Asymptotic behavior of $\frac{\int_{-n}^{n}\sum_{k=1}^{n}e^{-nx^{2}}dx}{\int_{-n}^{n}e^{-x^{2}}dx}$.

Question

I want to find the asymptotic behavior of :

$$\frac{\int_{-n}^{n}ne^{-nx^{2}}dx}{\int_{-n}^{n}e^{-x^{2}}dx}$$

Doing my research with Desmos I find that as $n\to \infty$ :

$$\frac{\int_{-n}^{n}ne^{-nx^{2}}dx}{\sqrt{n}\int_{-n}^{n}e^{-x^{2}}dx}\to 1$$

It recall me a bit the Stirling Formula but I recognize I don't know where the square root comes from because it's purely empirical.

As attempt I try Fubini's theorem without success and with no issue or other hint .

As it seems true I have a only question :

How to (dis)prove it ?

Thanks.

Answer 1

Some formulas:

$$\int_{-x}^x x\mathrm e^{-xt^2}\mathrm dt=\sqrt{ \pi x}\operatorname{erf}(x^{3/2}) \\ \int_{-x}^x\mathrm e^{-t^2}\mathrm dt=\sqrt \pi\operatorname{erf}(x)$$ Hence $$\frac{\int_{-x}^x x\mathrm e^{-xt^2}\mathrm dt}{\int_{-x}^x\mathrm e^{-t^2}\mathrm dt}=\sqrt x\frac{\operatorname{erf}(x^{3/2})}{\operatorname{erf}(x)}$$ For large $z$, erf has the asymptotic expansion $$\operatorname{erf}z\asymp1-\pi^{-1/2}\mathrm e^{-z^2}\sum_{m=0}^\infty (-1)^m \left(\prod_{j=0}^{m-1}(1/2+j)\right)z^{-2m-1} \\ \text{as}~z\to\infty$$ Taking only the first term (and taking the empty product to be $1$), we get $$\operatorname{erf}z\approx 1-~ \frac{\mathrm e^{-z^2}}{\sqrt \pi~z} \\ \text{as}~z\to\infty$$ Hence, $$\sqrt{x}\frac{\operatorname{erf}(x^{3/2})}{\operatorname{erf}(x)}\approx \sqrt{x}\frac{1-\frac{\exp(-x^3)}{\sqrt{\pi}~ x^{3/2}}}{1-\frac{\exp(-x^2)}{\sqrt{\pi}~x}} \\ =\sqrt{x}~\underbrace{\frac{\sqrt{\pi}~x-\frac{1}{\sqrt x}\exp(-x^3)}{\sqrt{\pi}~x-\exp(-x^2)}}_{\approx 1}$$ Which clearly behaves like $\sqrt x$ for large $x$.

Answer 2

Substituting $\sqrt n x= y$ in the integral in the numerator gives $$ \frac{\int_{-n}^{n}ne^{-nx^{2}}dx}{\sqrt{n}\int_{-n}^{n}e^{-x^{2}}dx} = \frac{\int_{-n^{3/2}}^{n^{3/2}}e^{-y^{2}}dy}{\int_{-n}^{n}e^{-x^{2}}dx} $$ and now both numerator and denominator converge to $\int_{-\infty}^{\infty}e^{-x^{2}}dx$, so that the quotient converges to $1$, confirming your conjecture.

Remark: Here we only need that $\int_{-\infty}^{\infty}e^{-x^{2}}dx$ is finite. The actual value (which is $\sqrt\pi$) is not needed to compute this limit.