Consider the function $$f(x)=e^{\frac{x^2}{2}}\int^\infty_0e^{-t^2}t^{2n}\cos(2xt)dt,\quad x\in\Bbb R$$
I find a lot difficultes to prove $$\forall (x,n)\in\Bbb R^+\times\Bbb N:|f(x)|\leq C\Gamma(n+\frac{1}{2})$$ for some const $C$. Thanks for any help.
My attempt:
Put $$f(x)=\frac{1}{\Gamma(n+\frac{1}{2})}e^{\frac{x}{2}}\int^\infty_0 e^{-t^2}t^{2n}\cos(2\sqrt{x}t)dt.$$
Let $x>0$, by the change of variable $t=\sqrt{n}s$ we get $$f(\frac{x}{n})=\frac{1}{\Gamma(n+\frac{1}{2})}n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^\infty_0 e^{-n\big(s^2-2\ln(s)\big)}\cos(2\sqrt{x}s)ds. $$ Now we apply Laplace's method wich says:\ \textbf{Theorem} Consider the integral $\int^b_a f(t)e^{-n g(t)}dt$ where $g$ is of class $\mathcal{C}^2$ on $[a,b[$ (here $a<b\leq+\infty)$ and $f$ is a continuous function on $[a,b[$ such that
- $f(a)\not=0$
- $\forall t\in[a,b[:g'(t)>0$
- $g'(a)=0$
- $\forall t\in[a,b[:g''(t)>0$
Then $$\int^b_a f(t)e^{-n g(t)}dt\sim\sqrt{\frac{\pi}{2g''(a)}}\frac{e^{-ng(a)}f(a)}{\sqrt{n}},\quad n\to+\infty.$$
So we can write $$f(\frac{x}{n})=\frac{1}{\Gamma(n+\frac{1}{2})}\Big(n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^0_{-1} e^{-n\big(s^2-2\ln(-s)\big)}\cos(2\sqrt{x}s)ds+n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^\infty_1 e^{-n\big(s^2-2\ln(s)\big)}\cos(2\sqrt{x}s)ds\Big).$$ Put $$I_1=n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^\infty_1 e^{-n\big(s^2-2\ln(s)\big)}\cos(2\sqrt{x}s)ds$$ with $g(s)=s^2-2\ln(s), \quad s\geq 1$ and $f(s)=\cos(2\sqrt{x}s)$, by the previous theorem we have $$I_1\sim \sqrt{\frac{\pi}{8}}n^{n}\sqrt{n}\frac{e^{-n}}{\sqrt{n}}\cos(2\sqrt{x})$$ (we have $g'(s)=2(s-\frac{1}{s})>0$ and $g''(s)=2(1+\frac{1}{s^2})>0$) Since $n!\sim(\frac{n}{e})^{n}\sqrt{2\pi n}$ we obtain $$I_1\sim\frac{n!}{2\sqrt{n}}\cos(2\sqrt{x})$$.
Using the same method we obtain $$I_2=n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^0_{-1} e^{-n\big(s^2-2\ln(-s)\big)}\cos(2\sqrt{x}s)ds\sim\frac{n!}{2\sqrt{n}}\cos(2\sqrt{x})$$
Now by using the well-known result $\frac{n!}{\sqrt{n}}\sim \Gamma(n+\frac{1}{2})$
we get $$\forall x>0,\qquad f(\frac{x}{n})\sim \cos(2\sqrt{x})$$
So, $\forall x>0,\qquad\exists n_0\in\Bbb N,\qquad n\geq n_0\Rightarrow \Big|f(\frac{x}{n})\Big|\leq C$ for some constant $C>0$.
Therefore if we replace $x$ by $nx$ we obtain:\
$\forall x>0,\qquad\exists n_0\in\Bbb N,\qquad n\geq n_0\Rightarrow \Big|f(x)\Big|\leq C$
And hence $$\sup_{(n,x)\in \{n\in\Bbb N; n\geq n_0\}\times\Bbb R^*_+}\Big|f(x)\Big|<+\infty. $$
It remains the case $(n,x)\in \{n\in\Bbb N; n\leq n_0\}\times\Bbb R^*_+$ and it is easy because if $n\leq n_0$ we have
$$f(x):=\Big(\frac{\Gamma(n+1)}{\Gamma(n+\frac{1}{2})}\Big)e^{\frac{-x}{2}}L^{\frac{-1}{2}}_{n}(x)\leq C' e^{\frac{-x}{2}}P(x) $$ for some constant $C'>0$ independent of $n$ and $P$ is polynomial with $\text{deg}(P)=n\leq n_0$.
Since $\lim_{x\to+\infty}e^{\frac{-x}{2}}P(x)=0$ we obtain $$\sup_{(n,x)\in \{n\in\Bbb N; n\leq n_0\}\times\Bbb R^*_+}\Big|f(x)\Big|<+\infty. $$
Finally: $$\sup_{(n,x)\in\Bbb N\times\Bbb R_+}\Big|f(x)\Big|<+\infty. $$
The case $x=0$ is easy because $\Psi(0)=c$.
This is a possible start.
Let $t^2=y$. $2t dt=dy$ so $dt=dy/(2\sqrt{y})$.
The integral becomes $\int_0^{\infty} e^{-y}y^{n-1/2}\cos(2x\sqrt{y})dy/2$ which is bounded by in absolute value by $\int_0^{\infty} e^{-y}y^{n-1/2}dy/2 =\Gamma(n+1/2)/2$.
To get rid of the $e^{x/2}$ we would have to consider how the cos affects the integral.