2.1: Gaussian Integral Revisited

As the first application, let’s revisit the Gaussian integral from Section 1.3,

\[\int_{-\infty}^{\infty} e^{-ax^{2}}dx. \label{2.1} \]

Question

Is the integral \(\sqrt{\pi α}\) or ­\(\sqrt{\pi/α}\)?

The correct choice must work for all \(α = 0\). At this range’s endpoints (\(α = \infty \text{ and } α = 0\)), the integral is easy to evaluate.

Question

What is the integral when \(α = \infty\)?

As the first easy case, increase \(α \text{ to } \infty\). Then −\(αx^{2}\) becomes very negative, even when \(x\) is tiny. The exponential of a large negative number is tiny, so the bell curve narrows to a sliver, and its area shrinks to zero. Therefore, as \(α \rightarrow \infty\) the integral shrinks to zero. This result refutes the option \(\sqrt{\pi α}\), which is infinite when \(α = \infty\); and it supports the option ­\(\sqrt{\pi/α}\), which is zero when \(α = \infty\).

Question

What is the integral when \(α = 0\)?

In the \(α = 0\) extreme, the bell curve flattens into a horizontal line with unit height. Its area, integrated over the infinite range, is infinite. This result refutes the \(\sqrt{\pi α}\) option, which is zero when \(α = 0\); and it supports the \(\sqrt{\pi/α}\) option, which is infinity when \(α = ­0\). Thus the πα option fails both easy-cases tests, and the \(\sqrt{\pi/α}\) option passes both easy-cases tests. If these two options were the only options, we would choose \(\sqrt{\pi/α}\). However, if a third option were \(\sqrt{2/α}\), how could you decide between it and \(\sqrt{\pi/α}\)? Both options pass both easy-cases tests; they also have identical dimensions. The choice looks difficult.

To choose, try a third easy case: \(α = 1\). Then the integral simplifies to

\[\int_{-\infty}^{\infty} e^{-x^{2}}dx. \label{2.2} \]

This classic integral can be evaluated in closed form by using polar coordinates, but that method also requires a trick with few other applications (textbooks on multivariable calculus give the gory details). A less elegant but more general approach is to evaluate the integral numerically and to use the approximate value to guess the closed form.

Therefore, replace the smooth curve \(e^{−x^{2}}\) with a curve having n line segments. This piecewise-linear approximation turns the area into a sum of \(n\) trapezoids. As
\(n\) approaches infinity, the area of the trapezoids more and more closely approaches the area under the smooth curve.

The table gives the area under the curve in the range \(x = −10 ... 10\), after dividing the curve into n line segments. The areas settle onto a stable value, and it looks familiar. It begins with 1.7, which might arise from \(\sqrt{3}\). However, it continues as 1.77, which is too large to be \(\sqrt{3}\). Fortunately, \(\pi\) is slightly larger than 3, so the area might be converging to \(\sqrt{\pi}\).

Let’s check by comparing the squared area against \(\pi\):

1.772453850905522 ≈ 3.14159265358980,

\[\pi ≈ 3.14159265358979. \label{2.3} \]

The close match suggests that the \(α = 1\) Gaussian integral is indeed \(\sqrt{π}\):

\[\int_{-\infty}^{\infty} e^{-x^{2}}dx = \sqrt{\pi}. \label{2.4} \]

Therefore the general Gaussian integral

\[\int_{-\infty}^{\infty} e^{-ax^{2}}dx \label{2.5} \]

must reduce to \(\sqrt{\pi}\) when \(α = 1\). It must also behave correctly in the other two easy cases \(α = 0 \text{ and } α = \infty\).

Among the three choices \(\sqrt{­2/α}\), \(\sqrt{­\pi/α}\), and \(\sqrt{\pi α}\), only \(\sqrt{\pi/α}\) passes all three tests \(α = 0, 1, \text{ and } \infty\). Therefore,

\[\int_{-\infty}^{\infty} e^{-ax^{2}}dx = \sqrt{\frac{\pi}{a}}. \label{2.6} \]

Easy cases are not the only way to judge these choices. Dimensional analysis, for example, can also restrict the possibilities (Section 1.3). It even eliminates choices like \(\sqrt{\pi/α}\) that pass all three easy-cases tests. However, easy cases are, by design, simple. They do not require us to invent or deduce dimensions for \(x, α\), and dx (the extensive analysis of Section 1.3). Easy cases, unlike dimensional analysis, can also eliminate choices like \(\sqrt{2/α}\) with correct dimensions. Each tool has its strengths.

Multiple problems

Problem 2.1 Testing several alternatives

For the Gaussian integral

\[\int_{-\infty}^{\infty} e^{-ax^{2}}dx, \label{2.7} \]

use the three easy-cases tests to evaluate the following candidates for its value.

(a) \(\sqrt{\pi/α}\)

(b) 1+(\(\sqrt{\pi}\)−1)/α

(c) 1/α\(^{2}\) +(\(\sqrt{\pi}\)−1)/α.

Problem 2.2 Plausible, incorrect alternative

Is there an alternative to \(\sqrt{\pi/α}\) that has valid dimensions and passes the three easy-cases tests?

Problem 2.3 Guessing a closed form

Use a change of variable to show that

\[\int_{0}^{\infty} \frac{dx}{1 + x^{2}} = 2\int_{0}^{1} \frac{dx}{1 + x^{2}}. \label{2.8} \]

The second integral has a finite integration range, so it is easier than the first integral to evaluate numerically. Estimate the second integral using the trapezoid approximation and a computer or programmable calculator. Then guess a closed form for the first integral.