Question

Determine whether the improper integrals converge or diverge. If possible, determine the value of the integrals that converge.\(\int_{0}^{\infty} e^{-x} \cos x d x\)

Short answer

The integral converges to \( \frac{1}{2} \).

Step-by-step solution

Check for Convergence or Divergence

To determine whether the integral \( \int_{0}^{\infty} e^{-x} \cos x \, dx \) converges, we examine the behavior of the integrand \( e^{-x} \cos x \) as \( x \to \infty \). The exponential term \( e^{-x} \) ensures that the integrand decreases rapidly to zero, suggesting the possibility of convergence.

Substitute and Simplify using Laplace Transform

To evaluate the integral, we can use a known transform property. The integral \( \int_{0}^{\infty} e^{-sx} \cos ax \, dx \) can be expressed as \( \frac{s}{s^2 + a^2} \) using the Laplace transform, where \( s = 1 \) and \( a = 1 \) in our case. Therefore, the integral becomes \( \frac{1}{1^2 + 1^2} = \frac{1}{2} \).

Conclusion for Convergence

Since we have applied the Laplace transform and found the integral evaluates to a finite value, it confirms that the integral converges. Thus, \( \int_{0}^{\infty} e^{-x} \cos x \, dx \) converges to \( \frac{1}{2} \).

Key concepts

Convergence

Convergence is an important concept in analyzing improper integrals, which are integrals with at least one infinite limit of integration or an unbounded integrand. For the purpose of this exercise, we are concerned with whether the integral from 0 to infinity of the function \( e^{-x} \cos x \) converges or diverges.

Convergence refers to the behavior of the integral as its limits approach infinity. Specifically, an integral is said to converge if it evaluates to a finite number. The main factor influencing the convergence of the given integral is the exponential function \( e^{-x} \). This function decreases rapidly to zero as \( x \to \infty \), which is crucial in ensuring that the integrand becomes negligible enough over a long interval to yield a finite area under the curve. By observing the decreasing nature of \( e^{-x} \), we can deduce that the integral likely converges, although this should be followed by a method of precise calculation, such as the Laplace transform.

Laplace transform

The Laplace transform is a valuable tool in solving improper integrals, particularly when dealing with functions that involve exponential decay and oscillatory components. It transforms complex functions into simpler algebraic forms that are easier to work with, especially in calculus and differential equations.

The function \( e^{-x} \cos x \) is an excellent candidate for evaluation using the Laplace transform. When the Laplace transformation is applied to an exponential function multiplied by a trigonometric function like \( \cos x \), it simplifies the integral into a form that can be straightforwardly evaluated. This transformation takes a function of a real variable into a function of a complex variable, often making it easier to find whether the integral converges and to compute its value. By transforming the original integral using the properties of the Laplace transform, the problem becomes more manageable and leads to a successful evaluation.

Laplace transform property

The key to using the Laplace transform lies in understanding its useful properties, which allow us to evaluate integrals like \( \int_{0}^{\infty} e^{-x} \cos x \, dx \). A specific property of the Laplace transform deals with functions of the form \( e^{-sx} \cos(ax) \), which can be simplified to \( \frac{s}{s^2 + a^2} \) where \( s \) and \( a \) are constants.

In the context of the exercise's integral, the property is used where \( s = 1 \) and \( a = 1 \). Hence, the integral becomes \( \frac{1}{1^2 + 1^2} \), which evaluates to \( \frac{1}{2} \). This provides a clear and concise result showing the utility of the Laplace transform's properties in determining both convergence and calculating the exact value of the integral, confirming that it converges to a finite value. Understanding and utilizing these properties can greatly simplify the processes involved in integral calculus, making them a crucial part of a mathematician's toolkit.