# 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 Commented 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
Commented Jun 3, 2019 at 17:06
• Ah ok. Then i can apply the theorem. So $\int lim_n f_n = 0$ Commented Jun 3, 2019 at 17:07
• Is that right then? Commented Jun 3, 2019 at 17:07
• @Leon1998 I’m explaining to you why it doesn’t matter. You don’t have to relax or alter the statement of dominated convergence in order to make this work. Commented Jun 3, 2019 at 17:29

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.