Problem 3
Question
From formulas (5) and (6) in the text $$\begin{aligned} A(\alpha) &=\int_{0}^{3} x \cos \alpha x d x=\left.\frac{x \sin \alpha x}{\alpha}\right|_{0} ^{3}-\frac{1}{\alpha} \int_{0}^{3} \sin \alpha x d x \\ &=\frac{3 \sin 3 \alpha}{\alpha}+\left.\frac{\cos \alpha x}{\alpha^{2}}\right|_{0} ^{3}=\frac{3 \alpha \sin 3 \alpha+\cos 3 \alpha-1}{\alpha^{2}} \end{aligned}$$ and $$\begin{aligned} B(\alpha) &=\int_{0}^{3} x \sin \alpha x d x=-\left.\frac{x \cos \alpha x}{\alpha}\right|_{0} ^{3}+\frac{1}{\alpha} \int_{0}^{3} \cos \alpha x d x \\ &=-\frac{3 \cos 3 \alpha}{\alpha}+\left.\frac{\sin \alpha x}{\alpha^{2}}\right|_{0} ^{3}=\frac{\sin 3 \alpha-3 \alpha \cos 3 \alpha}{\alpha^{2}} \end{aligned}$$ Hence $$\begin{aligned} f(x) &=\frac{1}{\pi} \int_{0}^{\infty} \frac{(3 \alpha \sin 3 \alpha+\cos 3 \alpha-1) \cos \alpha x+(\sin 3 \alpha-3 \alpha \cos 3 \alpha) \sin \alpha x}{\alpha^{2}} d \alpha \\ &=\frac{1}{\pi} \int_{0}^{\infty} \frac{3 \alpha(\sin 3 \alpha \cos \alpha x-\cos 3 \alpha \sin \alpha x)+\cos 3 \alpha \cos \alpha x+\sin 3 \alpha \sin \alpha x-\cos \alpha x}{\alpha^{2}} d \alpha \\ &=\frac{1}{\pi} \int_{0}^{\infty} \frac{3 \alpha \sin \alpha(3-x)+\cos \alpha(3-x)-\cos \alpha x}{\alpha^{2}} d \alpha \end{aligned}$$
Step-by-Step Solution
Verified Answer
The solution involves trigonometric integration and simplification using identities to evaluate the integral over \( \alpha \).
1Step 1: Understand the given formulas
The given formulas contain integrals of trigonometric functions \( \cos(\alpha x) \) and \( \sin(\alpha x) \) weighted by \( x \), both on the interval from 0 to 3. The task requires the integration over these trigonometric terms for function \( f(x) \).
2Step 2: Division of Integral into Parts
The integral for function \( f(x) \) consists of two main terms: one containing \( \cos(\alpha x) \) and another containing \( \sin(\alpha x) \). We need to evaluate each part separately through substitution of the corresponding terms from formulas for \( A(\alpha) \) and \( B(\alpha) \).
3Step 3: Simplifying the Argument
Note the expressions for \( A(\alpha) \) and \( B(\alpha) \) include a decomposition of the trigonometric functions into sine and cosine of the differences \( \alpha(3-x) \). This aids the integration by transforming the multiplied sine and cosine terms to a single sine function using angle sum identities.
4Step 4: Application of Trigonometric Identities
In function \( f(x) \), expressions \( \sin(3\alpha)\cos(\alpha x) - \cos(3\alpha)\sin(\alpha x) = \sin(\alpha(3-x)) \) due to angle subtraction formulas are used. Similarly, \( \cos(3\alpha)\cos(\alpha x) + \sin(3\alpha)\sin(\alpha x) = \cos(\alpha(3-x)) \).
5Step 5: Final Form of the Integral
The integral simplifies the terms to \( \frac{3 \alpha \sin(\alpha(3-x)) + \cos(\alpha(3-x)) - \cos(\alpha x)}{\alpha^2} \). This compact form is then integrated over \( \alpha \) from 0 to infinity.
6Step 6: Integral Evaluation
Integrate results utilizing properties of Fourier integrals and possibly Laplace transforms, considering integration limits from 0 to infinity and properties of improper integrals. Evaluate through integration techniques suitable for such transforms.
Key Concepts
Improper IntegralsFourier IntegralsTrigonometric Integrals
Improper Integrals
When working with improper integrals, we're dealing with integrals that have at least one infinite limit of integration or an integrand that becomes infinite within the integration limits. These integrals can seem a bit intimidating at first, but they are essential for evaluating functions over unbounded domains or where the integrand has infinite discontinuities.
