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

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

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

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

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

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