Problem 41 - University Calculus: Early Transcendentals

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

The value of I is \( \frac{\sqrt{\pi}}{2} \), and \( \lim_{x \to \infty} \operatorname{erf}(x) = 1 \).

Step 1: Convert to Polar Coordinates

In the double integral, we have \( \int_{0}^{\infty} \int_{0}^{\infty} e^{-(x^2+y^2)} \, dx \, dy \). To transform it into polar coordinates: Set \( x = r \cos \theta \) and \( y = r \sin \theta \). The Jacobian determinant for this coordinate transformation is \( r \, dr \, d\theta \). The limits for \( r \) are from 0 to \( \infty \), and for \( \theta \) from 0 to \( \pi/2 \).

Step 2: Simplify the Integral Expression

Substitute the polar coordinates into the integral to get \( \int_{0}^{\pi/2} \int_{0}^{\infty} e^{-r^2} r \, dr \, d\theta \). This simplifies to \( \int_{0}^{\pi/2} d\theta \int_{0}^{\infty} r e^{-r^2} \, dr \). The integral \( \int_{0}^{\infty} r e^{-r^2} \, dr \) is a standard result which equals \( \frac{1}{2} \).

Step 3: Evaluate the Angular Integral

The angular integral \( \int_{0}^{\pi/2} d\theta \) equals \( \frac{\pi}{2} \). Combine this with the radial integral to get \( \frac{\pi}{2} \times \frac{1}{2} = \frac{\pi}{4} \).

Step 4: Solve for \(I\)

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

Step 5: Evaluate the Limit

The integral \( \operatorname{erf}(x) = \int_{0}^{x} \frac{2e^{-t^2}}{\sqrt{\pi}} \, dt \) represents the error function. As \( x \to \infty \), the exponential term approaches zero, making the error function approach 1. Hence, \( \lim_{x \to \infty} \operatorname{erf}(x) = 1 \).

Key Concepts

Polar Coordinates

When we talk about polar coordinates, we're looking at an alternate way to express points in a plane compared to the usual Cartesian coordinates. In Cartesian coordinates, points are described with an x-coordinate (the distance along the x-axis) and a y-coordinate (the distance along the y-axis). However, polar coordinates describe the location using a radius and an angle.
This is especially useful in situations involving symmetry about a point, like circular patterns or when dealing with problems that cover entire circles or sectors. In polar coordinates, we represent each point by two values:

• The radius, usually denoted as \( r \), which is the distance from the origin to the point.
• The angle, commonly denoted as \( \theta \), which is the direction from the origin to the point, usually measured from the positive x-axis.

In the context of the problem, converting a double integral \( \int \int e^{-(x^2 + y^2)} \, dx \, dy \) into polar coordinates is beneficial because it simplifies the expression into a more manageable form. The conversion uses the formulae \( x = r \cos \theta \) and \( y = r \sin \theta \), along with a Jacobian determinant of \( r \, dr \, d\theta \), to integrate over radial and angular components separately. This can transform complex calculations into more straightforward ones.

Error Function (erf)

The error function, often represented as \( \operatorname{erf}(x) \), is a special mathematical function that arises frequently in probability, statistics, and partial differential equations. It's closely related to the cumulative distribution function of the standard normal distribution.
The error function is defined by the integral:
\[ \operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^2} \, dt \]
This measures the probability of a random variable with a standard normal distribution being between zero and \(x\). As \(x\) grows larger, the integral of the decaying exponential \(e^{-t^2}\) approaches a stable value.

• At \(x = 0\), \( \operatorname{erf}(0) = 0 \).
• As \(x\) goes to infinity, \( \operatorname{erf}(x) \) tends towards 1.
• Symmetry is a key characteristic where \( \operatorname{erf}(-x) = -\operatorname{erf}(x) \).

Understanding the behavior of the error function, especially its limit as \(x\) approaches infinity, is crucial as it informs about the distribution of probabilities in statistical contexts. In the given exercise, recognizing that this limit equals 1 provides insights into the nature of the bounds on certain integrals involving Gaussian functions.

Limits in Calculus

Limits are fundamental in calculus. They provide a way to understand the behavior of functions as inputs approach a certain value. Limits help us to analyze functions' results at points that might not be well-behaved, such as at infinity, or when a point is where the function isn’t explicitly defined.
In calculus, limits describe:

• How a function behaves as it approaches infinity or negative infinity.
• Truely understanding exact values a function nears as the variables in the function creep toward a specific number.

In the context of the exercise, limits compute the behavior of the error function as \( x \) becomes quite large. In the limit \(\lim_{x \rightarrow \infty} \operatorname{erf}(x)\), we explore how the error function stabilizes as its input grows to large values. Here, it's shown to approach 1 because the area under the curve of the standard normal distribution bounded by the limits of the integral becomes essentially the entire curve as \(x\) trends toward infinity.
Through limits, we gain insight into asymptotic behavior and the long-term tendencies of functions, which are crucial in higher-level mathematics and practical applications like engineering and physics.