Problem 27 - Advanced Engineering Mathematics

Question

$$\begin{aligned} \int_{-\infty}^{\infty} \frac{e^{2 i x}}{x^{4}+1} d x &=2 \pi i\left[\operatorname{Res}\left(f(z), \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} i\right)+\operatorname{Res}\left(f(z),-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} i\right)\right] \\ &=2 \pi i\left[\left(-\frac{\sqrt{2}}{8}-\frac{\sqrt{2}}{8} i\right) e^{(-\sqrt{2}+\sqrt{2} i)}+\left(\frac{\sqrt{2}}{8}-\frac{\sqrt{2}}{8} i\right) e^{(-\sqrt{2}-\sqrt{2} i)}\right] \\ &=\pi e^{-\sqrt{2}}\left[\frac{\sqrt{2}}{2} \cos \sqrt{2}+\frac{\sqrt{2}}{2} \sin \sqrt{2}\right] \end{aligned}$$ $$\int_{-\infty}^{\infty} \frac{\cos 2 x}{x^{4}+1} d x=\operatorname{Re}\left(\int_{-\infty}^{\infty} \frac{e^{2 i x}}{x^{4}+1} d x\right)=\pi e^{-\sqrt{2}}\left[\frac{\sqrt{2}}{2} \cos \sqrt{2}+\frac{\sqrt{2}}{2} \sin \sqrt{2}\right]$$ Therefore $$\int_{0}^{\infty} \frac{\cos 2 x}{x^{4}+1} d x=\pi e^{-\sqrt{2}} \frac{\sqrt{2}}{4}(\cos \sqrt{2}+\sin \sqrt{2})$$

Step-by-Step Solution

Answer

\( \int_{0}^{\infty} \frac{\cos 2x}{x^4+1} dx = \pi e^{-\sqrt{2}} \frac{\sqrt{2}}{4} (\cos \sqrt{2} + \sin \sqrt{2}) \).

Step 1: Understanding the Given Integral

The problem involves evaluating the integral \( \int_{-\infty}^{\infty} \frac{\cos 2x}{x^4+1} \, dx \). Since \( \cos 2x \) is the real part of \( e^{2ix} \), we can use contour integration and the residue theorem to evaluate this integral by considering the complex function \( \int_{-\infty}^{\infty} \frac{e^{2ix}}{x^4+1} \, dx \).

Step 2: Applying the Residue Theorem

We start by applying the residue theorem to evaluate \( \int_{-\infty}^{\infty} \frac{e^{2ix}}{x^4+1} \, dx \). This involves identifying the poles of the function \( f(z) = \frac{e^{2iz}}{z^4 + 1} \) that are located in the upper half-plane. The poles in the upper half-plane are \( z = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} \) and \( z = -\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} \).

Step 3: Calculating Residues

Calculate the residues of \( f(z) \) at the identified poles. For each pole, we find the residue using the formula for the residue at a simple pole. The residues are computed as follows: \( \operatorname{Res}(f(z), \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) = -\frac{\sqrt{2}}{8} - \frac{\sqrt{2}}{8}i \) and \( \operatorname{Res}(f(z), -\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) = \frac{\sqrt{2}}{8} - \frac{\sqrt{2}}{8}i \).

Step 4: Evaluating the Integral over Entire Real Line

The integral \( \int_{-\infty}^{\infty} \frac{e^{2ix}}{x^4+1} \, dx \) can be evaluated as \( 2\pi i \times \text{(sum of residues of poles in the upper half-plane)} \). This expression evaluates to \( \pi e^{-\sqrt{2}} \left( \frac{\sqrt{2}}{2} \cos \sqrt{2} + \frac{\sqrt{2}}{2} \sin \sqrt{2} \right) \).

