Showing that a Lebesgue integral tends to zero as $n\to\infty$ 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.
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 $\displaystyle \int_0^\infty \frac{1}{1+x^2} \, dx &lt; \infty$ you may apply LDCT: $$\lim_{n \to \infty} \int_0^\infty \frac{\sin (e^x)}{1 + nx^2} \, dx = \int_0^\infty \lim_{n \to \infty} \frac{\sin (e^x)}{1 + nx^2} \, dx = \int_0^\infty 0 \, dx =  0.$$
The direct route would be

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