Problem 20 - Advanced Engineering Mathematics

Question

$$\int_{0}^{\infty} \frac{1}{x^{6}+1} d x=\frac{1}{2} \int_{-\infty}^{\infty} \frac{1}{x^{6}+1} d x=\pi i\left[\operatorname{Res}\left(f(z), \frac{\sqrt{3}}{2}+\frac{1}{2} i\right)+\operatorname{Res}(f(z), i)+\operatorname{Res}\left(f(z), \frac{-\sqrt{3}}{2}+\frac{1}{2} i\right)\right]=\frac{\pi}{3}$$

Step-by-Step Solution

Answer

The value of the integral is \(\frac{\pi}{3}\).

Step 1: Recognize the Symmetry of the Function

Notice that the function \( f(x) = \frac{1}{x^6 + 1} \) is even. This allows us to write \( \int_{0}^{\infty} \frac{1}{x^{6}+1} dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{1}{x^{6}+1} dx \). This simplifies our calculations by focusing only on the positive half and later multiplying by two.

Step 2: Analyze the Complex Plane

To evaluate \( \int_{-\infty}^{fty} \frac{1}{x^6 + 1} dx \), we extend this integral to the complex plane as a contour integral. Identify the poles of \( f(z) = \frac{1}{z^6 + 1} \), which occur where \( z^6 + 1 = 0 \). Solve \( z^6 = -1 \) to find the poles: \( z = e^{i\frac{\pi}{6} + i\frac{\pi}{k}} \) for \( k = 0, 1, 2, 3, 4, 5 \).

Step 3: Locate Residues Inside the Upper Half-Plane

Only the poles in the upper half of the complex plane contribute to the contour integral. These poles are \( z = e^{i\frac{\pi}{6}}, e^{i\frac{3\pi}{6}}, e^{i\frac{5\pi}{6}} \). Calculate the residues at these points.

Step 4: Compute the Residues

The residues at the poles \( \frac{\sqrt{3}}{2} + \frac{1}{2}i \), \( i \), and \( \frac{-\sqrt{3}}{2} + \frac{1}{2}i \) are given. For a simple pole, the residue is \( \operatorname{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0)f(z) \). By calculation, the residues are respectively, \( \frac{1}{x^5} \) evaluated at each.

Step 5: Apply the Residue Theorem

Apply the residue theorem: \( \int_{-\infty}^{\infty} \frac{1}{x^{6}+1} dx = 2\pi i \sum \text{Residues in upper half-plane} \). Sum the residues give \( \pi (\frac{1}{e^{i\pi/6}}, \frac{1}{i}, \frac{1}{e^{5i\pi/6}} \) based on multiplication.

Step 6: Conclude with the Calculated Integral

Our calculation yielded \( \pi i (Res_1 + Res_2 + Res_3) = \frac{\pi}{3} \). Multiply by \( \frac{1}{2} \), which results in \( \frac{\pi}{3} \), confirming the integral is \( \frac{\pi}{3} \).

Key Concepts

Contour Integration

Contour integration is a technique used in complex analysis, a branch of mathematics focusing on functions of complex numbers. It is a powerful tool for evaluating integrals, especially those that are difficult to calculate using real variable methods. The idea is to extend the integral from the real line into the complex plane. This is done by treating the integral as an integral over a path or contour in the complex plane.

In our problem, the integral of the function \(\frac{1}{x^6 + 1}\) from \(-\infty\) to \(\infty\) is transformed into a contour integral in the complex plane. This involves defining a specific path or contour that encapsulates certain points, called poles, which contribute to the integral. The chosen contour usually extends into the upper or lower half of the complex plane but eventually returns to the real line. By applying complex analysis techniques, such as residue theorem, we can compute these complex integrals efficiently.

Residue Theorem

The Residue Theorem is a key concept in complex analysis that facilitates the evaluation of contour integrals by linking the integral to the residues of the poles enclosed within the chosen contour. A residue is essentially the coefficient of the \((z - z_0)^{-1}\) term in the Laurent series expansion of a function around a pole, \(z_0\).

In the example given, the residue theorem allows us to evaluate the contour integral of \(\frac{1}{x^6 + 1}\) by summing the residues of the function's poles that lie in the upper half-plane of the complex plane. According to the residue theorem:

• For a contour \(C\), the integral \(\oint_C f(z) \, dz\) equals \(2\pi i \) times the sum of the residues of \(f\) inside \(C\).

Thus, once we calculate the residues for the poles of our integrand within the semicircular contour, we can multiply this sum by \(2\pi i\) to find the value of the integral over the real line.

Even Functions

Even functions are symmetrical about the y-axis, meaning \(f(x) = f(-x)\). This property can be exploited when evaluating integrals, as it allows us to simplify calculations. For even functions, the integral over a symmetric interval (such as \(-\infty\) to \(\infty\)) can be calculated as twice the integral from 0 to \(\infty\).

In this problem, the function \(f(x) = \frac{1}{x^6 + 1}\) is even. Thus, we can express the integral over the entire real line as:

• \(\int_{-\infty}^{\infty} \frac{1}{x^6 + 1} \, dx = 2 \int_{0}^{\infty} \frac{1}{x^6 + 1} \, dx\).

This symmetry simplifies the initial steps by reducing the computational domain, allowing us to focus only on the positive half of the real line.

Poles in the Complex Plane

Poles in the complex plane are certain types of singularities of a function where the function tends to infinity. These poles play a crucial role in complex integration as they determine the contributions to integrals via residues.

For the function \(f(z) = \frac{1}{z^6 + 1}\), we identify poles by solving \(z^6 + 1 = 0\), which simplifies to \(z^6 = -1\). The poles are the sixth roots of \(-1\), given by:

• \(z = e^{i(\frac{\pi}{6} + \frac{k\pi}{3})}\) for \(k = 0, 1, 2, 3, 4, 5\).

Out of these, only those in the upper half-plane (i.e., with positive imaginary parts) contribute to the contour integrals in most applications. Identifying and calculating the residues at these poles provides the necessary inputs to use the residue theorem effectively.