I`m preparing for an exam and I have this question: how can I find the distributional limit for something like for $f\in L^1 (R)$ :
$$\lim_{t \rightarrow 0} \int^{\infty}_0 f(x) \cos \left(t \sqrt x\right) dx $$
$$\lim_{t\rightarrow \infty} \int^{\infty}_0 f(x) \cos\left(t \sqrt x\right) dx $$
Thanks.
Note that $|f(x)\cos(t\sqrt x)|\le |f(x)|$. Inasmuch as $\int_0^\infty |f(x)|\,dx<\infty$, the Dominated Convergence Theorem guarantees that
$$\begin{align} \lim_{t\to 0}\int_0^\infty f(x)\cos(t\sqrt x)\,dx&=\int_0^\infty \lim_{t\to 0}\left(f(x)\cos(t\sqrt x)\right)\,dx\\\\ &=\int_0^\infty f(x)\,dx \end{align}$$
So, the distributional limit is the classical limit
$$\lim_{t\to 0}\cos(t\sqrt x)=1$$
Note that since $\int_0^\infty |f(x)|\,dx<\infty$, then $\int_0^\infty x|f(x^2)|\,dx<\infty$ also. That is to say, $xf(x^2)\in L^1$.
Noting that $\lim_{t\to \infty} \int_0^\infty f(x)\cos(t\sqrt{x})\,dx=2\lim_{t\to \infty}\int_0^\infty xf(x^2)\cos(tx)\,dx$, the Riemann-Lebesgue Lemma guarantees that
$$\lim_{t\to \infty} \int_0^\infty f(x)\cos(t\sqrt{x})\,dx=0$$
Thus, the distributional limit is
$$\lim_{t\to \infty}\cos(t\sqrt x)\sim 0$$