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 $\phi_(0)=\int_{0}^{\infty} e^{-v^2} \cos(2uv) dv=\frac{\sqrt{\pi}}{2}e^{-u^2}$.
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, $f_n(x)≈\frac{n!}{\sqrt{n}}\cos(\sqrt{4xn}-\frac{2n\pi}{2})(1-\frac{x}{4n+1})^{-1/4}$.