Infuriating integral

The area below the curve $f(x) = e^{-x^2}$ is given by

$\begin{aligned} A & = \int_{-\infty}^\infty e^{-x^2} dx & \small \color{#3D99F6} \text{Since the integrand is even} \\ & = 2 \int_0^\infty e^{-x^2} dx & \small \color{#3D99F6} \text{Error function }\text{erf }(z) = \frac 2{\sqrt \pi} \int_0^z e^{-t^2} dt \\ & = \sqrt \pi \text{ erf }(\infty) & \small \color{#3D99F6} \text{and }\text{erf }(\infty) = 1 \\ & = \sqrt \pi \end{aligned}$

Therefore, $A^2 = \pi \approx \boxed{3.141593}$ .

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Reference: Error function

The antiderivative from negative infinity to infinity of e^-x^2 is (by definition) equal to sqrt(pi) erf(x). And the square of the limit of sqrt(pi) erf(x) as x—> infinity is pi.

(erf(x) is defined to be sqrt(pi)/2 x integral(0, x, e^-(x^2)))