# Calculating $\lim_{n \rightarrow \infty} \int_{[0, \infty)} \frac{\sin(e^x) }{1+nx^2}\,dx$

I want to calculate the limit of following Lebesgue-integral:

$$\lim_{n \rightarrow \infty} \int_{[0, \infty)} \frac{\sin(e^x) }{1+nx^2}\,\mathrm dx$$

Therefore I wanted to apply Lebesgue's dominated convergence theorem. $$f_n(x)$$ is measurable and $$f_n \rightarrow 0$$ pointwise. Now it holds:

$$\left|\frac{\sin(e^x) }{1+nx^2}\right| \leq \frac{1}{1+x^2} :=g(x)$$

The improper integral over does converge. That means f is lebesgue integrable. Therefore $$\lim_{n \rightarrow \infty} \int_{[0, \infty)} \frac{\sin(e^x) }{1+nx^2}\,\mathrm dx = \int_{[0, \infty)} \lim_{n \rightarrow \infty} \frac{\sin(e^x) }{1+nx^2}\,\mathrm dx =0$$ Consider $$f_n(0) = sin 1$$ does not converge to $$0$$. So I can't apply the theorem, can I ?

• you can bound by $(1+x^2)^{-1}$ instead Jun 3, 2019 at 17:05
• That doesn't mean that $f$ is not Lebesgue integrable; it means that your $g$ is not chosen properly.
– cmk
Jun 3, 2019 at 17:06
• Ah ok. Then i can apply the theorem. So $\int lim_n f_n = 0$ Jun 3, 2019 at 17:07
• Is that right then? Jun 3, 2019 at 17:07
• $f_n(0) = sin(1)$ cannot converge to 0. Does this matter? Jun 3, 2019 at 17:10

You are on the right track: apply Lebesgue's dominated convergence with $$g(x)=\frac{1}{1+x^2}$$ which is Lebesgue integrable in $$[0,+\infty)$$. Since $$f_n(x)=\frac{\sin(e^x) }{1+nx^2}\to 0$$ for all $$x>0$$ the sequence $$(f_n)_n$$ converges to zero almost everywhere on $$[0,+\infty)$$, that's enough for dominated convergence, and we may conclude that the limit of $$\int_0^{\infty} f_n(x)\,dx$$ is zero.
Alternative way (without dominated convergence): \begin{align}\left|\int_{[0, \infty)} \frac{\sin(e^x) }{1+nx^2}\, dx \right|&\leq \int_0^{+\infty} \frac{|\sin(e^x) |}{1+nx^2}\, dx\\ &\leq \int_0^{+\infty} \frac{dx}{1+nx^2}=\left[\frac{\arctan(\sqrt{n}x)}{\sqrt{n}}\right]_0^{+\infty}=\frac{\pi}{2\sqrt{n}}.\end{align} So, again, the limit as $$n\to\infty$$ is zero.