Proof inequality $\frac{\sqrt{\pi}}{2}\sqrt{1-e^{-a^2}} < \int_0^a e^{-x^2}dx < \frac{\sqrt{\pi}}{2}\sqrt{1-e^{-2a^2}}$

Question

I'm asked to prove the inequality

$$\frac{\sqrt{\pi}}{2}\sqrt{1-e^{-a^2}} < \int_0^a e^{-x^2}dx < \frac{\sqrt{\pi}}{2}\sqrt{1-e^{-2a^2}}$$

After playing around for a while I was able to find (without mistakes, hopefully) that the leftmost term in the inequality is equal to the square root of the following integral in polar coordinates:

$$\int_0^{\pi / 2}d\varphi \int_0^a e^{-r^2}rdr$$

and the rightmost equal to the square root of:

$$\int_0^{\pi / 2}d\varphi \int_0^{\sqrt{2}a} e^{-r^2}rdr$$

But I don't know how to proceed from this.

Any hint would be appreciated.

Answer 1

Short proof: a square is bounded between two quarter-circles.
Expanded proof: for any $a>0$ we have

$$ \left(\int_{0}^{a}e^{-x^2}\,dx\right)^2 = \iint_{[0,a]\times[0,a]} e^{-(x^2+y^2)}\,dx\,dy $$ and the RHS is greater than the integral over $\Gamma_{1}=\{(x,y):x,y>0, x^2+y^2\leq a^2\}$ and smaller than the integral over $\Gamma_{2}=\{(x,y):x,y>0, x^2+y^2\leq 2a^2\}$. These integrals can be easily computed by switching to polar coordinates. For instance $$ \iint_{\Gamma_1} e^{-(x^2+y^2)}\,dx\,dy = \int_{0}^{a}\int_{0}^{\pi/2}\rho e^{-\rho^2}\,d\theta\,d\rho=\left[-\frac{\pi}{4}e^{-\rho^2}\right]_{0}^{a}=\frac{\pi}{4}\left(1-e^{-a^2}\right).$$ The claim follows through a simple square root extraction.

Answer 2

Let \begin{equation} f(a) =\int_0^a e^{-x^2} \ dx - \frac{\sqrt{\pi}}{2}\sqrt{1-e^{-a^2}} \end{equation} then by Leibniz formula for integration (when deriving $\int_0^a e^{-x^2} \ dx $ with respect to $a$), we have \begin{equation} f'(a) = e^{-a^2} - \dfrac{\sqrt{{\pi}}a\mathrm{e}^{-a^2}}{2\sqrt{1-\mathrm{e}^{-a^2}}} = e^{-a^2}(1 - \frac{\sqrt{\pi}}{2} \frac{a}{\sqrt{1-e^{-a^2}}} ) \end{equation} Assuming $a > 0$ (Case where $a < 0$ is argued by symmetry), we can find $a_0$ for which $f(a_0) = 0$. We can therefore show that for all $a > a_0$ \begin{equation} 1 - \frac{\sqrt{\pi}}{2} \frac{a}{\sqrt{1-e^{-a^2}}} < 0 \end{equation} So $f(a) $ is decreasing on $[a_0,+\infty[$. Hence \begin{equation} f(a) > \lim_{a \rightarrow +\infty} f(a) \end{equation} Since $\int_0^{+\infty} e^{-x^2} = \frac{\sqrt{\pi}}{2}$ and $\lim_{a \rightarrow +\infty} \frac{\sqrt{\pi}}{2}\sqrt{1-e^{-a^2}} = \frac{\sqrt{\pi}}{2}$, you can show that \begin{equation} \lim_{a \rightarrow +\infty} f(a) = 0 \end{equation} Hence $f(a) > 0$ and so \begin{equation} \int_0^a e^{-x^2} \ dx > \frac{\sqrt{\pi}}{2}\sqrt{1-e^{-a^2}} \end{equation} The above is true only when $a \in [a_0,+\infty[$. But when $a \in ]0,a_0]$, $f(a)$ increasing hence the minimum is at $a = 0$, which is $f(a) = 0$, so also on $a \in ]0,a_0]$, we have $f(a) > 0$. Therefore, the above inequality is true for $a > 0$. Again, the case where $a < 0$ is argued by symmetry.

Can you do the other inequality ?