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

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.

Answer 1

Making the normalised change of variables $t = ns^2$, $x = 4ny^2$ (with $y \geq 0$ and $s$ of either sign) one can write $$ f_n(x) = e^{2ny^2} n^n \sqrt{n} \int_{{\bf R}} e^{-n\phi(s)}\ ds$$ where the phase $\phi(s)$ is given by $$ \phi(s) := s^2 - 2 \log s - 4iy s$$ using the standard branch of the complex logarithm. This phase has stationary points at $iy \pm \sqrt{1-y^2}$. In the "Bessel" regime $y < 1-\delta$ for some fixed $\delta>0$, the stationary points are non-degenerate, and $\mathrm{Re} \phi(s)$ attains a local minimum of $1 + 2y^2$ at these points (with respect to a horizontal contour), so one gets the required bound $f_n(x) \lesssim n^n/e^n$ (equivalent to the claimed bound by Stirling's formula) in this case by the usual saddle point method. Similarly, in the "exponential" regime $y > 1 + \delta$, the stationary points are again non-degenerate; if one makes the convenient substitution $y = \cosh \theta$ with $\theta>0$ then at the stationary point $i e^{\theta}$, $\mathrm{Re} \phi(s)$ attains a local minimum (again wrt a horizontal contour) of $1 + 2y^2 + (\sinh 2\theta - 2\theta) > 1 + 2y^2$, and stationary phase again gives the desired bound (with room to spare). However in the "Airy" regime in which $y$ is close to $1$ (or in your original coordinates, $x$ is close to $4n$) the stationary points coalesce and I don't think your claimed bounds actually hold (one can lose an additional factor of $n^{1/6}$, I think, and this is consistent with the standard Airy asymptotics given for instance here). You may want to double check your reference for the Laguerre polynomial bounds; they may not hold in the Airy regime $x \approx 4n$. For instance, the bound in Theorem 8.91.2 of Szego's book only gives a decay bound of $n^{-1/3}$ instead of $n^{-1/2}$ once one works in a range of parameters that includes the Airy regime.

Answer 2

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, $f_n(x)≈\frac{n!}{\sqrt{n}}\cos(\sqrt{4xn}-\frac{2n\pi}{2})(1-\frac{x}{4n+1})^{-1/4}$.