# Prove that $\lim_n \int_{\Bbb R} \frac{\sin(n^2 x^5)}{n^2 x^4} \chi_{(0,n]} d\lambda(x) = 0$

Prove that:

$$\lim_n \int_{\Bbb R} \frac{\sin(n^2 x^5)}{n^2 x^4} \chi_{(0,n]} d\lambda(x) = 0$$

I am self-learning these stuff, and I would like to check whether I did things right. Here's my work:

Call $f_n(x)$ the integrand. We have:

$$\left| f_n(x) \right| = \left| \frac{\sin(n^2 x^5)}{n^2 x^4}\chi_{(0,1]} + \frac{\sin(n^2 x^5)}{n^2 x^4}\chi_{(1,n]}\right| \le \frac{n^2 x^5}{n^2x^4} \chi_{(0,1]} + \frac{1}{n^2 x^4} \chi_{(1,n]} \\ \le x \chi_{(0,1]} + \frac1{x^4} \chi_{(1,n]} \le x \chi_{[0,1]} + \frac{1}{x^4}\chi_{[1,\infty)} := g(x)$$

for all $n \ge 1$ and $x \in \Bbb R$.

Note that $(f_n)$ is a sequence of measurable functions, and it converges pointwise to $0$.

The function $x \mapsto x$ is Riemann-integrable on $[0,1]$, hence it is Lebesgue-integrable and the integrals coincide.

Also, $x \mapsto 1/x^4$ is Riemann-integrable on every compact $[1,a]$, with $a > 1$, and its $\int_1^{\infty}$ is absolutely convergent, hence it is Lebesgue integrable and the integrals coincide.

Then,

$$\int_{\Bbb R} g d\lambda = \int_0^1 x dx + \int_1^{\infty} \frac{dx}{x^4} < \infty$$

Hence $g \in L^1$. Therefore, by LDCT,

$$\lim_n \int_{\Bbb R} f_n d\lambda = \int_{\Bbb R} \lim_n f_n d\lambda = 0$$

• All I can see a bit off is that in the longest displayed formula, the $=$ should be $\le$, because $1/(n^2x^4)\le 1/x^4$ but we don't have equality. – ForgotALot Jan 31 '16 at 4:28

Alternatively, one can use the fact that $|\sin(t)|\leqslant \min\{1,t\}$ for each non-negative $t$, we have for each $\delta$: $$\left|\int_{\Bbb R} \frac{\sin\left(n^2 x^5\right)}{n^2 x^4} \chi_{(0,n]} \mathrm d\lambda(x)\right|\leqslant \int_{[0,\delta]}x \mathrm d\lambda(x)+\frac 1{n^2}\int_{[\delta,+\infty)}\frac 1{x^4}\mathrm d\lambda(x)=\frac{\delta^2}2+\frac{3}{n^2\delta^3}.$$ This is enough to conclude the wanted result. By optimizing in $\delta$, we can see that the convergence is (at least) of order $n^{-2/5}$.