Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each improper integral or show that it diverges. \(\int_{0}^{\infty} e^{-x} \cos x d x\)

Step-by-Step Solution

Answer

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

Step 1: Identify the type of integral

The given integral \( \int_{0}^{\infty} e^{-x} \cos x \, dx \) is an improper integral because the upper limit of integration is infinity.

Step 2: Set up the integral for evaluation

To evaluate the improper integral \( \int_{0}^{\infty} e^{-x} \cos x \, dx \), we express it as a limit: \[ \lim_{b \to \infty} \int_{0}^{b} e^{-x} \cos x \, dx \]

Step 3: Integration by parts

Use integration by parts to solve \( \int e^{-x} \cos x \, dx \). Let \( u = \cos x \) and \( dv = e^{-x} dx \). Then, \( du = - \sin x \, dx \) and \( v = -e^{-x} \). The integration by parts formula is:\[ \int u \, dv = uv - \int v \, du\]Applying it gives:\[ \int e^{-x} \cos x \, dx = -e^{-x} \cos x - \int (-e^{-x})(-\sin x) \, dx\]which simplifies to:\[ -e^{-x} \cos x + \int e^{-x} \sin x \, dx\]

Step 4: Repeat integration by parts for the second integral

Apply integration by parts again to \( \int e^{-x} \sin x \, dx \). Let \( u = \sin x \) and \( dv = e^{-x} dx \) which gives \( du = \cos x \, dx \) and \( v = -e^{-x} \). Applying the formula again:\[ \int e^{-x} \sin x \, dx = -e^{-x} \sin x - \int (-e^{-x})(\cos x) \, dx\]This simplifies to:\[ -e^{-x} \sin x - \int e^{-x} \cos x \, dx\]

Step 5: Solve the resulting equation

Now, using the results from Steps 3 and 4, you have:\[ \int e^{-x} \cos x \, dx = -e^{-x} \cos x + \int e^{-x} \sin x \, dx\]From the second integration, substitute back:\[ \int e^{-x} \sin x \, dx = -e^{-x} \sin x - \int e^{-x} \cos x \, dx\]You get:\[ \int e^{-x} \cos x \, dx = -e^{-x} \cos x - e^{-x} \sin x - \int e^{-x} \cos x \, dx\]Rearranging gives:\[ 2 \int e^{-x} \cos x \, dx = -e^{-x} (\cos x + \sin x)\]so:\[ \int e^{-x} \cos x \, dx = -\frac{1}{2}e^{-x} (\cos x + \sin x)\]

Step 6: Apply the limits and evaluate the improper integral

Evaluate \( \lim_{b \to \infty} \int_{0}^{b} e^{-x} \cos x \, dx = \lim_{b \to \infty} \left[ -\frac{1}{2}e^{-x}(\cos x + \sin x) \right]_{0}^{b} \).When evaluated at \(b\) as \(\infty\), the term \(-\frac{1}{2}e^{-b}(\cos b + \sin b)\) approaches zero because \(e^{-b} \to 0\) as \(b \to \infty\).When evaluated at 0, you plug in 0:\[ -\frac{1}{2}e^{0}(\cos 0 + \sin 0) = -\frac{1}{2}(1 + 0) = -\frac{1}{2}\]

Step 7: Conclude the result

Combining the results of the limits, the final value of the integral is:\[ -\frac{1}{2} - 0 = -\frac{1}{2}\]Thus, the improper integral evaluates to \(-\frac{1}{2}\).

Key Concepts

Integration by Parts

Integration by parts is a technique that helps us integrate products of functions. It is particularly useful when dealing with integrals involving two different types of functions, such as a polynomial and an exponential function, or exponential and trigonometric functions as in the exercise here. The formula for integration by parts is derived from the product rule of differentiation:\[ \int u \, dv = uv - \int v \, du \]Here's a simple breakdown:

• Choose which part of the integrand to be \( u \) and which part to be \( dv \). Ideally, \( u \) should be a function that becomes simpler when differentiated.
• Differentiate \( u \) to get \( du \), and integrate \( dv \) to get \( v \).
• Substitute into the formula to find the integral.

In the given problem, we applied integration by parts twice due to the recurring nature of the terms resulting from the product of the exponential function \( e^{-x} \) and trigonometric functions \( \cos x \) and \( \sin x \). By doing so, we were able to solve the integral.

Limits in Calculus

Limits are central to calculus and are used to define continuity, derivatives, and integrals formally. In the context of improper integrals, limits help us make sense of integrals with infinite boundaries or indeterminable integrals.

• An improper integral involves at least one of its limits of integration being infinite, or the function being integrated has an infinite discontinuity.
• To evaluate an improper integral, replace the infinity with a variable. This variable approaches infinity through a limit.

In our original exercise, the upper limit of \( \int_{0}^{\infty} e^{-x} \cos x \, dx \) is infinite. Therefore, we expressed it as a limit:\[ \lim_{b \to \infty} \int_{0}^{b} e^{-x} \cos x \, dx \]This step converts the problem into a format where calculus tools, including differentiation and integration methods like integration by parts, can be applied, treating \( b \) as a finite number that approaches infinity.

Convergence of Integrals

The concept of convergence is crucial when evaluating improper integrals. Just like a series, an integral converges if it has a finite value as its limit approaches infinity (or whichever boundary is improper).

• An integral converges if, when evaluated over its range, it results in a finite number.
• Conversely, if the integral does not result in a finite number, it diverges.

In the given exercise, after evaluating the integral using integration by parts and calculating the limit, we established that:\[ \int_{0}^{\infty} e^{-x} \cos x \, dx = -\frac{1}{2} \]This shows the integral converges to \(-\frac{1}{2}\). Therefore, the improper integral of this function over the range from 0 to infinity converges to this definite value.