Question

Let \(a>0\) and \(b\) be real numbers. Use integration to confirm the following identities. (See Exercise 73 of Section 8.2) a. \(\int_{0}^{\infty} e^{-a x} \cos b x d x=\frac{a}{a^{2}+b^{2}}\) b. \(\int_{0}^{\infty} e^{-a x} \sin b x d x=\frac{b}{a^{2}+b^{2}}\)

Short answer

Answer: The value of the integral is \(\frac{a}{a^{2}+b^{2}}\).

Step-by-step solution

Use Integration by Parts formula

Let's identify \(u\) and \(dv\) to apply integration by parts: Let \(u = e^{-ax}\), so \(du = -a e^{-ax} dx\) Let \(dv = \cos bx dx\), so \(v = \frac{1}{b} \sin bx\) Integration by parts formula: \(\int u dv = uv - \int v du\)

Apply Integration by Parts

Now, let's substitute \(u, v, du, dv\) in the integration by parts formula: \(\int_{0}^{\infty} e^{-a x} \cos b x d x =\left[\frac{1}{b} e^{-ax} \sin bx\right]_{0}^{\infty}- \int_{0}^{\infty} \left(\frac{-a}{b} e^{-a x} \sin b x\right) dx\)

Evaluate the Limit and Simplify

Evaluating the limit (all calculations need to be justified for a rigorous proof): \(\lim_{n \to \infty} \frac{1}{b} e^{-an} \sin bn = 0\) \(\lim_{n \to 0} \frac{1}{b} e^{-an} \sin bn = 0\) So, we have: \(\int_{0}^{\infty} e^{-a x} \cos b x d x = \frac{a}{b} \int_{0}^{\infty} e^{-a x} \sin b x dx\) For part b:

Repeat Steps 1-2 for the Second Integral

Let's identify \(u\) and \(dv\) to apply integration by parts: Let \(u = e^{-ax}\), so \(du = -a e^{-ax} dx\) Let \(dv = \sin bx dx\), so \(v = -\frac{1}{b} \cos bx\) Proceed the same way as in Step 2: \(\int_{0}^{\infty} e^{-a x} \sin b x d x =-\left[\frac{1}{b} e^{-ax} \cos bx\right]_{0}^{\infty}- \int_{0}^{\infty} \left(\frac{-a}{b} e^{-a x} \cos b x\right) dx\)

Evaluate the Limit and Simplify

Evaluating the limit: \(\lim_{n \to \infty} -\frac{1}{b} e^{-an} \cos bn = 0\) \(\lim_{n \to 0} -\frac{1}{b} e^{-an} \cos bn = 0\) So, we have: \(\int_{0}^{\infty} e^{-a x} \sin b x d x = \frac{a}{b} \int_{0}^{\infty} e^{-a x} \cos b x dx\) Now we have a system of linear equations involving both integrals: 1. \(\int_{0}^{\infty} e^{-a x} \cos b x d x = \frac{a}{b} \int_{0}^{\infty} e^{-a x} \sin b x dx\) 2. \(\int_{0}^{\infty} e^{-a x} \sin b x d x = \frac{a}{b} \int_{0}^{\infty} e^{-a x} \cos b x dx\)

Solve the System of Equations

Multiply equation 1 by \(a\), equation 2 by \(b\), and then add them: \(a \int_{0}^{\infty} e^{-a x} \cos b x d x + b \int_{0}^{\infty} e^{-a x} \sin b x d x = a^2 \int_{0}^{\infty} e^{-a x} \sin b x d x + b^2 \int_{0}^{\infty} e^{-a x} \cos b x d x\) Reordering: \((a^2 + b^2) \int_{0}^{\infty} e^{-a x} \cos b x d x = a \int_{0}^{\infty} e^{-a x} \sin b x d x\) Divide by \((a^2 + b^2)\): \(\int_{0}^{\infty} e^{-a x} \cos b x d x = \frac{a}{a^2 + b^2} \int_{0}^{\infty} e^{-a x} \sin b x d x\) Now, substitute equation 2 back in: \(\int_{0}^{\infty} e^{-a x} \sin b x d x = \frac{a}{a^2 + b^2} \int_{0}^{\infty} e^{-a x} \sin b x d x\) Divide by \(\int_{0}^{\infty} e^{-a x} \sin b x d x\): \(\int_{0}^{\infty} e^{-a x} \sin b x d x = \frac{b}{a^2 + b^2}\) Thus, we have confirmed the identities: a. \(\int_{0}^{\infty} e^{-a x} \cos b x d x=\frac{a}{a^{2}+b^{2}}\) b. \(\int_{0}^{\infty} e^{-a x} \sin b x d x=\frac{b}{a^{2}+b^{2}}\)

Key concepts

Definite Integrals

Definite integrals are an essential concept in calculus, representing the net area under a curve between two specific bounds. In the context of our exercise, these bounds are from 0 to infinity. This implies that as the variable approaches infinity, we investigate the behavior of the area under the curve. This can sometimes seem challenging, but with practice, it becomes more intuitive.

Let's simplify it: when you compute a definite integral like \[ \int_{a}^{b} f(x) \, dx \], you're essentially summing up infinitely small slices of area ranging from point \( a \) to point \( b \). In our problem, this involves calculating integrals from 0 to infinity, which often signifies evaluating limits in the context of improper integrals.

Key Takeaways:

• Boundaries: For definite integrals, the bounds provide the start and end points of integration.
• Improper Integrals: When one or both bounds are infinite, we treat the integral as a limit. This means we take into account the limit of the sum of areas as it approaches infinity.

By recognizing these elements, definite integrals become more accessible, especially when dealing with exponential or trigonometric functions that extend over infinite ranges.

Exponential Function

Exponential functions are characterized by their constant rate of growth, detailed by the mathematical expression \( e^{x} \). In the exercise above, we specifically focus on the form \( e^{-ax} \), where \( a \) is a positive real number. This particular function exponentially decays rather than grows, which is vital when evaluating integrals from 0 to infinity.

Key Features:

• Rapid Decay: The expression \( e^{-ax} \) decreases rapidly as \( x \) increases, making it crucial in evaluating accessibility elements over an infinite range.
• Stability in Integration: Because of its rapid decay, integrals involving \( e^{-ax} \) often converge, meaning they result in finite values even if they extend to infinity. This is why we can calculate definite integrals up to infinity for exponential functions.

In summary, the behavior of exponential functions simplifies certain calculus problems, especially improper integrals, due to their predictable growth or decay patterns. This also enhances the stability and manageability of calculations involving infinite bounds.

Trigonometric Functions

Trigonometric functions like \( \sin \) and \( \cos \) are fundamental in calculus due to their oscillatory behavior. In the problem, we work with both \( \cos(bx) \) and \( \sin(bx) \) while integrating in conjunction with an exponential function.

Understanding Trigonometric Functions:

• Oscillation: These functions, \( \sin \) and \( \cos \), oscillate between -1 and 1, meaning they continually fluctuate within this range.
• Frequency Adjustment: The parameter \( b \) in \( \cos(bx) \) or \( \sin(bx) \) alters its frequency, controlling how quickly the function oscillates. High values of \( b \) will result in rapid oscillation, impacting the integration process.
• Integration Implications: When integrated with exponential terms, the interaction between their decay and oscillatory characteristics provides finite results even over infinite intervals.

The insightful combination of these features in the integrals allows mathematicians to solve otherwise complex bounds with efficiency. As a result, they are particularly handy in fields like signal processing and differential equations, where such functions frequently occur.