I was able to find it, $\displaystyle\lim_{n\rightarrow \infty}\displaystyle\int_{1}^{\infty}{\dfrac{\sqrt{x}\log{nx}\sin{nx}}{1+nx^{3}}}=\displaystyle\int_{1}^{\infty}{\displaystyle\lim_{n\rightarrow \infty}\dfrac{\sqrt{x}\log{nx}\sin{nx}}{1+nx^{3}}}=0$.

## 1 Answer

For $x\in [0,1]$ and $n\in \Bbb N$, $|\log nx \sin nx| = \log nx |\sin nx| \le nx$, and $1 + nx^3 \ge nx^3$. So $$\left|\frac{\sqrt{x}\log nx \sin nx}{1 + nx^3}\right| \le \sqrt{x}\frac{nx}{nx^3} = \frac{1}{x^{3/2}} \quad (x \ge 1, n\in \Bbb N)$$

Since $\int_1^\infty \frac{1}{x^{3/2}}\, dx < \infty$, you can proceed as you did.