I'm sort of stuck on evaluating the limit (as $n \to \infty$) of this Lebesgue integral, I hope someone here can help me out. The integral is:

$I_n=\int_{\mathbb{R}} \frac{1}{x^2+n^2} \cos(\sqrt{x^2+n^2})~d\lambda$

Let $u_n = \frac{1}{x^2+n^2} \cos(\sqrt{x^2+n^2})$

First of, it must be well-defined for each $n \in \mathbb{N}$, seeing as $u_n$ is continuous for each $n$ and thus measurable. Furthermore, we have that

$|I_n| \leq \int_{\mathbb{R}} \frac{1}{x^2+1} = \pi$

Now, here's my problem: I'm pretty sure that $\lim\limits_{n \to \infty} I_n = 0$, but I can't really show it - I can't use Lebesgue monotone convergence theorem, since $u_n$ is neither increasing nor decreasing in $n$. So, I'm all out of ideas - Any hints?

Much appreciated.

  • 1
    $\begingroup$ Can you use the Dominated Convergence Theorem? $\endgroup$ – Joey Zou Nov 1 '15 at 14:27
  • $\begingroup$ oh, of course, thank you! It's of course bounded by $\frac{1}{x^2+1}$ - thank you very much, I can't believe I didn't think to use dominated convergence :-) $\endgroup$ – Phillip Bredahl Nov 1 '15 at 14:32

Since $|\cos(\sqrt{x^2 + n^2})| \le 1$, then

$$|I_n| \le \int_{\Bbb R} \frac{1}{x^2 + n^2}\, d\lambda.$$

Compute the integral $\int_{\Bbb R} \frac{1}{x^2 + n^2}\, d\lambda$ directly and show that it converges to $0$.


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