Editing Gaussian integral (section)

===By Cartesian coordinates===
A different technique, which goes back to Laplace (1812),<ref name="york.ac.uk" /> is the following. Let
<math display="block">\begin{align}
y & = xs \\
dy & = x\,ds.
\end{align}</math>

Since the limits on {{mvar|s}} as {{math|''y'' → ±∞}} depend on the sign of {{mvar|x}}, it simplifies the calculation to use the fact that {{math|''e''<sup>−''x''<sup>2</sup></sup>}} is an [[even function]], and, therefore, the integral over all real numbers is just twice the integral from zero to infinity. That is,

<math display="block">\int_{-\infty}^{\infty} e^{-x^2} \, dx = 2\int_{0}^{\infty} e^{-x^2}\,dx.</math>

Thus, over the range of integration, {{math|''x'' ≥ 0}}, and the variables {{mvar|y}} and {{mvar|s}} have the same limits. This yields:
<math display="block">\begin{align}
I^2 &= 4 \int_0^\infty \int_0^\infty e^{-\left(x^2 + y^2\right)} dy\,dx \\[6pt]
&= 4 \int_0^\infty \left( \int_0^\infty e^{-\left(x^2 + y^2\right)} \, dy \right) \, dx \\[6pt]
&= 4 \int_0^\infty \left( \int_0^\infty e^{-x^2\left(1+s^2\right)} x\,ds \right) \, dx \\[6pt]
\end{align}</math>
Then, using [[Fubini's theorem]] to switch the [[order of integration (calculus)|order of integration]]:
<math display="block">\begin{align}
I^2 &= 4 \int_0^\infty \left( \int_0^\infty e^{-x^2\left(1 + s^2\right)} x \, dx \right) \, ds  \\[6pt]
&= 4 \int_0^\infty \left[ \frac{e^{-x^2\left(1+s^2\right)} }{-2 \left(1+s^2\right)} \right]_{x=0}^{x=\infty} \, ds \\[6pt]
&= 4 \left (\frac{1}{2} \int_0^\infty \frac{ds}{1+s^2} \right) \\[6pt]
&= 2 \arctan(s)\Big |_0^\infty \\[6pt]
&= \pi.
\end{align}</math>

Therefore, <math>I = \sqrt{\pi}</math>, as expected.