The derivation

We are interested in the integral

$$I := \int_0^\infty {\rm e}^{-x^2} {\rm d}x{\rm .}$$

In order to apply the above formula, we need to transform the integrand. We do this by substituting $s := x^2$. Then,

$$I = \frac12 \int_0^\infty \frac{{\rm e}^{-s}}{\sqrt{s}} {\rm d}s{\rm .}$$

Now take

$$h(s) := \frac1{\sqrt{s}}$$

and find the inverse Laplace transform. We note that

$$\mathcal{L}\left[t^{-\frac12}\right](s) = \frac{\Gamma\left(... ...2\right)}{s^\frac12} = \Gamma\left(\frac12\right) h(s) {\rm .}$$

It turns out that the constant, $\Gamma(1/2)$, works out nicely:

$$\Gamma\left(\frac12\right) = \int_0^\infty {\rm e}^{-y} y^{-\frac12} {\rm d}y = 2I{\rm .}$$

We now use the linearity of the Laplace transform to obtain

$$h(s) = \mathcal{L}\left[\frac{t^{-\frac 12}}{2 I}\right](s){\rm ,}$$

so that

$$\mathcal{L}^{-1}[h(s)](t) = \frac1{2 I\sqrt{t}} {\rm .}$$

Now substitute into the formula obtained in the beginning, with $g(s) := {\rm e}^{-s}$:

$$\begin{eqnarray*} \int_0^\infty g(s) h(s) {\rm d}s & = & \int_0^\infty \frac{{\... ... e}^{-st} {\rm e}^{-s}}{2I\sqrt{t}} {\rm d}s {\rm d}t {\rm .} \end{eqnarray*}$$

Now rearrange this to obtain

$$4I^2 = \int_0^\infty \int_0^\infty \frac{{\rm e}^{-s(t + 1)}... ...rm d}t = \int_0^\infty \frac{{\rm d}t}{\sqrt{t}(t + 1)}{\rm .}$$

To work out the final integral, substitute $t := u^2$:

$$\int_0^\infty \frac{{\rm d}t}{\sqrt{t}(t + 1)} = \int_0^\inf... ...c{{\rm d}u}{1 + u^2} = 2 \left(\frac\pi2\right) = \pi {\rm .}$$

Finally,

$$4I^2 = \pi$$

so that

$$I = \int_0^\infty {\rm e}^{-t^2} {\rm d}t = \frac{\sqrt\pi}2{\rm .}$$