Problem 80 - Thomas Calculus

Question

a. Show that if \(f\) is even and the necessary integrals exist, then $$\int_{-\infty}^{\infty} f(x) d x=2 \int_{0}^{\infty} f(x) d x$$ b. Show that if \(f\) is odd and the necessary integrals exist, then $$\int_{-\infty}^{\infty} f(x) d x=0$$

Step-by-Step Solution

Answer

a. The integral of an even function is twice the integral from 0 to infinity. b. The integral of an odd function over an even domain is zero.

Step 1: Understand Even Function Properties

An even function satisfies the property \( f(x) = f(-x) \). This means if you evaluate the function at any positive number, it will yield the same result as if you evaluate it at the negative of that number.

Step 2: Split the Integral for Even Functions

Given the integral \( \int_{-\infty}^{\infty} f(x) \, dx \), recognize it can be split into two integrals: \( \int_{-\infty}^{0} f(x) \, dx \) and \( \int_{0}^{\infty} f(x) \, dx \).

Step 3: Apply Properties of Even Functions

For the integral from \(-\infty\) to 0, use the property of even functions:\[\int_{-\infty}^{0} f(x) \, dx = \int_{-\infty}^{0} f(-x) \, dx = \int_{0}^{\infty} f(x) \, dx.\]

Step 4: Solve Part a

Add the integrals: \( \int_{-\infty}^{\infty} f(x) \, dx = \int_{0}^{\infty} f(x) \, dx + \int_{0}^{\infty} f(x) \, dx = 2 \int_{0}^{\infty} f(x) \, dx.\)

Step 5: Understand Odd Function Properties

An odd function satisfies the property \( f(-x) = -f(x) \). This means if you evaluate the function at a positive number, it will be the negative of the function evaluated at the negative of that number.

Step 6: Split the Integral for Odd Functions

Given the integral \( \int_{-\infty}^{\infty} f(x) \, dx \), similarly split it into two parts: \( \int_{-\infty}^{0} f(x) \, dx \) and \( \int_{0}^{\infty} f(x) \, dx \).

Step 7: Apply Properties of Odd Functions

For the integral from \(-\infty\) to 0, use the property of odd functions:\[\int_{-\infty}^{0} f(x) \, dx = \int_{-\infty}^{0} f(-x) \, dx = \int_{0}^{\infty} -f(x) \, dx.\]

Step 8: Solve Part b

Combine the integrals: \( \int_{-\infty}^{\infty} f(x) \, dx = \int_{0}^{\infty} f(x) \, dx + \int_{0}^{\infty} -f(x) \, dx = 0.\) Since these two integrals cancel each other out.

Key Concepts

Even Function

An even function is a special type of mathematical function that exhibits symmetry about the vertical axis. This characteristic is formally expressed by the equation: \( f(x) = f(-x) \). What this means in simple terms is that the function's graph looks the same on the right side of the vertical axis as it does on the left. This property is crucial when evaluating integrals over symmetric intervals.

• If you take any number \( x \) and substitute it into the function, the output will be the same as if you substitute \( -x \).
• This unique feature of even functions is leveraged to simplify computations, especially in integral calculus, as seen when extending functions to the entire real line.

When dealing with integrals of even functions over symmetric limits, this symmetry is particularly helpful. If an integral of an even function extends from negative infinity to positive infinity, you can actually compute it as twice the integral from zero to infinity. This is because the areas under the curve from 0 to infinity and from minus infinity to 0 are equal, simplifying calculations considerably.

Odd Function

An odd function is another important type of mathematical function with its own defining characteristics and symmetries. It satisfies the property \( f(-x) = -f(x) \), which means its graph is symmetric about the origin. This indicates that if you take a positive number and evaluate the function, it will be the negative of the function evaluated at the corresponding negative number.

• Odd functions have the interesting trait that for every point on the function, if you reflect it across both the x- and y-axes, you'll get another point on the function.
• This unique symmetry about the origin provides significant simplification benefits in calculus problems involving integrals.

In integral calculus, when you compute the integral of an odd function over an interval symmetric about the origin, the result is always zero. This happens because the positive and negative regions of the function's graph cancel each other out. Thus, for intervals extending from \(-a\) to \(a\), this property of odd functions can be utilized to quickly determine that the integral equals zero.

Definite Integrals

Definite integrals are a fundamental concept in integral calculus. They compute the net area under a curve between two specified limits and are represented by the symbol \( \int_{a}^{b} f(x) \, dx \). Unlike an indefinite integral, which represents a family of functions, a definite integral provides a specific numerical value.

• The upper and lower limits in a definite integral denote the bounds within which the area under the curve is calculated.
• The result represents the accumulated quantity, such as distance, area, volume, etc., depending on the context of the function.

Definite integrals are crucial for solving a variety of practical problems in physics, engineering, and other fields requiring quantifiable measures. They help in determining quantities like total displacement, the area between curves, and the average value of a function, among others.

Integral Calculus

Integral calculus is a branch of mathematics focused on integrals, the inverse process of differentiation. Its primary concerns are the accumulation of quantities and the area beneath curves. Integral calculus is divided into two main parts: indefinite integrals and definite integrals.

• Indefinite integrals calculate a general form of the antiderivative of a function without specific limits, thus including a constant of integration \( C \).
• Definite integrals, on the other hand, calculate the total accumulation within specified limits and result in a numerical value.

An integral is represented by the integral sign \( \int \), and the process of finding integrals is known as integration. This process is widely applied in various fields to find volumes, areas, central points, and in many real-world applications like physics and engineering. Integral calculus also plays a critical role in solving differential equations, which model numerous natural phenomena.