Problem 58
Question
Approximate each integral. $$ \int_{-\infty}^{\infty} \frac{1}{1+x^{2}} d x $$
Step-by-Step Solution
Verified Answer
The integral evaluates to \( \pi \).
1Step 1: Recognize the Integral Type
The given integral \( \int_{-\infty}^{\infty} \frac{1}{1+x^2} \, dx \) is a standard form and is recognized as an improper integral over the whole real line.
2Step 2: Identify the Function
The function \( f(x) = \frac{1}{1+x^2} \) represents a standard form of a Lorentzian (Cauchy) distribution, which is symmetric about the y-axis.
3Step 3: Use the Standard Integral Result
The integral of \( \frac{1}{1+x^2} \) over the interval \([-\infty, \infty]\) is a standard result. This integral gives \( \pi \). This stems from recognizing the function as the derivative of the inverse tangent function, \( \tan^{-1}(x) \).
4Step 4: Confirm the Evaluation
Evaluate the indefinite integral as \( \tan^{-1}(x) + C \). For the definite integral \( \int_{-\infty}^{\infty} \frac{1}{1+x^2} \, dx \), we have limits of \( \tan^{-1}(\infty) = \frac{\pi}{2} \) and \( \tan^{-1}(-\infty) = -\frac{\pi}{2} \). Thus, the solution is \( \pi \).
Key Concepts
Lorentzian DistributionDefinite IntegralInverse Tangent Function
Lorentzian Distribution
The Lorentzian distribution, also known as the Cauchy distribution, is an important concept in probability and statistics. It describes a family of distributions resembling the bell shape of a Gaussian distribution, but with heavier tails. This means that the probability of observing extreme values is higher compared to a Gaussian distribution.
The function \( f(x) = \frac{1}{1+x^2} \) is a standard representation of the Lorentzian distribution. This function is symmetric about the y-axis, indicating that it looks the same to the left and right of the y-axis.
The function \( f(x) = \frac{1}{1+x^2} \) is a standard representation of the Lorentzian distribution. This function is symmetric about the y-axis, indicating that it looks the same to the left and right of the y-axis.
- **Properties of the Lorentzian Distribution:**
- The peak of the distribution occurs at \( x = 0 \), where it achieves its maximum value of 1.
- Unlike the Gaussian distribution, the Lorentzian has undefined mean and variance due to its heavy tails.
Definite Integral
A definite integral is a fundamental concept in calculus that provides the net area under a curve defined by a function, within certain bounds. For the integral \( \int_{a}^{b} f(x) \ dx \), \( a \) and \( b \) are the limits of integration, and \( f(x) \) is the integrand.
In the case of improper integrals, like \( \int_{-\infty}^{\infty} \frac{1}{1+x^2} \ dx \), these bounds approach infinity. Such integrals require understanding how the function behaves as \( x \) approaches infinity or negative infinity.
In the case of improper integrals, like \( \int_{-\infty}^{\infty} \frac{1}{1+x^2} \ dx \), these bounds approach infinity. Such integrals require understanding how the function behaves as \( x \) approaches infinity or negative infinity.
- The improper integral is calculated by finding the limit of the definite integral as one or both of the bounds approach infinity.
- Improper integrals are crucial in physics and engineering, where they model phenomena like wave functions in quantum mechanics.
Inverse Tangent Function
The inverse tangent function, commonly denoted \( \tan^{-1}(x) \), is a function that serves as the inverse of the tangent function \( y = \tan(x) \). In calculus, it's crucial for evaluating integrals involving trigonometric functions.
The derivative of \( \tan^{-1}(x) \) gives the function \( \frac{1}{1+x^2} \), which assists in solving integrals like \( \int_{-\infty}^{\infty} \frac{1}{1+x^2} \ dx \) by identifying the primitive function.
The derivative of \( \tan^{-1}(x) \) gives the function \( \frac{1}{1+x^2} \), which assists in solving integrals like \( \int_{-\infty}^{\infty} \frac{1}{1+x^2} \ dx \) by identifying the primitive function.
- **Properties of the Inverse Tangent Function:**
- It returns angles in the range \((-\frac{\pi}{2}, \frac{\pi}{2})\).
- \( \tan^{-1}(\infty) = \frac{\pi}{2} \) and \( \tan^{-1}(-\infty) = -\frac{\pi}{2} \), which helps resolve the improper integral with limits at infinity.
Other exercises in this chapter
Problem 58
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