Problem 24 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(1-34\) without using tables. $$ \int_{-\infty}^{\infty} 2 x e^{-x^{2}} d x $$

Step-by-Step Solution

Answer

The integral evaluates to 0 because it is the integral of an odd function over a symmetric interval.

Step 1: Recognize the Integral as Gaussian

The integral \( \int_{-\infty}^{\infty} e^{-x^2} dx \) is known as the Gaussian integral, which evaluates to \( \sqrt{\pi} \). The given integral, \( \int_{-\infty}^{\infty} 2x e^{-x^2} dx \), is similar but contains an additional \( x \) multiplier, which suggests it is an odd function.

Step 2: Analyze the Function's Symmetry

The integrand \( 2x e^{-x^2} \) is an odd function because \( f(-x) = -f(x) \) due to the linear component \( x \). The exponential function \( e^{-x^2} \) is even, so the overall product is odd.

Step 3: Apply the Property of Odd Functions in Integrals

For any odd function \( f(x) \), we have \( \int_{-a}^{a} f(x)\, dx = 0 \). Therefore, the integral of any odd function over a symmetric interval about the origin (such as \(-\infty\) to \(\infty\)) will result in zero. Thus: \( \int_{-\infty}^{\infty} 2x e^{-x^2} dx = 0 \).

Key Concepts

Odd Functions

In mathematics, an odd function is a function that satisfies the condition that \( f(-x) = -f(x) \) for all \( x \) in the domain. This symmetry of odd functions around the origin has significant implications, especially in the context of calculus and integration:
When you integrate an odd function over a symmetric interval centered around zero (like from \(-a\) to \(a\)), the positive and negative contributions of the function cancel each other out. Thus, the integral evaluates to zero.
For example, in the integral \( \int_{-\infty}^{\infty} 2x e^{-x^2} \, dx \), the function \( 2x e^{-x^2} \) is odd. The term \( 2x \) is linear and changes sign with \( x \). Importantly, when multiplied by the even function \( e^{-x^2} \), the result is still odd:

• The linear term \( 2x \) flips sign, and
• The even function \( e^{-x^2} \) retains its value whether \( x \) is positive or negative,

indicating that the entire product is odd, leading to zero when integrated over a symmetric interval.

Symmetry in Integration

The concept of symmetry is crucial in simplifying integrals and understanding their outcomes. In particular, symmetry can often tell us whether an integral evaluates to zero or not without every needing to compute it explicitly. Here’s a closer look:
When dealing with integrals over symmetric intervals, recognizing symmetry allows us to make predictions about the outcome. This insight is especially useful with functions that exhibit even or odd symmetry.

• **Even functions** \( f(x) \) satisfy \( f(-x) = f(x) \) and contribute symmetrically so that the areas under the curve to the left and right of the y-axis are equal.
• **Odd functions** \( g(x) \) satisfy \( g(-x) = -g(x) \), leading to the areas under the curve cancelling each other out in symmetric bounds.

The importance of symmetry comes into play strongly when calculating definite integrals. With the integral \( \int_{-\infty}^{\infty} 2x e^{-x^2} \, dx \), the odd symmetry simplifies the process: since it is zero across a symmetric interval about the origin, you can identify it as such without further calculations.

Exponential Functions

Exponential functions, characterized by the presence of \( e \) (Euler's number), are incredibly important in mathematics due to their diverse applications. The function \( e^{-x^2} \) serves as a classic example that frequently appears in problems involving Gaussian integrals.
This specific exponential function is even because it satisfies \( e^{-(-x)^2} = e^{-x^2} \). Even functions have reflective symmetry over the y-axis.
In the exercise at hand, the integral includes \( e^{-x^2} \) multiplied by an odd function \( 2x \). This characteristic of the exponential becoming zero at positive and negative infinities (because \( e^{-x^2} \to 0 \) as \( x \to \pm\infty \)) supports the assessment that this integrand will also have a net result of zero when performed over symmetric limits.
Additionally, exponential functions enjoy powerful properties such as rapid decay and continuous differentiability, which make them relatively easy to handle in integration calculus, although they present complex behaviors worthy of study.