## Solving The Gaussian Integral

The Gaussian function is an extremely important function in statistics, forming the baseline for the normal distribution and the central limit theorem. This means that finding the area under this function can be very important.

Unfortunately for us, the Gaussian function doesn’t have an antiderivative, making it impossible to solve using any normal integration method. It is therefore necessary to make some pretty unusual choices to find this elusive answer.

To begin, it is important to see what we are working with, so let’s take a look at the Gaussian function and integral:

f(x)=e^{-x^2} Gussian Function

\int_{-\infty}^{\infty} e^{-x^2} dx Gaussian Integral

The Guassian function forms a typical “bell shape” when graphed, with the area under this graph being the value of the Gaussian integral.

To begin, we are going to set the value of the Gaussian integral to be I, the reason for this will become clearer as we progress.

To find the value for I , we will first need to find the value of I^2 , noting that this is just the integral being multiplied by itself.

What we do next, is change the variable for the second integral to y (note that this will not change the value for the answer). This is then followed by manipulating the expression to get the following integral:

I^2= \int_{-\infty}^{\infty} e^{-x^2} dx \int_{-\infty}^{\infty} e^{-x^2} dx \\   \\   \\  I^2= \int_{-\infty}^{\infty} e^{-x^2} dx \int_{-\infty}^{\infty} e^{-y^2} dy
\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2)} dydx

What this actually is, is the volume under the graph of z=e^{-(x^2+y^2)} , which looks like a 3D bell curve with rotational symmetry:

The aspect of rotational symmetry is actually very important, as this means that we can do a transformation of coordinates to the polar world. There are a few important notes to make when doing this transformation, as it is not as simple as subbing in variables:

1) The limits of integration will change, as \theta can only vary from 0 to 2\pi and r can only vary from  0 to  \infty

2) An extra factor of  r must be multiplied by the integrand. This is the Jacobian of the transformation, and this factor takes into account any scaling issues that occur.

\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2)} dydx = \int_{0}^{2\pi} \int_{0}^{\infty} re^{-r^2} drd\theta

This is now a double integral that can be solved from the inside out using u-substitution, letting u=-r^2 . The result that you get is:

I^2=\int_{0}^{2\pi} \int_{0}^{\infty} re^{-r^2} drd\theta = \pi
I=\int_{-\infty}^{\infty} e^{-x^2} dx =\sqrt{\pi}

This is an unbelievably amazing result, as Pi shows up out of seemingly nowhere. This example shows the extreme power that a transformation of coordinates holds, allowing an impossible-to-solve integral to become a relatively easy one to solve.