The exercise provided involved evaluating an integral with an upper limit approaching infinity. Proper techniques, such as treating the integral as a limit, help solve these types of integrals. For example, if you have an integral of the form \[ \int_{a}^{b} f(x) \, dx \]where either \(a\) or \(b\) is infinite, we express it as a limit, \[ \lim_{c \to \infty} \int_{a}^{c} f(x) \, dx \] This approach helps us handle infinite limits and analyze the behavior of the function as it stretches out to infinity. Always ensure to check the convergence of the integral, as improper integrals can diverge if not properly evaluated. Our task involved using such techniques to evaluate an integral resulting from Fourier and trigonometric components over an infinite domain.
The exercise provided involved evaluating an integral with an upper limit approaching infinity. Proper techniques, such as treating the integral as a limit, help solve these types of integrals. For example, if you have an integral of the form \[ \int_{a}^{b} f(x) \, dx \]where either \(a\) or \(b\) is infinite, we express it as a limit, \[ \lim_{c \to \infty} \int_{a}^{c} f(x) \, dx \] This approach helps us handle infinite limits and analyze the behavior of the function as it stretches out to infinity. Always ensure to check the convergence of the integral, as improper integrals can diverge if not properly evaluated. Our task involved using such techniques to evaluate an integral resulting from Fourier and trigonometric components over an infinite domain.
Fourier Integrals
Fourier integrals are a tool used to represent a function as a sum of sine and cosine terms. This is particularly useful for signal processing, where these integrals can express a signal in terms of its frequency components.
In this exercise, Fourier integrals were critical, being employed to decompose the functions into their frequency components. This involves an integral of the form \[ f(x) = \int_{-\infty}^{\infty} F(\alpha) e^{i\alpha x} \, d\alpha \]where \( F(\alpha) \) represents the Fourier transform of \( f(x) \). However, when dealing with real-valued functions, it's often simplified using sine and cosine terms as shown in the formulas provided.
A key step in the solution process was rewriting trigonometric terms to express the function using its Fourier representation. This transforms complex trigonometric combinations into simpler forms which are easier to integrate, typically resulting in easier management of complex exponential terms.
Understanding the connection between Fourier integrals and transformations helps in evaluating the transforms of periodic signals, optimizing analysis in practical and theoretical applications.
In this exercise, Fourier integrals were critical, being employed to decompose the functions into their frequency components. This involves an integral of the form \[ f(x) = \int_{-\infty}^{\infty} F(\alpha) e^{i\alpha x} \, d\alpha \]where \( F(\alpha) \) represents the Fourier transform of \( f(x) \). However, when dealing with real-valued functions, it's often simplified using sine and cosine terms as shown in the formulas provided.
A key step in the solution process was rewriting trigonometric terms to express the function using its Fourier representation. This transforms complex trigonometric combinations into simpler forms which are easier to integrate, typically resulting in easier management of complex exponential terms.
Understanding the connection between Fourier integrals and transformations helps in evaluating the transforms of periodic signals, optimizing analysis in practical and theoretical applications.
Trigonometric Integrals
Trigonometric integrals involve integrating functions that contain trigonometric functions like sine and cosine. In the provided exercise, integrals such as \[ \int x \cos(\alpha x) \, dx \] play a central role.
The strategy is often to use integration by parts, which allows one to reduce a complex integral into more manageable parts. Integration by parts is represented by the formula:\[ \int u \, dv = uv - \int v \, du \]For trigonometric integrals, this method can transform an integral of a product of two functions into more straightforward integrals.
Throughout the solution, trigonometric identities also provided simplifications, particularly in turning products of trigonometric functions into sums or differences. Identities like \(\cos(a)\cos(b) + \sin(a)\sin(b) = \cos(a-b)\) were pivotal.
These identities reduce the complexity of integrating products of sins and cosines, allowing the exploiting of their periodic properties and symmetry effectively, which is fundamental in deriving final solutions for trigonometric integrals.
The strategy is often to use integration by parts, which allows one to reduce a complex integral into more manageable parts. Integration by parts is represented by the formula:\[ \int u \, dv = uv - \int v \, du \]For trigonometric integrals, this method can transform an integral of a product of two functions into more straightforward integrals.
Throughout the solution, trigonometric identities also provided simplifications, particularly in turning products of trigonometric functions into sums or differences. Identities like \(\cos(a)\cos(b) + \sin(a)\sin(b) = \cos(a-b)\) were pivotal.
These identities reduce the complexity of integrating products of sins and cosines, allowing the exploiting of their periodic properties and symmetry effectively, which is fundamental in deriving final solutions for trigonometric integrals.
Other exercises in this chapter
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