By polar coordinates
A standard way to compute the Gaussian integral, the idea of which goes back to Poisson,[3] is to make use of the property that:
Consider the function on the plane , and compute its integral two ways:
- on the one hand, by double integration in the Cartesian coordinate system, its integral is a square:
- on the other hand, by shell integration (a case of double integration in polar coordinates), its integral is computed to be
Comparing these two computations yields the integral, though one should take care about the improper integrals involved.
where the factor of r is the Jacobian determinant which appears because of the transform to polar coordinates (r dr dθ is the standard measure on the plane, expressed in polar coordinates Wikibooks:Calculus/Polar Integration#Generalization), and the substitution involves taking s = −r2, so ds = −2r dr.
Combining these yields
so
Complete proof
To justify the improper double integrals and equating the two expressions, we begin with an approximating function:
If the integral
were absolutely convergent we would have that its Cauchy principal value, that is, the limit
would coincide with
To see that this is the case, consider that
So we can compute
by just taking the limit
Taking the square of yields
Using Fubini's theorem, the above double integral can be seen as an area integral
taken over a square with vertices {(−a, a), (a, a), (a, −a), (−a, −a)} on the xy-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 , and similarly the integral taken over the square's circumcircle must be greater than . The integrals over the two disks can easily be computed by switching from Cartesian coordinates to polar coordinates:
(See to polar coordinates from Cartesian coordinates for help with polar transformation.)
Integrating,
By the squeeze theorem, this gives the Gaussian integral
By Cartesian coordinates
A different technique, which goes back to Laplace (1812),[3] is the following. Let
Since the limits on s as y → ±∞ depend on the sign of x, it simplifies the calculation to use the fact that e−x2 is an even function, and, therefore, the integral over all real numbers is just twice the integral from zero to infinity. That is,
Thus, over the range of integration, x ≥ 0, and the variables y and s have the same limits. This yields:
Then, using Fubini's theorem to switch the order of integration:
Therefore, , as expected.
In Laplace approximation, we deal only with up to second-order terms in Taylor expansion, so we consider .
In fact, since for all , we have the exact bounds:
Then we can do the bound at Laplace approximation limit:
That is,
By trigonometric substitution, we exactly compute those two bounds: and
By taking the square root of the Wallis formula,
we have , the desired lower bound limit. Similarly we can get the desired upper bound limit.
Conversely, if we first compute the integral with one of the other methods above, we would obtain a proof of the Wallis formula.