Limit of Lebesgue integral over the real line For every $n \geq 1$ consider the integral:
$I_n=\int_{{\mathbb{R}}}^{}  \frac{\sin(x^2+n^2)}{x^2+n^2}  \,d\lambda$
I have to explain why $I_n$ is well-defined for every such $n$ and I want to compute $\lim_{n \to \infty} I_n$.
The part about explaining the well-definedness really confuses me. However, I tried to come up with an explanation. First I will find a majorant for the integrand:
$|\frac{sin(x^2+n^2)}{x^2+n^2}|\leq |\frac{1}{x^2+1}|=\frac{1}{x^2+1} $
I have a question to this: Are you allowed to choose a majorant depending on n?
I know that the Lebesgue integral is an extension of the proper Riemann integral. Hence this equality:
$\int_{[-k,k]}^{}  \frac{1}{x^2+1}  \,d\lambda = \int_{-k}^{k}  \frac{1}{x^2+1}  \,dx =2arctan(k)$
Now I will take a look at the limits:
$ \int_{\mathbb{R}}^{}  \frac{1}{x^2+1}d\lambda=\lim_{k \to \infty}\int_{[-k,k]}^{}  \frac{1}{x^2+1}  \,d\lambda = \lim_{k \to \infty}\int_{-k}^{k}  \frac{1}{x^2+1}  \,dx =\lim_{k \to \infty}2arctan(k)=\pi$
So by monotonicity it must hold that:
$ I_n=\int_{{\mathbb{R}}}^{}  |\frac{sin(x^2+n^2)}{x^2+n^2}|  \,d\lambda \leq \int_{\mathbb{R}}^{}  \frac{1}{x^2+1}d\lambda =\pi  $
Since $|\frac{sin(x^2+n^2)}{x^2+n^2}|$ is Lebesgue integrable, $\frac{sin(x^2+n^2)}{x^2+n^2}$ also is. Therefore $I_n$ is well-defined for every $n$. The conditions for using Lebesgue's dominated convergence theorem clearly are met, so I can simply push the limit into the integral:
$\lim_{n \to \infty}I_n=\lim_{n \to \infty}\int_{{\mathbb{R}}}^{}  \frac{sin(x^2+n^2)}{x^2+n^2}d\lambda=\int_{{\mathbb{R}}}^{}  \lim_{n \to \infty}\frac{sin(x^2+n^2)}{x^2+n^2}  \,d\lambda=\int_{{\mathbb{R}}}^{} 0  \,d\lambda=0$
I think the last part is right, but I am not sure about the first part since it is an improper integral.
 A: 
I have a question to this: Are you allowed to choose a majorant depending on $n$?

If your goal is to just show that $x\mapsto \frac{\sin(x^2+n^2)}{x^2+n^2}$ is Lebesgue-integrable on $\Bbb{R}$, then yes you can. For instance, this is bounded by $\frac{1}{x^2+n^2}$, and this latter function is clearly Lebesgue-integrable on $\Bbb{R}$. So such a reasoning will suffice to explain the well-definition of $I_n$.
However, your goal is two-fold: you want to show well-definition of $I_n$, and compute its limit. Thus, what you've done is better: you found an upper bound which is independent of $n$, and is still integrable on $\Bbb{R}$. This is good because it allows you to deduce well-definition of each $I_n$ as above; furthermore, you can now compute the limit $\lim\limits_{n\to\infty}I_n$ using dominated convergence, as you argued. And yes, the limit is indeed $0$ because of dominated convergence (Lebesgue integration and DCT holds for any measure space, so the fact that you're working on all of $\Bbb{R}$ is not at all a problem here).
