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 ?

  • 2
    $\begingroup$ you can bound by $(1+x^2)^{-1}$ instead $\endgroup$ Jun 3, 2019 at 17:05
  • $\begingroup$ That doesn't mean that $f$ is not Lebesgue integrable; it means that your $g$ is not chosen properly. $\endgroup$
    – cmk
    Jun 3, 2019 at 17:06
  • $\begingroup$ Ah ok. Then i can apply the theorem. So $ \int lim_n f_n = 0$ $\endgroup$
    – Leon1998
    Jun 3, 2019 at 17:07
  • $\begingroup$ Is that right then? $\endgroup$
    – Leon1998
    Jun 3, 2019 at 17:07
  • $\begingroup$ $f_n(0) = sin(1) $ cannot converge to 0. Does this matter? $\endgroup$
    – Leon1998
    Jun 3, 2019 at 17:10

1 Answer 1


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.

  • $\begingroup$ @CalvinKhor Good point! I revised my answer. $\endgroup$
    – Robert Z
    Jun 4, 2019 at 6:10

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