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.