# 18 Gaussian Integral

The Gaussian Integral is defined by

$\int\limits_{-\infty}^\infty e^{-x^2} \cdot dx$

The Gaussian Integral may be generalized to the form

$\int\limits_{-\infty}^\infty e^{-a(x + b)^2} \cdot dx$

## 18.1 Theorems for the Gaussian Integral

### 18.1.1 Theorem

The Gaussian Integral is an even function.

Proof:

Recall that an even function is a function $$f(x)$$ such that $$f(-x) = f(x)$$.

Let $$f(x)$$ be the Gaussian Integral

$f(x) = \int\limits_{-\infty}^\infty e^{-x^2} \cdot dx$

\begin{aligned} f(-x) &= \int\limits_{-\infty}^\infty e^{-(-x)^2} \cdot dx \\ &= \int\limits_{-\infty}^\infty e^{-x^2} \cdot dx \\ &= f(x) \end{aligned}

### 18.1.2 Theorem

$\int\limits_{-\infty}^\infty e^{-x^2} \cdot dx = \sqrt{\pi}$

Proof:

Let $$y = x$$, and let $$I = \int\limits_{-\infty}^\infty e^{-x^2} \cdot dx$$. This permits the equation

$I = \int\limits_{-\infty}^\infty e^{-x^2} \cdot dx = \int\limits_{-\infty}^\infty e^{-y^2} \cdot dy$

We use this equality to define the double integral for $$I^2$$.

\begin{aligned} I^2 &= I \cdot I \\ &= \Big(\int\limits_{-\infty}^\infty e^{-x^2} \cdot dx\Big) \cdot \Big(\int\limits_{-\infty}^\infty e^{-y^2} \cdot dx\Big) \\ &= \int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty e^{-x^2} e^{-y^2} \cdot dx\ dy\\ &= \int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty e^{-(x^2 + y^2)} \cdot dx\ dy\\ ^{[1]} &= \int\limits_0^{2\pi} \int\limits_{0}^\infty e^{-r^2} \cdot r \cdot dr\ d\theta \\ &= \int\limits_0^{2\pi} -\frac{1}{2} e ^{-r^2} \Big\rvert_{r = 0}^{r = \infty} \cdot d\theta \\ &= \int\limits_0^{2\pi} - 0 - \big(-\frac{1}{2}\big) \cdot d\theta\\ &= \int\limits_0^{2\pi} \frac{1}{2} \cdot d\theta\\ &= \frac{\theta}{2} \Big\rvert_{\theta = 0}^{\theta = 2\pi} \\ &= \frac{2\pi}{2} - \frac{0}{2} \\ &= \frac{2\pi}{2} \\ &= \pi \end{aligned}

1. Conversion to polar coordinates. Let the radius be $$r = x^2 + y^2$$ on the domain of $$[0, \infty]$$ and let the angle be $$\theta$$ on the domain of $$[0, 2\pi]$$. $$dx\ dy = r dr\ d\theta$$.

This establishes that $$I^2 = \pi$$. It follows:

\begin{aligned} I^2 &= \pi \\ I &= \sqrt{\pi} \\ \int\limits_{-\infty}^\infty e^{-x^2} &= \sqrt{\pi} \end{aligned}

### 18.1.3 Theorem

$\int\limits_{-\infty}^\infty e^{-a(x + b)^2} \cdot dx = \sqrt{\frac{\pi}{a}}$

Proof:

Let $$y = x$$, and let $$I = \int\limits_{-\infty}^\infty e^{-a(x+b)^2} \cdot dx$$. This permits the equation

$I = \int\limits_{-\infty}^\infty e^{-a(x+b)^2} \cdot dx = \int\limits_{-\infty}^\infty e^{-a(y+b)^2} \cdot dy$

We use this equality to define the double integral for $$I^2$$.

\begin{aligned} I^2 &= I \cdot I \\ &= \Big(\int\limits_{-\infty}^\infty e^{-a(x+b)^2} \cdot dx\Big) \cdot \Big(\int\limits_{-\infty}^\infty e^{-a(y+b)^2} \cdot dx\Big) \\ &= \int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty e^{-a(x+b)^2} e^{-a(y+b)^2} \cdot dx\ dy\\ &= \int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty e^{-(a(x + b)^2 + a(y + b)^2)} \cdot dx\ dy\\ ^{[1]} &= \int\limits_0^{2\pi} \int\limits_{0}^\infty e^{-2a(r+b)^2} \cdot r \cdot dr\ d\theta \\ &= \int\limits_0^{2\pi} -\frac{1}{2a} e ^{-r^2} \Big\rvert_{r = 0}^{r = \infty} \cdot d\theta \\ &= \int\limits_0^{2\pi} - 0 - \big(-\frac{1}{2a}\big) \cdot d\theta\\ &= \int\limits_0^{2\pi} \frac{1}{2a} \cdot d\theta\\ &= \frac{\theta}{2a} \Big\rvert_{\theta = 0}^{\theta = 2\pi} \\ &= \frac{2\pi}{2a} - \frac{0}{2} \\ &= \frac{2\pi}{2a} \\ &= \frac{\pi}{a} \end{aligned}

1. Conversion to polar coordinates. Let the radius be $$r = a(x+b)^2 + a(y+b)^2$$ on the domain of $$[0, \infty]$$ and let the angle be $$\theta$$ on the domain of $$[0, 2\pi]$$. $$dx\ dy = r dr\ d\theta$$.

This establishes that $$I^2 = \frac{\pi}{a}$$. It follows:

\begin{aligned} I^2 &= \frac{\pi}{a} \\ I &= \sqrt{\frac{\pi}{a}} \\ \int\limits_{-\infty}^\infty e^{-a(x+b)^2} &= \sqrt{\frac{\pi}{a}} \end{aligned}

## 18.2 References

• Theodore Hatch Whitfield, Lecture Notes, E156 Mathematical Foundations of Statistical Software.