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The derivation

We are interested in the integral

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

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

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

Now take

\begin{displaymath}
h(s) := \frac1{\sqrt{s}}
\end{displaymath}

and find the inverse Laplace transform. We note that

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

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

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

We now use the linearity of the Laplace transform to obtain

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

so that

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

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

\begin{displaymath}
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 .}
\end{displaymath}

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

\begin{displaymath}
\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 .}
\end{displaymath}

Finally,

\begin{displaymath}
4I^2 = \pi
\end{displaymath}

so that

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


next up previous
Next: About this document ... Up: A derivation of the Previous: The transform of
Alexander Frolkin 2002-02-14