The evaluation of the Gaussian integral goes:
$$ \int_{-\infty}^\infty e^{-x^2} dx = I$$
$$ \int_{-\infty}^\infty e^{-y^2} \left(\int_{-\infty}^\infty e^{-x^2}dx \right) dy = I^2 $$
$$ \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)}dx dy = I^2 $$
Switching into polar coordinates,
$$\int_{\theta=0}^{2\pi} \int_{r=0}^\infty e^{-r^2} \color{red}{r dr d\theta} = I^2 $$
$$ (2\pi) \int_0^\infty e^{-r^2}r dr = I^2 $$
$$ (2\pi) \left[\frac{e^{-r^2}}{-2}\right]_0^\infty = I^2 $$
$$ -\pi \left[0 - 1\right] = I^2 $$
$$ \pi = I^2$$
Therefore
$$\boxed{\int_{-\infty}^\infty e^{-x^2} dx = \sqrt \pi}$$
I do not quite understand why $dx dy = (r) dr d\theta$ after a long derivation, but not simply $dx dy = (r \cos^2 \theta )dr d\theta$, since
$$x = r \cos\theta$$ $$y = r \sin\theta$$
Then
$$dx = \cos \theta dr$$ $$dy = r \cos \theta d\theta$$
So
$$dx dy = (r \cos^2 \theta) dr d\theta$$
?
Variations:
$$\int_{-\infty}^\infty e^{-2bx^2} dx= \sqrt{\frac{\pi}{2b}} $$ $$\frac{d}{db}\int_{-\infty}^\infty e^{-2bx^2} dx= \frac{d}{db}\sqrt{\frac{\pi}{2}} \left( b^{-\frac{1}{2}}\right)$$ $$\int_{-\infty}^\infty \frac{d}{db} e^{-2bx^2} dx= \sqrt{\frac{\pi}{2}}\left( -\frac{1}{2}\right) \left( b^{-\frac{3}{2}}\right) $$ $$\int_{-\infty}^\infty -2x^2 e^{-2bx^2} dx= \left( -\frac{1}{2}\right)\sqrt{\frac{\pi}{2}} \left( b^{-\frac{3}{2}}\right) $$ $$b \int_{-\infty}^\infty -2x^2 e^{-2bx^2} dx= b \left( -\frac{1}{2}\right)\sqrt{\frac{\pi}{2}} \left( b^{-\frac{3}{2}}\right) $$
Therefore
$$ \boxed{\int_{-\infty}^\infty -2bx^2 e^{-2bx^2} dx = \left( -\frac{1}{2}\right)\sqrt{\frac{\pi}{2b}}}$$
