Show that: $\left[\underset{n\to \infty }{\text{lim}}\int_1^{\infty } \frac{\sin (x)}{x^{\{n+1\}}} \, dx\right] = 0 $

Question

Show that: $$\left[\underset{n\to \infty }{\text{lim}}\int_1^{\infty } \frac{\sin (x)}{x^{\{n+1\}}} \, dx\right] = 0 $$

My attemp: I using Taylor series: $$ \sin(x)= x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\frac{x^9}{362880}+O(x^{11}) $$

$$\left[\underset{n\to \infty }{\text{lim}}\int_1^{\infty } \frac{x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\frac{x^9}{362880}+O(x^{11})}{x^{\{n+1\}}} \, dx\right] = 0 $$ We have main part $$ \left[\underset{n\to \infty }{\text{lim}}\int_1^{\infty } \frac{1}{x^{\{n\}}} \, dx\right] = \frac{x^{1-n}}{1-n} $$ But it's not convergence.

Do you have better a idea?

Answer 1

No need to use the Taylor series. For $x > 0$, $$\left|\frac{\sin(x)}{x^{n+1}}\right| \leq \frac{1}{x^{n+1}}$$ so that $$\left|\int_1^\infty \frac{\sin(x)}{x^{n+1}}\,dx\right| \leq \int_1^\infty \left|\frac{\sin(x)}{x^{n+1}}\right|\,dx \leq \int_1^\infty \frac{1}{x^{n+1}}\,dx = \frac{1}{n} \xrightarrow{n\to\infty} 0$$

Answer 2

We have on $[1,\infty)$: $$\left|\frac{\sin(x)}{x^{n+1}}\right| \le \frac{1}{x^2}$$ and $$\int_1^\infty \frac{1}{x^2} dx = 1 < \infty$$

So we can use dominated convergence theorem to get:

$$\lim_{n\to\infty} \int_1^\infty \frac{\sin(x)}{x^{n+1}} dx = \int_1^\infty \lim_{n\to\infty} \frac{\sin(x)}{x^{n+1}} dx = \int_1^\infty 0 \;dx = 0$$