I want to calculate the limit of following Lebesgue-integral:
lim
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 ?
Answer
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.
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