I was reading this article which derives the value for integral of a negative exponent.
I follow the derivation, from circular symmetry, to integration by substitution, and this is the resulting equation: $$(l(a))^2=\pi\int_0^\infty e^{(-ax)} \, dx=\frac \pi a\tag1$$ (where $l(a)$ is some function on $a$.)
And we take the $\sqrt{}$ of equation (1), which produces: $$l(a)=\int_{-\infty}^\infty e^{(-ax^2)} \, dx=\sqrt{\frac \pi a} \text{ for } \operatorname{Re}(a)>0\tag2$$
I understand that the bound became negative infinity because taking the square root gives negative solutions.
However, I don't understand why the exponential factor became $x^2$, and I have no idea where the $\pi$ from equation (1) went. Why did the integral eat the $\pi$?