Problem 12 - Advanced Engineering Mathematics

Question

The function is odd. Thus from formula (11) in the text $$B(\alpha)=\int_{0}^{\infty} x e^{-x} \sin \alpha x d x$$ Now recall $$\mathscr{L}\\{t \sin k t\\}=-\frac{d}{d s} \mathscr{L}\\{\sin k t\\}=2 k s /\left(s^{2}+k^{2}\right)^{2}$$ If we set \(s=1\) and \(k=\alpha\) we obtain $$B(\alpha)=\frac{2 \alpha}{\left(1+\alpha^{2}\right)^{2}}$$ Hence from formula (10) in the text $$f(x)=\frac{4}{\pi} \int_{0}^{\infty} \frac{\alpha \sin \alpha x}{\left(1+\alpha^{2}\right)^{2}} d \alpha$$

Step-by-Step Solution

Answer

The function \( f(x) \) is determined by \( f(x) = \frac{4}{\pi} \int_{0}^{\infty} \frac{\alpha \sin \alpha x}{(1+\alpha^{2})^{2}} \, d\alpha \).

Step 1: Understanding the Problem

The problem involves determining a function based on its integral representation and relates to the Laplace transform. We have a function \(B(\alpha)\) that is expressed in terms of an integral, where \(\alpha\) is a parameter.

Step 2: Reviewing Given Formulae

We are given the integral \(B(\alpha) = \int_{0}^{\infty} x e^{-x} \sin \alpha x \, dx\) and the Laplace transform \(\mathscr{L}\{t \sin k t\}=-\frac{d}{ds} \mathscr{L}\{\sin k t\}=2 k s /\left(s^{2}+k^{2}\right)^{2}\).

Step 3: Substituting Values

Substitute \(s = 1\) and \(k = \alpha\) into the given Laplace transform formula to find \(B(\alpha)\). This substitution yields \(B(\alpha) = \frac{2 \alpha}{(1+\alpha^{2})^{2}}\).

Step 4: Using Formula (10) to Find f(x)

Formula (10) suggests that \( f(x) = \frac{4}{\pi} \int_{0}^{\infty} \frac{\alpha \sin \alpha x}{(1+\alpha^{2})^{2}} \, d\alpha\). This relates the integral result to a specific function \( f(x) \) of interest, using the known \( B(\alpha) \) expression.

Step 5: Interpreting the Final Expression

The function \( f(x) \) is expressed in terms of an integral involving \(\alpha\) and uses the previously determined expression for \(B(\alpha)\) within this integral formulation. This is a common approach in Fourier transform and inverse Laplace transform calculations.

Key Concepts

Fourier Transform

The Fourier Transform is a powerful mathematical tool that transforms a time domain signal into its corresponding frequency domain representation. This helps analyze the frequencies that comprise a given function. In the context of the exercise, while it directly discusses the Laplace transform, Fourier transforms are closely related as they both handle frequency components of functions.

Fourier transforms are particularly useful for solving differential equations and analyzing systems. Imagine you have a complex signal. The Fourier Transform breaks it down into a sum of sine and cosine functions. This can greatly simplify many mathematical problems and is widely used in engineering, physics, and applied mathematics.

The transform of a function \( f(t) \) is given by:

• \[ F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-j \omega t} \, dt \]

Here, \( \omega \) represents the angular frequency, and \( j \) is the imaginary unit. This representation highlights how frequencies are distributed in a signal.

Integral Representation

Integral representation is a technique used to express functions or solutions to problems through integrals. It’s particularly useful in handling problems that seem too complex to solve directly. In this exercise, the function \( B(\alpha) \) is expressed as \( \int_{0}^{\infty} x e^{-x} \sin \alpha x \, dx \).

This form of representation allows us to use known integral solutions or transform techniques, like Laplace or Fourier, to evaluate or simplify the function. The integral representation plays a crucial role in many areas such as solving differential equations, evaluating complex functions, and finding solutions that satisfy boundary conditions.

For newcomers, understanding how to interpret \( \int \) notations and practice solving these integrals can shed light on a wide array of applications that utilize this approach.

Differential Equations

Differential equations are equations that involve derivatives of a function. They are essential in describing natural phenomena, engineering systems, and many scientific fields. The exercise indirectly pertains to solving differential equations as both Laplace and Fourier transforms are fundamental tools used in this process.

Solving a differential equation often looks like discovering a function or set of functions that satisfy the relationship specified by the derivatives. In the realm of Fourier and Laplace transforms, these tools help transform the differential equations into algebraic equations, which are easier to handle.

For example, if you have a simple differential equation, you might use the Laplace transform to transition it into an algebraic equation. After solving it in this transformed domain, the inverse transform is applied to return to the time domain solution.

Understanding differential equations and their solutions is paramount, as they are the language through which much of science communicates patterns and predictions.