Problem 30 - Advanced Engineering Mathematics

Question

$$\begin{aligned} &\int_{-\infty}^{\infty} \frac{x e^{i x}}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x=2 \pi i[\operatorname{Res}(f(z), i)+\operatorname{Res}(f(z), 2 i)]=2 \pi i\left[\frac{1}{6} e^{-1}-\frac{1}{6} e^{-2}\right]=\frac{\pi}{3}\left(e^{-1}-e^{-2}\right) i\\\ &\int_{-\infty}^{\infty} \frac{x \sin x}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x=\operatorname{Im}\left(\int_{-\infty}^{\infty} \frac{x e^{i x}}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x\right)=\frac{\pi}{3}\left(e^{-1}-e^{-2}\right) \end{aligned}$$ Therefore, $$\int_{0}^{\infty} \frac{x \sin x}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x=\frac{1}{2}\left[\frac{\pi}{3}\left(e^{-1}-e^{-2}\right)\right]=\frac{\pi}{6}\left(e^{-1}-e^{-2}\right)$$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{6}(e^{-1}-e^{-2}) \).

Step 1: Understanding the Integral

The task is to evaluate the definite integral \( \int_{0}^{\infty} \frac{x \sin x}{ (x^{2}+1)(x^{2}+4)} dx \). This involves using properties of complex functions and residues.

Step 2: Use of Complex Functions

We first consider the integral \( \int_{-\infty}^{\infty} \frac{x e^{ix}}{(x^{2}+1)(x^{2}+4)} dx \) to solve the target integral. The given solution employs the relation of complex functions to simplify the calculation, hinting at using residue theory.

Step 3: Contribution of Residues

The integral along the entire real line is handled by evaluating residues at poles of the function \( f(z) = \frac{z e^{iz}}{(z^{2}+1)(z^{2}+4)} \). The poles occur at \( z = i \) and \( z = 2i \) since they lie in the upper half-plane.

Step 4: Calculation of Residues

The residues are calculated as follows: \( \operatorname{Res}(f(z), i) \) and \( \operatorname{Res}(f(z), 2i) \), which are given to yield \( \frac{1}{6} e^{-1} \) and \(-\frac{1}{6} e^{-2} \), respectively.

Step 5: Evaluate the Definite Integral Using Imaginary Part

The imaginary part of the integral \( \int_{-\infty}^{\infty} \frac{x e^{ix}}{(x^{2}+1)(x^{2}+4)} dx \) equals \( \int_{-\infty}^{\infty} \frac{x \sin x}{(x^{2}+1)(x^{2}+4)} dx \). It results in \( \frac{\pi}{3}(e^{-1}-e^{-2}) \).

Step 6: Compute Half-Integral from Zero to Infinity

Finally, because the sine function is odd, the integral over \([0, \infty)\) is half the integral over \([-\infty, \infty)\), leading to \( \frac{1}{2} \left( \frac{\pi}{3}(e^{-1}-e^{-2}) \right) = \frac{\pi}{6}(e^{-1}-e^{-2}) \).

Key Concepts

Complex Analysis

Complex analysis is a branch of mathematics focusing on functions of complex numbers. These functions hold intriguing properties. One key aspect is how they can extend the concept of calculus from real numbers to complex numbers. This extension allows us to explore integrals of complex-valued functions and uncover behaviors not seen in real-valued functions.

• Complex Numbers: These numbers have the form \( a + bi \), where \( a \) is the real part, and \( bi \) is the imaginary part. Imaginary numbers arise from the square roots of negative numbers.
• Function Behavior: Complex functions often exhibit unique traits like differentiability everywhere within their domain, where they are termed analytic.
• Integration over Contours: In complex analysis, integration can be performed over contours in the complex plane. This is vital for evaluating integrals like those encountered with the use of the Residue Theorem.

The basis of using complex analysis in definite integrals, as seen in our problem, centers on transforming a real integral into a contour integral, which can sometimes be simpler to evaluate.

Definite Integrals

Definite integrals are a fundamental concept in calculus. They provide a way to calculate the area under a curve defined by a function within set limits. Unlike an indefinite integral, which introduces a constant of integration, the definite integral results in a numerical value. This represents the enclosed area between the function, the limits of integration, and the x-axis.

• Limits of Integration: These denote the interval over which the integration occurs, such as \( [0, \infty) \) seen in the given problem.
• Real-Valued Functions: Definite integrals typically apply to real-valued functions, but complex analysis takes them into the realm of complex numbers.
• Application in Real Life: Definite integrals are used in various fields, such as physics and engineering, to compute quantities like the area, distance, and volume.

In the problem outlined, the definite integral transforms through complex analysis to enable the evaluation using residues. This approach changes the integration path, allowing the influence of poles in the complex plane.

Residue Calculation

Residue calculation is a central element of the Residue Theorem in complex analysis. It involves finding the residue of a function at its poles, which are points where a complex function becomes undefined. Understanding residues helps in evaluating contour integrals, especially when dealing with definite integrals that stretch over an infinite range.

• Pole Identification: A pole is a type of singularity where a function takes on an infinite value. This exercise involves poles at \( z = i \) and \( z = 2i \).
• Residue Theorem: This theorem allows us to evaluate integrals by summing the residues of a function's poles within a contour. It simplifies complex integral evaluations significantly.
• Calculation Method: The residues at \( z = i \) and \( z = 2i \) were calculated using known formulas, resulting in \( \frac{1}{6} e^{-1} \) and \(-\frac{1}{6} e^{-2} \) respectively.

Residue calculation helps convert an integral from a real to a complex analysis form, making the evaluation of seemingly difficult real integrals more approachable.