Problem 21 - Advanced Engineering Mathematics

Question

(a) From the identity $$\sin A \cos B=\frac{1}{2}[\sin (A+B)+\sin (A-B)]$$ we have $$\begin{aligned} \sin \alpha \cos \alpha x &=\frac{1}{2}[\sin (\alpha+\alpha x)+\sin (\alpha-\alpha x)] \\ &=\frac{1}{2}[\sin \alpha(1+x)+\sin \alpha(1-x)] \\ &=\frac{1}{2}[\sin \alpha(x+1)-\sin \alpha(x-1)] \end{aligned}$$ Then $$\frac{2}{\pi} \int_{0}^{\infty} \frac{\sin \alpha \cos \alpha x}{\alpha} d \alpha=\frac{1}{\pi} \int_{0}^{\infty} \frac{\sin \alpha(x+1)-\sin \alpha(x-1)}{\alpha} d \alpha$$ (b) Noting that $$\begin{aligned} F_{b} &=\frac{1}{\pi} \int_{0}^{b} \frac{\sin \alpha(x+1)-\sin \alpha(x-1)}{\alpha} d \alpha \\ &=\frac{1}{\pi}\left[\int_{0}^{b} \frac{\sin \alpha(x+1)}{\alpha} d \alpha-\int_{0}^{b} \frac{\sin \alpha(x-1)}{\alpha} d \alpha\right] \end{aligned}$$ and letting \(t=\alpha(x+1)\) so that \(d t=(x+1) d \alpha\) in the first integral and \(t=\alpha(x-1)\) so that \(d t=(x-1) d \alpha\) in the second integral we have $$F_{b}=\frac{1}{\pi}\left[\int_{0}^{b(x+1)} \frac{\sin t}{t} d t-\int_{0}^{b(x-1)} \frac{\sin t}{t} d t\right]$$ since \(\operatorname{Si}(x)=\int_{0}^{x}[(\sin t) / t] d t,\) this becomes $$F_{b}=\frac{1}{\pi}[\operatorname{Si}(b(x+1))-\operatorname{Si}(b(x-1))]$$ (c) In Mathematica we define \(\mathbf{f}\left[\mathbf{b}_{-}\right]:=(\mathbf{1} / \mathbf{P} \mathbf{i})(\sin \operatorname{Integral}[\mathbf{b}(\mathbf{x}+\mathbf{1})]-\sin \operatorname{Integral}[\mathbf{b}(\mathbf{x}-\mathbf{1})] .\) Graphs of \(F_{b}(x)\) for \(b=4,6,15,\) and 75 are shown below.

Step-by-Step Solution

Answer

\( F_b(x) = \frac{1}{\pi} [ \operatorname{Si}(b(x+1)) - \operatorname{Si}(b(x-1)) ] \).

Step 1: Understand the Given Identity

We are given the trigonometric identity \( \sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)] \). This identity allows us to express the product \( \sin A \cos B \) as the average of two sine functions.

Step 2: Apply the Identity to the Integrand

Using the identity, we transform \( \sin \alpha \cos \alpha x \) into \( \frac{1}{2}[\sin \alpha(x+1) + \sin \alpha(x-1)] \). This replaces the original product in the integral with a sum of two integrals, allowing us to proceed with integration.

Step 3: Set Up the Required Integral

Transform the original integral \( \int_{0}^{\infty} \frac{\sin \alpha \cos \alpha x}{\alpha} d\alpha \) using the transformed expression from Step 2. We get \( \frac{1}{\pi} \int_{0}^{\infty} \frac{\sin \alpha(x+1) - \sin \alpha(x-1)}{\alpha} \).

Step 4: Evaluate the Integral Using Substitution

Introducing substitution for each integral, let \( t = \alpha(x+1) \) for the first part and \( t = \alpha(x-1) \) for the second part. Change the differential \( d\alpha \) accordingly; this simplifies each integral to \( \frac{1}{\pi} [ \int_{0}^{b(x+1)} \frac{\sin t}{t} dt - \int_{0}^{b(x-1)} \frac{\sin t}{t} dt ] \).

Step 5: Use the Sine Integral Function

Recognize that the integrals represent the definition of the sine integral function \( \operatorname{Si}(x) = \int_{0}^{x} \frac{\sin t}{t} dt \). Therefore, the expression simplifies to \( \frac{1}{\pi} [ \operatorname{Si}(b(x+1)) - \operatorname{Si}(b(x-1)) ] \).

Step 6: Abstract the Result to a Function

The final expression for \( F_b(x) \) is given by \( F_b(x) = \frac{1}{\pi} [ \operatorname{Si}(b(x+1)) - \operatorname{Si}(b(x-1)) ] \). This presents us with a solvable function of \( x \) and \( b \), suitable for graphing or further analysis in computational tools.

Key Concepts

Trigonometric Identities

A key part of solving problems involving trigonometric functions is understanding and applying trigonometric identities. These identities help simplify expressions and make complex problems more manageable. The identity \( \sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)] \) is especially useful when dealing with products of sine and cosine functions. Instead of multiplying these functions directly, the identity allows us to transform the product into a sum of sine functions.
This simplification is invaluable when tackling integrals involving trigonometric functions. In the given problem, this identity was used to rewrite \( \sin \alpha \cos \alpha x \) as \( \frac{1}{2}[\sin \alpha(x+1) + \sin \alpha(x-1)] \). This transformation changed the original integral into a problem that could be approached using the properties of sine functions.
By breaking down the product into these sum terms, we open up the way for using other mathematical tools like integration and transformation techniques. Learning to recognize when and how to apply trigonometric identities is a fundamental skill for anyone studying calculus, math, or physics.

Integration Techniques

Integration is a crucial mathematical technique that allows us to find the area under curves, among other applications. In this context, integration transforms a problem by aggregating values over a continuous range. However, integrating trigonometric functions, especially with complicated expressions, requires specific techniques.
Substitution is one of the core techniques used in this exercise. When given the integral \( \int_{0}^{\infty} \frac{\sin \alpha(x+1) - \sin \alpha(x-1)}{\alpha} d \alpha \), substitution helps simplify the expression. By letting \( t = \alpha(x+1) \) for the first integral and \( t = \alpha(x-1) \) for the second, the variables become more manageable, transforming each part of the integral into a standard form that can be solved using known methods.

• This change of variables aligns the integral with the sine integral function, allowing simplification.
• It effectively reduces the complexity of the integral by removing dependencies on \( \alpha \), leaving only the new variable \( t \) to confront.

Such techniques as substitution are fundamental parts of calculus, often serving as the bridge between complex trigonometric expressions and solvable integrals.

Sine Integral Function

The sine integral function, denoted as \( \operatorname{Si}(x) \), plays a significant role in evaluating integrals involving sine functions. It is defined as \( \operatorname{Si}(x) = \int_{0}^{x} \frac{\sin t}{t} dt \), providing a succinct way to express these otherwise complex integrals.
The sine integral appears in many physics and engineering problems, often describing oscillations and waveforms. In our problem, recognizing the integral as a sine integral function allows a direct path to the solution. This is because the integral \( \int_{0}^{b(x+1)} \frac{\sin t}{t} dt \) naturally aligns with the definition of \( \operatorname{Si}(x) \).

• The transformation of the original integral via substitution leads directly to the sine integral form.
• The relationship between the bounds of integration and the function inside correlates to how we apply the sine integral definition.

Understanding and employing special functions like the sine integral can simplify the calculation process, especially for those involved in higher-level math and science. Mastery of these concepts can enhance problem-solving capabilities significantly.