But hey, $x$ is just a dummy variable. If we change it to $y$ this integration will still be the same.

\begin{align} I=\int_{-\infty}^{\infty}e^{-my^2}dy \label{eq:2} \end{align} Just for fun, let's multiply $\eqref{eq:1}$ and $\eqref{eq:2}$ \begin{align} I^2=\int_{-\infty}^{\infty}e^{-mx^2}dx \int_{-\infty}^{\infty}e^{-my^2}dy \end{align} Using the rule of integration we can write it as \begin{align} I^2=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}e^{-m(x^2+y^2)}dx dy \end{align} This equation looks like something that could be evaluated in polar coordinates. That's what we are going to do next.... We need the integrating factor to Jacobian $r$. We know the regular old polar coordinate statement is $$r^2=x^2+y^2$$ So, let's go ahead and convert this integral into polar coordinates. The limit for $r$ will be $(0,\infty)$ but the limit for $\theta$ will be $(0,2\pi)$

So \begin{align} I^2=\int_{0}^{2\pi} \int_{0}^{\infty}re^{-mr^2}dr d\theta \end{align} Look, the term $re^{-mr^2}$ has no independence on $\theta$. So we get \begin{align} I^2=2\pi \int_{0}^{\infty}re^{-mr^2}dr \end{align} Hmm, looks better. May be we can try using substitution method this time..... Let $$u=r^2$$ \begin{align*} \dfrac{du}{dr}&=2r\\ \Rightarrow dr &=\dfrac{du}{2r} \end{align*} And the limit will still be $(0,\infty)$. So \begin{align*} I^2 &=2\pi \int_{0}^{\infty}re^{-mu} \dfrac{du}{2r}\\ &=\pi \int_{0}^{\infty}e^{-mu} du\\ \end{align*} Looks awesome. Now this integral is really easy to evaluate, isn't it? Let's evaluate it..... \begin{align*} I^2 &=\pi \dfrac{1}{m}(-e^{-mu}\bigg|_0^{\infty})\\ \Rightarrow I^2 &=\dfrac{\pi}{m}\\ \Rightarrow I &=\sqrt{\dfrac{\pi}{m}} \end{align*} And that's how you evaluate the Gaussian integral.