Editing Gaussian integral (section)

===By polar coordinates===
A standard way to compute the Gaussian integral, the idea of which goes back to Poisson,<ref name="york.ac.uk">{{cite web |title=The Probability Integral | last=Lee | first=Peter M. |url=https://www.york.ac.uk/depts/maths/histstat/normal_history.pdf }}</ref> is to make use of the property that:

<math display="block">\left(\int_{-\infty}^{\infty} e^{-x^2}\,dx\right)^2 = \int_{-\infty}^{\infty} e^{-x^2}\,dx \int_{-\infty}^{\infty} e^{-y^2}\,dy = \int_{-\infty}^{\infty}  \int_{-\infty}^{\infty} e^{-\left(x^2+y^2\right)}\, dx\,dy. </math>

Consider the function <math>e^{-\left(x^2 + y^2\right)} = e^{-r^{2}}</math>on the plane <math>\mathbb{R}^2</math>, and compute its integral two ways:
# on the one hand, by [[double integration]] in the [[Cartesian coordinate system]], its integral is a square: <math display="block">\left(\int e^{-x^2}\,dx\right)^2;</math>
# on the other hand, by [[shell integration]] (a case of double integration in [[polar coordinates]]), its integral is computed to be <math>\pi</math>

Comparing these two computations yields the integral, though one should take care about the [[improper integral]]s involved.

<math display="block">\begin{align}
   \iint_{\R^2} e^{-\left(x^2 + y^2\right)}dx\,dy
   &= \int_0^{2\pi} \int_0^{\infty} e^{-r^2}r\,dr\,d\theta\\[6pt]
   &= 2\pi \int_0^\infty re^{-r^2}\,dr\\[6pt]
   &= 2\pi \int_{-\infty}^0 \tfrac{1}{2} e^s\,ds && s = -r^2\\[6pt]
   &= \pi \int_{-\infty}^0 e^s\,ds \\[6pt]
   &= \pi \, \left[ e^s\right]_{-\infty}^{0} \\[6pt]
   &= \pi \,\left(e^0 - e^{-\infty}\right) \\[6pt]
   &= \pi \,\left(1 - 0\right) \\[6pt]
   &=\pi,
 \end{align}</math>
where the factor of {{mvar|r}} is the [[Jacobian determinant]] which appears because of the [[list of canonical coordinate transformations|transform to polar coordinates]] ({{math|''r'' ''dr'' ''dθ''}} is the standard measure on the plane, expressed in polar coordinates [[Wikibooks:Calculus/Polar Integration#Generalization]]), and the substitution involves taking {{math|1=''s'' = −''r''<sup>2</sup>}}, so {{math|1=''ds'' = −2''r'' ''dr''}}.

Combining these yields
<math display="block">\left ( \int_{-\infty}^\infty e^{-x^2}\,dx \right )^2=\pi,</math>
so
<math display="block">\int_{-\infty}^\infty e^{-x^2} \, dx = \sqrt{\pi}.</math>

====Complete proof====
To justify the improper double integrals and equating the two expressions, we begin with an approximating function:
<math display="block">I(a) = \int_{-a}^a e^{-x^2}dx.</math>

If the integral
<math display="block">\int_{-\infty}^\infty e^{-x^2} \, dx</math>
were [[absolutely convergent]] we would have that its [[Cauchy principal value]], that is, the limit
<math display="block">\lim_{a\to\infty} I(a) </math>
would coincide with
<math display="block">\int_{-\infty}^\infty e^{-x^2}\,dx.</math>
To see that this is the case, consider that

<math display="block">\int_{-\infty}^\infty \left|e^{-x^2}\right| dx < \int_{-\infty}^{-1} -x e^{-x^2}\, dx + \int_{-1}^1 e^{-x^2}\, dx+ \int_{1}^{\infty} x e^{-x^2}\, dx < \infty .</math>

So we can compute
<math display="block">\int_{-\infty}^\infty e^{-x^2} \, dx</math>
by just taking the limit
<math display="block">\lim_{a\to\infty} I(a).</math>

Taking the square of <math>I(a)</math> yields

<math display="block">\begin{align}
I(a)^2 & = \left ( \int_{-a}^a e^{-x^2}\, dx \right ) \left ( \int_{-a}^a e^{-y^2}\, dy \right ) \\[6pt]
& = \int_{-a}^a \left ( \int_{-a}^a e^{-y^2}\, dy \right )\,e^{-x^2}\, dx \\[6pt]
& = \int_{-a}^a \int_{-a}^a e^{-\left(x^2+y^2\right)}\,dy\,dx.
\end{align}</math>

Using [[Fubini's theorem]], the above double integral can be seen as an area integral
<math display="block">\iint_{[-a, a] \times [-a, a]} e^{-\left(x^2+y^2\right)}\,d(x,y),</math>
taken over a square with vertices {{math|{(−''a'', ''a''), (''a'', ''a''), (''a'', −''a''), (−''a'', −''a'')}<nowiki/>}} on the ''xy''-[[Cartesian plane|plane]].

Since the exponential function is greater than 0 for all real numbers, it then follows that the integral taken over the square's [[incircle]] must be less than <math>I(a)^2</math>, and similarly the integral taken over the square's [[circumcircle]] must be greater than <math>I(a)^2</math>. The integrals over the two disks can easily be computed by switching from Cartesian coordinates to [[list of canonical coordinate transformations|polar coordinates]]:

<math display="block">\begin{align}
x &= r \cos \theta, &
y &= r \sin\theta
\end{align}</math>
<math display="block">
\mathbf J(r, \theta) = 
\begin{bmatrix}
  \dfrac{\partial x}{\partial r} & \dfrac{\partial x}{\partial\theta}\\[1em]
  \dfrac{\partial y}{\partial r} & \dfrac{\partial y}{\partial\theta} \end{bmatrix}
= \begin{bmatrix}
  \cos\theta &           -  r\sin \theta \\
  \sin\theta & \hphantom{-} r\cos \theta
\end{bmatrix}
</math>
<math display="block">d(x,y) = \left|J(r, \theta)\right| d(r,\theta) = r\, d(r,\theta).</math>
<math display="block">\int_0^{2\pi} \int_0^a re^{-r^2} \, dr \, d\theta < I^2(a) < \int_0^{2\pi} \int_0^{a\sqrt{2}} re^{-r^2} \, dr\, d\theta.</math>

(See [[list of canonical coordinate transformations|to polar coordinates from Cartesian coordinates]] for help with polar transformation.)

Integrating,
<math display="block">\pi \left(1-e^{-a^2}\right) <  I^2(a) < \pi \left(1 - e^{-2a^2}\right). </math>

By the [[squeeze theorem]], this gives the Gaussian integral
<math display="block">\int_{-\infty}^\infty e^{-x^2}\, dx = \sqrt{\pi}.</math>