# A Simple Evaluation of the Gaussian Integral

The Gaussian Integral is as follows:

$I = \int_0^\infty e^{-x^2} \:dx$

The value of I matters in statistics, for example, and it is necessarily to know its value when working with the Normal Distribution. The problem is that the usual way of evaluating it is too difficult for an A-Level student (or similar) to understand, at it changes coordinate systems in a double integral and so requires the use of the Jacobian. Even people who know about the Jacobian will probably have only seen it derived informally, and so it is best avoided.

What follows is a much simpler proof. It uses a double integral again, but only makes ordinary changes of variable. It is a common trick to evaluate a double integral in two different ways in order to prove something, but it is hard to come up with integrals where this actually works. This is one of those rare ones! The method is due to Hirokazu Iwasawa.

The integral to consider is as follows:

$A = \int_0^\infty \left(\int_0^\infty xe^{-x^2(y^2+1)}\:dx\right)\:dy$

Evaluting it the way suggested there, we get that:

$A = \int_0^\infty \left(\frac{1}{2(1+y^2)}\right)\:dy$

This can again by integrated using a usual trigometric substitution:

$A = \int_0^\infty \left(\frac{1}{2(1+y^2)}\right)\:dy = \left[\frac{1}{2}\tan^{-1}(y)\right]_0^\infty = \frac{\pi}{4}$

Now we swap the order of the integration, so that instead we evaluate the following:

$A = \int_0^\infty \left( \int_0^\infty xe^{-x^2(y^2+1)}\:dy \right)\:dx$

We can write this as:

$A = \int_0^\infty x e^{-x^2} \left( \int_0^\infty e^{-x^2y^2}\:dy \right)\:dx$

Using the substitution t = xy in the nested integral, we get the following:

$A = \int_0^\infty x e^{-x^2} \left( \frac{1}{x}\int_0^\infty e^{-t^2}\:dt \right)\:dx$

And we get:

$A = \int_0^\infty x e^{-x^2} \left( \frac{1}{x} I \right)\:dx = I^2$

Therefore we get the result:

$\int_0^\infty e^{-x^2}\: dx = \frac{\sqrt{\pi}}{2}$

So every step makes sense with A-Level mathematics, proving you convince yourself that it is valid to swap the order of integration, but I think that is rather obvious. I’ve tried to put in brackets and things to show that there is nothing weird going on with double integrals, and that they really are just embedded integrals.

Possibly the biggest problem with it is that it is not obvious why you would consider that integral in the first place. But who cares? Somebody has thought of it for you.