Problem 35 - Advanced Engineering Mathematics

Question

The improper integral \(\int_{0}^{\infty} e^{-x^{2}} d x\) is important in the theory of probability, statistics, and other areas of applied mathematics. If \(I\) denotes the integral, then $$ I=\int_{0}^{\infty} e^{-x^{2}} d x \quad \text { and } \quad I=\int_{0}^{\infty} e^{-y^{2}} d y $$ and consequently $$ I^{2}=\left(\int_{0}^{\infty} e^{-x^{2}} d x\right)\left(\int_{0}^{\infty} e^{-y^{2}} d y\right)=\int_{0}^{\infty} \int_{0}^{\infty} e^{-\left(x^{2}+y^{2}\right)} d x d y $$ Discuss how to use polar coordinates to evaluate the last integral. Find the value of \(I\).

Step-by-Step Solution

Answer

The value of \( I \) is \( \frac{\sqrt{\pi}}{2} \).

Step 1: Express Integral in Terms of Two Variables

The given expression is \( I^2 = \int_{0}^{\infty} \int_{0}^{\infty} e^{-(x^2+y^2)} \, dx \, dy \). This is a double integral representing the area under the curve of \( e^{-(x^2+y^2)} \) over the first quadrant of the Cartesian plane. To simplify the evaluation, we will convert this to polar coordinates.

Step 2: Convert to Polar Coordinates

In polar coordinates, \( x = r\cos\theta \) and \( y = r\sin\theta \), where \( r^2 = x^2 + y^2 \). The area element \( dx \, dy \) is converted to \( r \, dr \, d\theta \). Therefore, the integral becomes \( \int_{0}^{\pi/2} \int_{0}^{\infty} e^{-r^2} \, r \, dr \, d\theta \).

Step 3: Evaluate the Integral with Respect to \( r \)

The inner integral is \( \int_{0}^{\infty} e^{-r^2} \, r \, dr \). Letting \( u = r^2 \) gives \( du = 2r \, dr \) or \( r \, dr = \frac{1}{2} \, du \). The limits of \( u \) go from 0 to \( \infty \). Substituting, the integral becomes \( \frac{1}{2} \int_{0}^{\infty} e^{-u} \, du \), which evaluates to \( \frac{1}{2} \).

Step 4: Integrate with Respect to \( \theta \)

Now, using the result from the previous step, the original double integral becomes \( \int_{0}^{\pi/2} \frac{1}{2} \, d\theta \). Evaluating this integral gives \( \frac{1}{2} \times \frac{\pi}{2} = \frac{\pi}{4} \).

Step 5: Find the Value of \( I \)

Since \( I^2 = \frac{\pi}{4} \), we need to find \( I \). Thus, \( I = \sqrt{\frac{\pi}{4}} = \frac{\sqrt{\pi}}{2} \).

Key Concepts

Polar Coordinates

Polar coordinates offer a different way to describe points on a plane using a radius and an angle. Instead of Cartesian coordinates which use

• X (horizontal distance)
• Y (vertical distance)

polar coordinates use:

• \( r \): the distance from the origin
• \( \theta \): the angle from the positive x-axis

The conversion between these systems follows these formulas:

• \( x = r \cos(\theta) \)
• \( y = r \sin(\theta) \)
• \( r^2 = x^2 + y^2 \)

This system is especially helpful for integrating over circular regions. In our problem, converting to polar coordinates streamlined the computation of the integral \( \int_0^{\infty} \int_0^{\infty} e^{-(x^2+y^2)} \, dx \, dy \), because the function's symmetry around the origin fits naturally in polar coordinates.

Double Integral

A double integral is like adding up values across a two-dimensional region. You can think of it as an extension of a one-dimensional integral. In our original problem, it is used to evaluate the total area under the function \( e^{-(x^2+y^2)} \) in the first quadrant.
Double integrals can be computed in Cartesian or polar coordinates:

• In Cartesian coordinates, they appear as \( \int \int \, f(x, y) \, dx \, dy \).
• In polar coordinates, it turns into \( \int \int \, f(r \cos\theta, r \sin\theta) \, r \, dr \, d\theta \).

The change to polar coordinates from Cartesian was crucial in solving our problem, involving symmetry and repeated patterns.

Probability Theory

The improper integral \( \int_{0}^{\infty} e^{-x^2} \, dx \) has strong ties to probability theory. It is connected to the Gaussian distribution, which is a cornerstone of statistics, describing how values are distributed across a mean in datasets.
In probability,

• The Gaussian or normal distribution has the formula \( f(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \).
• The integral \( \int_{-\infty}^{\infty} e^{-x^2} \, dx \) is essential in defining aspects of this distribution.
• It's the foundational piece in deriving properties like standard deviation.

Our solution demonstrates how converting the function into a double integral helps us calculate an expression related to this distribution.

Applied Mathematics

In applied mathematics, improper integrals like \( \int_{0}^{\infty} e^{-x^2} \, dx \) show up frequently. This integral is fundamental in fields like physics and engineering due to the appearance of Gaussian functions.
Areas of application include:

• Calculus: Evaluating challenging integrals not easily solved with standard methods.
• Physics: Modeling processes like diffusion and heat conduction.
• Engineering: Signal processing, where Gaussian functions can help in analyzing real-world signals.

Improper integrals give us tools to solve problems involving infinite limits, and understanding them can open doors to powerful analytical methods.