Limits of Gaussian error function

Question

Show that $$\lim_{x\to\infty} \int_{1}^x e^{-y^2} dy$$ exist and is in $[e^{-4},1]$

I cannot find a good minoration, I must show it whitout using the double integration, I totally don't know how to solve it

Answer 1

Existence: Note that $e^{-x^2} \le e^{-x}$ for all $x \in [1,\infty)$

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Bounds: $$\begin{align} \int_1^x e^{-t^2} dt &=\int_1^3 e^{-t^2} dt +\int_3^\infty e^{-t^2} dt\\ &<\int_1^3 e^{-t^2} dt +\int_3^\infty e^{-t} dt \end{align}$$ Estimate the finite integral in your favorite manner, and the second integral is trivial. For tighter bounds replace $3$ with any larger real number.

Answer 2

I have a funny answer, if you find Fourier transform without integrating fun. :D

Note that $e^{-y^2}$ is monotonically decreasing as $|y|\rightarrow\infty$ and that the function is always positive. That means \begin{equation} 0<\int_{1}^{\infty}e^{-y^2}dy<\int_{0}^{\infty}e^{-y^2}dy. \end{equation} If we find the value of the right integral and find that is indeed smaller than 1 and if we find a piece of area larger than $e^{-4}$ we are done. So let's do it!

I will give the function a name: $f:\mathbb{R}\rightarrow\mathbb{R},\quad f(y)=e^{-y^2/2}$.

First note that the function is even $f(-y)=f(y)$, so $$\ \int_{\mathbb{R}}f(y)dy=2\ \int_{0}^{\infty}f(y)dy.$$ With the Fourier transform (=FT) of it (which exists because it is a function of Schwarz space) $$\tilde{f}(k)=F(f)(k)=a\int_{\mathbb{R}}f(y)e^{-iky}dy$$ and the inverse Fourier transform of it $$f(x)=F^{-1}\left(\tilde{f}\right)(x)=a\int_{\mathbb{R}}\tilde{f}(k)e^{iky}dk$$ where $a=\frac{1}{\sqrt{2\pi}}$. OK, let's note that $$f'(y)=-yf(y).$$ We know that $\tilde{f}(k)$ exists. So let me calculate, weirdly enough its derivative $$ \frac{d}{dk}\tilde{f}(k)=\frac{d}{dk}a\int_{\mathbb{R}}f(y)e^{-iky}dy=a\int_{\mathbb{R}}\frac{\partial}{\partial k}f(y)e^{-iky}dy=a\int_{\mathbb{R}}f(y)e^{-iky}(-iy)dy=(*). $$ From the differential equation above we substitute to get $f'(y)$ in the integrand:$$ (*)=ai\int_{\mathbb{R}}f'(y)e^{-iky}dy. $$ Now with the product rule/parital integration we get$$ (*)=ai\left(f(y)e^{-iky}\vert_{-\infty}^{\infty}-\int_{\mathbb{R}} f(y)e^{-iky}(-ik)dy\right)=-k\tilde{f}(k)$$ There! The FT of $f$ satisfies the same linear differential equation as $f$. So we can conclude that $f=\tilde{f}$. I.e. $$\tilde{f}(k)=e^{-k^2/2}$$ If you are pedantic, you would say the integration constant of the solution to $$\tilde{f}'(k)=-k\tilde{f}(k)$$ has nothing to do with the FT. That's true but with the inverse FT, you will see the constant is 1. Luckily, I am not pedantic.

Moving on, let's calculate the integral using $u=\sqrt{2}y$ substitution $$\int_{0}^{\infty}e^{-y^2}dy=\frac{1}{2}\int_{\mathbb{R}}e^{-y^2}dy=\frac{1}{2}\int_{\mathbb{R}}e^{-u^2/2}\frac{1}{\sqrt{2}}du=\frac{\sqrt{2\pi}}{\sqrt{2\pi}}\frac{1}{2}\int_{\mathbb{R}}e^{-u^2/2}\frac{1}{\sqrt{2}}e^{-i0u}du=\frac{\sqrt{\pi}}{2}\tilde{f}(0)=\frac{\sqrt{\pi}}{2}<1. $$ Now let's vertify the lower bound is still $\geq e^{-4}$: The rectangle with length 1 and height $e^{-4}$ fits underneath the graph of $e^{-y^2}$ for $y\in[1,2],$ so $$ e^{-4}<\int_{1}^{\infty}e^{-y^2}dy<1. $$