Step 5: Extracting the Real Part

The integral \( \int_{-\infty}^{\infty} \frac{\cos 2x}{x^4+1} \, dx \) is the real part of the previously evaluated integral. Thus, \( \int_{-\infty}^{\infty} \frac{\cos 2x}{x^4+1} \, dx = \pi e^{-\sqrt{2}} \left( \frac{\sqrt{2}}{2} \cos \sqrt{2} + \frac{\sqrt{2}}{2} \sin \sqrt{2} \right) \).

Step 6: Adjusting for Half-Range of Integration

To find the integral from 0 to infinity, we observe that the function is even, so the integral over the entire real line is twice the integral from 0 to infinity. Thus, \( \int_{0}^{\infty} \frac{\cos 2x}{x^4+1} \, dx = \frac{1}{2} \times \pi e^{-\sqrt{2}} \left( \frac{\sqrt{2}}{2} \cos \sqrt{2} + \frac{\sqrt{2}}{2} \sin \sqrt{2} \right) = \pi e^{-\sqrt{2}} \frac{\sqrt{2}}{4} (\cos \sqrt{2} + \sin \sqrt{2}) \).

Key Concepts

Residue Theorem

The residue theorem is a powerful tool in complex analysis for evaluating contour integrals. It is especially useful when dealing with improper integrals involving complex functions. The theorem states that for a function with isolated singularities inside a closed contour, the integral along that contour is dictated by the sum of residues at these singularities.
This is expressed mathematically as:

• \( \int_C f(z) \, dz = 2 \pi i \sum \operatorname{Res}(f, z_k) \)

Here, \(C\) is the contour, \(f(z)\) is the function being integrated, and \(z_k\) are the poles of \(f\) inside \(C\). To apply this, we:

• Identify the poles of \(f(z)\) inside the contour.
• Calculate the residue at each pole, which involves finding the coefficient of \((z - z_k)^{-1}\) in the Laurent series expansion of \(f\) around \(z_k\).
• Sum the residues and multiply by \(2\pi i\) to find the integral.

This allows us to evaluate the integral without directly computing it over the complex plane, greatly simplifying problems with symmetries or specific singularities.

Complex Analysis

Complex analysis is the study of complex numbers and functions of a complex variable. It explores not only arithmetic operations and polynomials but also extends to more complex topics such as contour integration, residues, and even applications like fluid dynamics and signal processing.
A function \(f\) is complex if it has the form \(f(z) = u(x, y) + iv(x, y)\), where \(u\) and \(v\) are real-valued functions of real variables \(x\) and \(y\) with \(z = x + iy\).
Some key concepts include:

• Analytic Functions: Functions that are differentiable at all points in their domain, leading to the property of being expressible as power series. Examples include \(e^z\), \(sin(z)\), and \(cos(z)\).
• Singularities: Points where a function ceases to be analytic, such as poles where a function goes to infinity.
• Contour Integrals: Integrals of complex functions over a path or contour, essential in evaluating real integrals using complex methods.
  • Contour integrals exploit the geometrical nature of complex functions, allowing us to encircle singularities to extract information.

Complex analysis provides a robust framework for solving problems involving functions with complex variables, making it a key area of study in both theoretical and applied mathematics.

Improper Integrals

Improper integrals involve integration over an infinite interval or when the integrand has a singularity in its domain. These integrals are pivotal in various applications, especially when working with Fourier transforms or probability distributions.
To handle these integrals, we often turn to techniques such as contour integration, where complex analysis becomes an ally:

• Evaluation Strategy: Convert the improper integral into a contour integral using a function in the complex plane.
• Choosing a Path: Select a path or contour that simplifies the integral and encloses singularities (such as poles) relevant to the problem.
• Residues Application: Use the residue theorem to evaluate the integral along the contour, especially for integrals from \(-\infty\) to \(\infty\).

Contour integration becomes instrumental in transforming improper integrals into forms that are solvable using established theorems and methods. This is exemplified by the use of the residue theorem to indirectly evaluate real-valued improper integrals by considering their complex counterparts.