Gaussian Integral using single integration So the Gaussian integral basically states that:
$$ I = \int_{-\infty}^{\infty} e^{-x^2} \ dx =\sqrt{\pi}$$
So the way to solve this is by converting to polar co-ordinates and doing a double integration.
Since I haven’t learnt double integration, I have searched a lot to solve this kind of integral using single integration. But to no avail. I have even tried it myself a couple of times but have been unsuccessful.
So here’s my question, is it possible to integrate the above using only single integration and if so how? If this is not possible to integrate using single integration then what is the reason behind it.
 A: A famous idea is to approximate the integral $\int_{0}^{+\infty}e^{-x^2}\,dx$ with $\int_{0}^{+\infty}\frac{dt}{(1+t^2/n)^n}$ and $\int_{0}^{\sqrt{n}}(1-s^2/n)^n\,ds$, with $n\in\mathbb{N}$ tending to $+\infty$. These integrals are elementary: they can be computed through the substitutions $t=\sqrt{n}\tan\theta$, $s=\sqrt{n}\sin\varphi$ and repeated integration by parts.
The outcome is the double bound
$$ L(n)=\frac{\sqrt{n}4^n}{\binom{2n}{n}(2n+1)}\leq \int_{0}^{+\infty}e^{-x^2}\,dx \leq \frac{\pi n \sqrt{n}\binom{2n}{n}}{(2n-1) 4^n}=R(n)\tag{1}$$
holding for any $n\geq 1$. The statement
$$ \lim_{n\to +\infty}\frac{R(n)}{L(n)}=1\tag{2} $$
is equivalent to Wallis' product and the statement
$$ \lim_{n\to +\infty} R(n)L(n)=\frac{\pi}{4}\tag{3} $$
is trivial. By squeezing it follows that $\int_{0}^{+\infty}e^{-x^2}\,dx=\frac{\sqrt{\pi}}{2}$ and $\int_{-\infty}^{+\infty}e^{-x^2}\,dx = \sqrt{\pi}$.

Truth to be told, this is a bit of a fraud, too. We are not really avoiding the $\Gamma$ function, we are just relying on the reflection/duplication formulas for $\Gamma$ without making an explicit mention of $\Gamma$. Wallis' product itself is an instance of $\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin(\pi s)}$.
On the other hand, why to avoid the $\Gamma$ function? To say the least, it plays a major role in many relevant probability distributions, and as one of my menthors (C.Viola) likes to say, "a good mathematician or phycisist should not be afraid to manipulate the $\Gamma$ function or the sine function, also because they are not that different. The sooner one gets introduced to $\Gamma$, the better".

A (very!) shortened approach is to notice that $\Gamma\left(\frac{1}{2}\right)$ is the value we are looking for and $\Gamma\left(\frac{1}{2}\right)^2$ is the area of the unit circle.
A: Here's a proof using the gamma function.
The function $$e^{-x^2}=e^{-(-x)^2}$$ is even, so $$\int_{-\infty}^\infty e^{-x^2}\,dx=2\int_0^\infty e^{-x^2}\,dx$$ Let $t=x^2\implies\dfrac{dx}{dt}=\dfrac12t^{-1/2}$. Hence $$\int_{-\infty}^\infty e^{-x^2}\,dx=2\int_0^\infty e^{-t}\cdot\frac12t^{-1/2}\,dt=\int_0^\infty e^{-t}t^{-1/2}\,dt=\Gamma\left(\frac12\right)=\sqrt\pi$$
