Inequality with Gamma. 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 $\Psi(0)=c$.
 A: 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.
A: Assuming $n >0$
$$f(x)=e^{\frac{x}{2}}\int^\infty_0e^{-t^2}t^{2n}\cos(2xt)\,dt$$
$$f(x)=\frac{1}{2} e^{\frac{x}{2}}\, \Gamma \left(n+\frac{1}{2}\right) \,
   _1F_1\left(n+\frac{1}{2};\frac{1}{2};-x^2\right)$$ So, it suffices to show that $\forall  x\in\Bbb R$
$$g(x)= e^{\frac{x}{2}}\,  \,
   _1F_1\left(n+\frac{1}{2};\frac{1}{2};-x^2\right) < 2$$ If you work a little the function $g(x)$ you should notice that its maximum value is  close to $1$ (we already have $g(0)=1$).
A: According to 22.10.15 in [1], we have
$$H_{2n}(x) = (-1)^n\mathrm{e}^{x^2}\frac{2^{2n+1}}{\sqrt{\pi}}\int_0^\infty
\mathrm{e}^{-t^2}t^{2n} \cos (2xt) \mathrm{d} t$$
where $H_{n}(x)$ is the Hermite polynomial satisfying
$$\int_{-\infty}^\infty \mathrm{e}^{-x^2}H_n(x) H_m(x) \mathrm{d} x = \sqrt{\pi} 2^n n!\delta_{nm}, \
n, m = 0, 1, 2, \cdots$$
Thus, we have
$$\mathrm{e}^{x^2/2}\int_0^\infty \mathrm{e}^{-t^2}t^{2n} \cos (2xt) \mathrm{d} t
= \frac{1}{2} \mathrm{e}^{-x^2/2}\Gamma(n + 1/2) (-1)^n \frac{n!}{(2n)!} H_{2n}(x)$$
where we have made use of $\Gamma(n + \tfrac{1}{2}) = \frac{(2n)!}{n!} \frac{\sqrt{\pi}}{2^{2n}}$.
The question is: Is there absolute constant $C$ such that
$$\mathrm{e}^{-x^2/2} \frac{n!}{(2n)!} |H_{2n}(x)| \le C ?$$
Theorem 1 in [2] says the answer is NO: It holds that
$$\max_{x> 0} \mathrm{e}^{-x^2/2} \frac{n!}{(2n)!} |H_{2n}(x)|
> C_1 n^{1/6}$$
for some absolute constant $C_1$.
The same estimate is given in Theorem 1 in [3] where
$p_n(x) = (\sqrt{\pi}\, 2^n n!)^{-1/2}H_n(x)$.
Reference
[1] M. Abramowitz and I. A. Stegun (Eds.), “Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables”, 1964.
[2] Ilia Krasikov, “New Bounds On The Hermite Polynomials”, https://arxiv.org/pdf/math/0401310.pdf
[3] "Estimates of the Hermite and the Freud polynomials", Journal of Approximation Theory,
Volume 63, Issue 2, November 1990, Pages 210-224.
