Problem 53 - Calculus Single Variable

Question

Determine whether the given improper integral converges or diverges. If it converges, then evaluate it. $$ \int_{0}^{\infty} e^{-x} \cos (x) d x $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Type of Integral

The given integral is an improper integral of the form \( \int_{0}^{\infty} e^{-x} \cos(x) \, dx \), which extends to infinity. To solve it, we should first assess its convergence.

Step 2: Use Limit Definition for Improper Integrals

We express the integral using limits as: \[\lim_{b \to \infty} \int_{0}^{b} e^{-x} \cos(x) \, dx.\]This allows us to consider convergence as \( b \to \infty \).

Step 3: Integrate by Parts

We perform integration by parts where \( u = \cos(x) \) and \( dv = e^{-x} \, dx \), implying \( du = -\sin(x) \, dx \) and \( v = -e^{-x} \):\[\int e^{-x} \cos(x) \, dx = -e^{-x} \cos(x) + \int e^{-x} \sin(x) \, dx.\]This creates a new integral, \( \int e^{-x} \sin(x) \, dx \), which we will solve using integration by parts again.

Step 4: Integrate the New Integral by Parts Again

Apply integration by parts again on \( \int e^{-x} \sin(x) \, dx \) with \( u = \sin(x) \) and \( dv = e^{-x} \, dx \). Then, \( du = \cos(x) \, dx \) and \( v = -e^{-x} \):\[\int e^{-x} \sin(x) \, dx = -e^{-x} \sin(x) - \int e^{-x} \cos(x) \, dx.\]

Step 5: Solve for the Original Integral

By combining results from the integration by parts steps, we get:\[I = \int e^{-x} \cos(x) \, dx = -e^{-x} \cos(x) - e^{-x} \sin(x) + I.\]So,\[2I = -e^{-x} (\cos(x) + \sin(x)).\]Solving for \( I \):\[I = -\frac{1}{2} e^{-x} (\cos(x) + \sin(x)).\]

Step 6: Evaluate the Limit

Now, evaluate this at the bounds from 0 to \( b \):\[\lim_{b \to \infty} \left( -\frac{1}{2} e^{-b} (\cos(b) + \sin(b)) - (-\frac{1}{2}(\cos(0) + \sin(0))) \right).\]As \( b \to \infty \), \( e^{-b} \) approaches 0, so the first term vanishes. At \( x = 0 \), the expression evaluates to \(-\frac{1}{2} (1 + 0) = -\frac{1}{2}\).

Step 7: Conclude Convergence and Value

Since the integral evaluates to a finite number \(-\frac{1}{2}\), the improper integral converges. The final result is \(-\frac{1}{2}\).

Key Concepts

Integration by Parts

Integration by parts is a handy technique used to find integrals of products of functions. It is particularly useful when straightforward integration is difficult. The formula for integration by parts is derived from the product rule of differentiation and is given by:
\[ \int u \, dv = uv - \int v \, du \]This method involves choosing parts of the integrand: one as \( u \) and the other as \( dv \). The intention is to simplify the integration.

To illustrate, take the example from the exercise: \( \int e^{-x} \cos(x) \, dx \). We choose:

• \( u = \cos(x) \), which gives \( du = -\sin(x) \, dx \).
• \( dv = e^{-x} \, dx \), which gives \( v = -e^{-x} \).

Substitute these into the formula and solve step by step. If the resulting integral is still challenging, you may need to apply integration by parts again, as was required in the exercise. This iterative process is often useful for oscillatory functions or when the integral repeats its form after partial integration.

Improper Integrals

Improper integrals are integrals with one or more infinite limits or with integrands that become infinite within the limits of integration. In these cases, standard integration techniques do not apply directly, as the function behaves in a complex manner where it approaches infinity.

The integral in the exercise is improper because it extends to infinity: \( \int_{0}^{\infty} e^{-x} \cos(x) \, dx \). To determine whether such an integral converges or diverges, we generally use limits to redefine the integral:

• Replace the infinite limit with a variable, say \( b \), so \( \int_{0}^{\infty} \) becomes \( \int_{0}^{b} \).
• Then evaluate the limit as \( b \to \infty \).

If the limit exists and is finite, the integral converges. If not, it diverges. It's critical to correctly address the behavior of the function at both ends of the integration range to ensure a rigorous interpretation of convergence.

Limit Definition for Integrals

The limit definition for integrals is crucial in dealing with improper integrals, especially those with infinite bounds or an infinite discontinuity. This approach allows us to interpret these otherwise problematic integrals within a framework similar to regular definite integrals.

The process involves expressing the improper integral as a limit of a proper integral. The idea is to approximate the area under the curve by considering integrals with finite bounds that extend to the original limits:

• Use \( \int_{0}^{b} \) instead of \( \int_{0}^{\infty} \).
• Evaluate \( \lim_{b \to \infty} \int_{0}^{b} f(x) \, dx \).

This method lets us determine whether the area converges to a specific number or not. For the integral in question, the function \( e^{-x} \cos(x) \) diminishes to zero as \( x \) approaches infinity, aiding in establishing convergence. Thus, using limits plays a pivotal role in determining the convergence behavior of the function over infinite intervals.