Showing that a Lebesgue integral tends to zero as $n\to\infty$

Question

I want to show that $\displaystyle\int_{0}^{\infty} \frac{\sin(e^x)}{ 1+nx^2}dx \to 0 $ as $n\to\infty$ from the point of view of Lebesgue integration. Is this as simply as bounding the numerator and saying the denominator tends to zero. Any help would be appreciated.

Answer 1

The integrand satisfies $\displaystyle \left| \frac{\sin (e^x)}{1 + nx^2} \right| \le \frac{1}{1+x^2}$ for all $n \ge 1$. Since $\displaystyle \int_0^\infty \frac{1}{1+x^2} \, dx < \infty$ you may apply LDCT: $$\lim_{n \to \infty} \int_0^\infty \frac{\sin (e^x)}{1 + nx^2} \, dx = \int_0^\infty \lim_{n \to \infty} \frac{\sin (e^x)}{1 + nx^2} \, dx = \int_0^\infty 0 \, dx = 0.$$

Answer 2

The direct route would be

$$ \left| \int_0^\infty \frac{\sin e^x}{1+nx^2}\,dx \right| \leq \int_0^\infty \frac{dx}{1+n x^2} = \frac{1}{\sqrt{n}} \int_0^\infty \frac{dt}{1+t^2}\,dt. $$