Problem 39 - Calculus for Biology and Medicine

Question

In Problems 39-42, find a comparison function for each integrand and determine whether the integral is convergent. $$ \int_{-\infty}^{\infty} e^{-x^{2} / 2} d x $$

Step-by-Step Solution

Answer

The integral is convergent by comparison with the Gaussian integral.

Step 1: Recognizing the Gaussian Integral Form

The given integral is \[ \int_{-\infty}^{\infty} e^{-x^2/2} \, dx \] which resembles the Gaussian integral, generally of the form \( \int_{-\infty}^{\infty} e^{-ax^2} \, dx \). Here \(a = \frac{1}{2}\). The Gaussian integral is known to be convergent.

Step 2: Finding the Comparison Function

For the integrand \(e^{-x^2/2}\), the function is always positive and decreases as \(|x|\) increases. A probable comparison function is the similar Gaussian \(e^{-x^2}\), because it is a simpler and well-known convergent function, and satisfies \(0 \leq e^{-x^2/2} \leq e^{-x^2}\).

Step 3: Applying the Comparison Test

We know that \[ \int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi} \] which is finite and convergent. Since \(0 \leq e^{-x^2/2} \leq e^{-x^2}\) and \( \int_{-\infty}^{\infty} e^{-x^2} \, dx \) converges, by the Comparison Test, \( \int_{-\infty}^{\infty} e^{-x^2/2} \, dx \) is also convergent.

Key Concepts

Gaussian integral

The Gaussian integral is a fundamental element in calculus and probability theory. It is expressed as \[ \int_{-\infty}^{\infty} e^{-ax^2} \, dx = \sqrt{\frac{\pi}{a}} \]when\( a > 0 \). The Gaussian integral is important because it deals with functions having exponential decay, often appearing in statistical distributions like the normal distribution.

- In the exercise, the form is \( \int_{-\infty}^{\infty} e^{-x^2/2} \, dx \). Here, the coefficient \( a \) is \( \frac{1}{2} \).- Understanding the Gaussian integral helps in recognizing the integral as convergent.

This result is widely used in various fields due to the symmetry and properties of the Gaussian function, making it calculable even over infinite bounds.

Convergence

Convergence is a crucial concept in calculus when dealing with integrals, especially over infinite intervals. An integral is said to be convergent if it approaches a finite limit as the interval extends to infinity.

- For example, the integral \( \int_{-\infty}^{\infty} e^{-x^2/2} \, dx \) is determined to be convergent by recognizing its Gaussian form.

Convergence can be established through various tests, such as the Comparison Test, which is applied in this exercise. By comparing the given function with a known function that converges, we demonstrate that the original function also converges.

Comparison function

A comparison function is a vital tool in determining the convergence of integrals. It serves as a benchmark to evaluate whether another function will behave in the same way over an interval, particularly an infinite one.

- In this exercise, the function \( e^{-x^2} \) is chosen as a comparison function for \( e^{-x^2/2} \).- The key criterion for a suitable comparison function is satisfying \( 0 \leq f(x) \leq g(x) \) for all relevant \( x \). Here, \( f(x) = e^{-x^2/2} \) and \( g(x) = e^{-x^2} \).

Using \( e^{-x^2} \) as it is well-known for being convergent helps simplify the process of showing that \( e^{-x^2/2} \) also converges.

Integrand

The integrand is the function being integrated in an integral expression. Understanding its behavior is essential to solve and evaluate integrals, especially when dealing with infinite bounds.

- In this problem, the integrand is \( e^{-x^2/2} \).- This particular form indicates exponential decay, meaning it gets smaller as \( x \) moves away from zero, which is an important characteristic in determining convergence.

Recognizing the characteristics and comparison potential of an integrand helps in applying integration techniques efficiently. By thoroughly understanding the integrand, one can accurately find a suitable comparison function and properly utilize tests for convergence.