Introduction

The Gauss integral shows up in statistical mechanics, in kinetic theory of gases, and in elementary statistics, and it is worth learning how to do it. Define the Gauss integral as

$$G = \int_{-\infty}^\infty e^{-\alpha x^2} dx$$ (1)

($\alpha$ is assumed constant) and recognize that I (which is G/2) is, by the evenness (is there such a word?) of the integrand

$$I = \int_0^\infty e^{-\alpha x^2} dx$$ (2)

sufficient, since if we can obtain I, we have G.

$$I = \int_0^\infty e^{-\alpha x^2} dx$$ (3)

$$I = \int_0^\infty e^{-\alpha y^2} dy$$ (4)

so the product of these two is

$$I^2 = \int_0^\infty e^{-\alpha x^2} dx \int_0^\infty e^{-\alpha y^2} dy$$ (5)

 

Figure 1: The Gauss integral.

which we rewrite as

$$I^2 = \int_0^\infty dx \left \{ \int_0^\infty dy \left [ e^{-\alpha x^2} e^{-\alpha y^2} \right ] \right \}$$ (6)

where we have used many theorems from our freshman calculus days.

Switching to polar coordinates we then have

$$I^2 = \int_0^\infty r dr \left \{ \int_0^{\pi/2} d\theta \left [ e^{-\alpha r^2} \right ] \right \}$$ (7)

which is

$$I^2 = \frac{\pi}{2}\int_0^\infty r dr \left [ e^{-\alpha r^2} \right ]$$ (8)

after integration over theta. This last integral is elementary. Let $\alpha r^2 = u$, so that $2 \alpha r dr = du$ and $r dr = \frac{du}{2 \alpha}$ so our remaining integral becomes

$$I^2 = \frac{\pi}{2} \frac{1}{2 \alpha}\int_0^\infty du \left [ e^{-u} \right ]$$ (9)

This means that

$$I^2 = \frac{\pi}{2} \frac{1}{2 \alpha} \left . \frac{e^{-u}}{-1} \right \vert _0^\infty$$ (10)

$$I^2 = \frac{\pi}{2} \frac{1}{2 \alpha} \left . \frac{e^{-\alpha r^2}}{-1} \right \vert _0^\infty = \frac{\pi} {4 \alpha}$$ (11)

so, taking the square root, we have

$$I = \sqrt{\frac{\pi} {4 \alpha}}$$ (12)

and since 2I = G, we have

$$I = \sqrt{\frac{\pi} {\alpha}}$$ (13)