Problem 37 - Thomas Calculus

Question

a. The usual way to evaluate the improper integral \(I=\int_{0}^{\infty} e^{-x^{2}} d x\) is first to calculate its square: $$I^{2}=\left(\int_{0}^{\infty} e^{-x^{2}} d x\right)\left(\int_{0}^{\infty} e^{-y^{2}} d y\right)=\int_{0}^{\infty} \int_{0}^{\infty} e^{-\left(x^{2}+y^{2}\right)} d x d y$$ Evaluate the last integral using polar coordinates and solve the resulting equation for \(I .\) b. Evaluate $$\lim _{x \rightarrow \infty} \operatorname{erf}(x)=\lim _{x \rightarrow \infty} \int_{0}^{x} \frac{2 e^{-t^{2}}}{\sqrt{\pi}} d t$$

Step-by-Step Solution

Answer

\( I = \frac{\sqrt{\pi}}{2} \); \( \lim_{x \to \infty} \operatorname{erf}(x) = 1 \).

Step 1: Interpret the Double Integral

The expression given in the problem is \( I^2 = \int_{0}^{\infty} \int_{0}^{\infty} e^{-(x^2 + y^2)} \, dx \, dy \). We recognize this as an integral over the first quadrant of the plane, dealing with exponential decay.

Step 2: Convert to Polar Coordinates

Change variables to polar coordinates where \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \). The integral becomes \( \int_{0}^{\infty} \int_{0}^{\frac{\pi}{2}} e^{-r^2} r \, dr \, d\theta \) due to the transformation \( dx \, dy = r \, dr \, d\theta \).

Step 3: Evaluate the Polar Integral

Evaluate the inner integral: \( \int_{0}^{\infty} e^{-r^2} r \, dr \). Use the substitution \( u = r^2 \), then \( du = 2r \, dr \), giving \( \frac{1}{2}\int_{0}^{\infty} e^{-u} \, du = \frac{1}{2}\left[-e^{-u}\right]_{0}^{\infty} = \frac{1}{2} \).

Step 4: Integrate with Respect to Theta

Now integrate \( \int_{0}^{\frac{\pi}{2}} \frac{1}{2} \, d\theta \), which simplifies to \( \frac{\pi}{4} \).

Step 5: Solve for \( I \)

Since \( I^2 = \frac{\pi}{4} \), take the square root of both sides to find \( I = \frac{\sqrt{\pi}}{2} \).

Step 6: Evaluate the Limit of the Error Function

The limit \( \lim_{x \rightarrow \infty} \operatorname{erf}(x) = \lim_{x \rightarrow \infty} \int_{0}^{x} \frac{2 e^{-t^2}}{\sqrt{\pi}} \, dt \) approaches 1, due to the integral spanning the entire Gaussian distribution.

Key Concepts

Polar Coordinates

Integrating over distinct regions in the Cartesian coordinate system can get tricky, especially when dealing with circular or radial symmetries. This is where polar coordinates come into play. By using polar coordinates, we can simplify double integrals as they relate to circular areas.
Polar coordinates are defined by

• An angle \(\theta\), how far around the circle you are from the initial line (usually the positive x-axis).
• A radius \(r\), the distance away from the origin (0,0).

When converting to polar coordinates, the area element \(dx \, dy\) transforms into \(r \, dr \, d\theta\). This multiplied with the integrand often simplifies the integration of exponential decay terms like \(e^{-(x^2 + y^2)}\), especially because \(x^2 + y^2 = r^2\) in polar coordinates. For our problem, this conversion notably reduces complexity, transforming a cumbersome pair of integrals into a more manageable form.

Error Function

The error function, often denoted as \(\operatorname{erf}(x)\), is a special function that arises frequently in statistics, probability, and partial differential equations. It is intimately connected to the Gaussian distribution and is defined as:
\[\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^2} \, dt\]
This integral cannot be expressed in terms of elementary functions, making \(\operatorname{erf}(x)\) essential for calculations involving Gaussian distributions and normal probabilities. The function represents the probability of a random variable, with a normal distribution, falling between the mean and \(x\) standard deviations away from the mean. As \(x\) approaches infinity, \(\operatorname{erf}(x)\) approaches 1, which is consistent with the concept that nearly all the possible values fall within a few standard deviations from the mean in a Gaussian setting.

Limits of Integration

Determining the limits of integration is pivotal when solving integrals, as they define the boundaries for the area or volume being calculated. With improper integrals, these limits can include infinity, making evaluation a bit more complex due to their infinite span.
For the double integral \(\int_{0}^{\infty} \int_{0}^{\infty} e^{-(x^2 + y^2)} \, dx \, dy\), the limits from 0 to infinity indicate that we're assessing the entire first quadrant unboundedly. By switching to polar coordinates, the radial integration from 0 to \(\infty\) covers the entire distance outward from the origin, while the angular limit from 0 to \(\frac{\pi}{2}\) covers a quarter of the full rotation (90 degrees), capturing the entire first quadrant. Defining these limits correctly is crucial to derive accurate results in integration.

Gaussian Distribution

The Gaussian distribution, also known as the normal distribution, is a statistical distribution characterized by its bell-shaped curve. This distribution is defined by the probability density function:
\[f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}\]
where \(\mu\) represents the mean and \(\sigma\) the standard deviation. This distribution describes many natural phenomena, where data tends to cluster around a central mean value with symmetrical spread on either side.
In the context of this exercise, the Gaussian form is evident in the \\(e^{-(x^2+y^2)}\) term. This term essentially represents a two-dimensional Gaussian centered at the origin, which when integrated over its entire domain results in the value \\(\pi\), revealing a characteristic property of Gaussians: they integrate to 1 over their range when appropriately normalized. This concept explains why the error function asymptotically approaches 1 as its upper limit tends toward infinity, reflecting the distribution's cumulative property.