TITLE: How to show simply that $e^{\frac{x}{2}}\int^\infty_0 e^{-t}t^{n-\frac{1}{2}}\cos(2\sqrt{xt})dt=O\big(\frac{n!}{\sqrt{n}}\big)$?
QUESTION [6 upvotes]: Can we prove, without using Laguerre polynomials, that
$f_n(x)=O(\frac{n!}{\sqrt{n}})$ i.e. that
$$
\exists C>0, \exists N\in\mathbb N, \forall x\geq0, \forall n\geq N :\ \big| f_n(x) \big|\leq C \frac{n!}{\sqrt{n}},
$$
where
$$
f_n(x)=e^{\frac{x}{2}}\int^\infty_0 e^{-t}t^{n-\frac{1}{2}}\cos(2\sqrt{xt})dt,\quad x  \geq 0\;?
$$
Proof (by using Laguerre polynomials): it's easy to show that
$$
f_n(x)=\sqrt \pi e^{-x/2} n! L^{(-1/2)}_n(x)
$$
and we know that
$$
L^{(-1/2)}_n(x)=O\Big(e^{x/2}\frac{1}{\sqrt{n}}\Big).
$$
My reference is page 9 formula 18 $$L^{\alpha }_n(x)=O\Big(e^{\frac x2}x^{\frac{-\alpha}2  -\frac 14}n^{\frac{\alpha}2 -\frac 14}     \Big) .$$ or   see The polynomials' asymptotic behaviour for large n 
However, I'd like to not use this simple argument.

REPLY [4 votes]: Let, $u^2=x$ and $v^2=t$, and let,
$f_n(x)e^{-\frac{x}{2}}=\phi_{n}(u)=2\int_{0}^{\infty} e^{-v^2}v^{2n}\cos(2uv) dv$
Hence, $\phi''_{n}(u)=-2.4\int_{0}^{\infty} e^{-v^2}v^{2(n+1)} \cos(2uv)dv =(-4)\phi_{n+1}{u}$
Using this we get, $\phi_{n}(u)=2.\frac{(-1)^n}{2^{2n}}\frac{d^{2n}}{du^{2n}}\phi_0(n)$
Now, using contour integral we can find $\phi_{0}(u)=\int_{0}^{\infty} e^{-v^2} \cos(2uv) dv=\frac{\sqrt{\pi}}{2}e^{-u^2}$.
From Rodrigue's formula of Hermite polynomial and the previous expression we get
$\phi_n(u)=\sqrt{\pi}\frac{e^{-u^2}}{2^{2n}}H_{2n}(u)$
$\Rightarrow f_{n}(x=u^2)=\sqrt{\pi}e^{\frac{-u^2}{2}}2^{-2n}H_{2n}(u)$.
Now, we have an asymptotic of Hermite polynomial like this,
$e^{\frac{-x^2}{2}}H_n(x) ≈\frac{2^n}{\sqrt{\pi}}\Gamma(\frac{n+1}{2})\cos(x\sqrt{2n}-\frac{n\pi}{2})(1-\frac{x^2}{2n+1})^{-1/4}$
https://en.m.wikipedia.org/wiki/Hermite_polynomials
We get, $f_n(x)≈\frac{n!}{\sqrt{n}}\cos(\sqrt{4xn}-\frac{2n\pi}{2})(1-\frac{x}{4n+1})^{-1/4}$.