Problem 19 - Advanced Engineering Mathematics

Question

$$\int_{0}^{\infty} \frac{x^{2}+1}{x^{4}+1} d x=\frac{1}{2} \int_{-\infty}^{\infty} \frac{x^{2}+1}{x^{4}+1} d x=\pi i \operatorname{Res}(f(z), i)=\frac{\pi}{\sqrt{2}}$$

Step-by-Step Solution

Answer

The integral \( \int_{0}^{\infty} \frac{x^2+1}{x^4+1} \, dx \) is \( \frac{\pi i}{2} \).

Step 1: Recognize the Integral Form

The problem involves evaluating the improper integral \( \int_{0}^{\infty} \frac{x^{2}+1}{x^{4}+1} \, dx \). Due to its structure, the integrand \( \frac{x^2+1}{x^4+1} \) is rational, suggesting a potential use of complex analysis techniques, such as residue theory.

Step 2: Symmetry and Substitution

Notice that \( \int_{-\infty}^{\infty} \frac{x^{2}+1}{x^{4}+1} \, dx = 2 \int_{0}^{\infty} \frac{x^{2}+1}{x^{4}+1} \, dx \), by the function's even symmetry. Therefore, solving either integral means solving the other. The symmetry simplifies evaluations using methods such as residue calculus.

Step 3: Move to the Complex Plane

To use residue calculus, convert the real integral into a complex contour integral. Consider the function \( f(z) = \frac{z^2+1}{z^4+1} \). The poles of \( f(z) \) in the upper half-plane are relevant since they contribute to the evaluation of the integral from \( -\infty \) to \( \infty \).

Step 4: Determine the Poles

Find where \( z^4 + 1 = 0 \), which gives the poles at \( z = e^{\pi i/4}, \) \( z = e^{3\pi i/4}, \) \( z = e^{5\pi i/4}, \) and \( z = e^{7\pi i/4} \). We focus only on the poles in the upper half-plane: \( z = e^{\pi i/4} \) and \( z = e^{3\pi i/4} \).

Step 5: Calculate Residues

Use the formula for residues at simple poles to compute \( \operatorname{Res}(f(z), e^{\pi i/4}) \) and \( \operatorname{Res}(f(z), e^{3\pi i/4}) \). Recall that for simple poles, \( \operatorname{Res}(f, z_0) = \lim_{z \to z_0} (z-z_0)f(z) \). Here, \( \operatorname{Res}(f(z), e^{\pi i/4}) = \frac{\pi}{\sqrt{2}} \).

Step 6: Apply Residue Theorem

The Residue Theorem relates a contour integral to the sum of residues: \( \int_C f(z) \, dz = 2\pi i \sum_{\text{poles in } C} \text{Res}(f, \text{pole}) \). Here, only the poles at \( e^{\pi i/4} \) and \( e^{3\pi i/4} \) contribute to the integral over the entire real line.

Step 7: Solve the Real Integral

By the Residue Theorem, the integral \( \int_{-\infty}^{\infty} \frac{x^2+1}{x^4+1} \, dx = 2\pi i\left( \frac{\pi}{\sqrt{2}}\right) = \pi i \). Hence, \( \int_{0}^{\infty} \frac{x^2+1}{x^4+1} \, dx = \frac{\pi i}{2} \).

Key Concepts

Residue Theorem

The Residue Theorem is a powerful tool in complex analysis. It allows us to evaluate contour integrals by connecting them to the sum of residues within a given contour. When analyzing complex functions, poles play a crucial role in finding residues. To understand this, consider a complex function \( f(z) \) that is integrable within a closed contour \( C \). If \( f(z) \) has isolated singularities (or poles) within \( C \), then the contour integral of \( f(z) \) around \( C \) is expressed as:

• \( \int_C f(z) \, dz = 2\pi i \sum_{\text{poles in } C} \operatorname{Res}(f, \text{pole}) \)

This remarkable theorem reduces the complex task of evaluating a contour integral down to the relatively simple problem of finding and summing residues at the poles within the contour. In practice, applying the Residue Theorem can greatly simplify the evaluation of definite integrals on the real line by converting them into contour integrals in the complex plane.

Contour Integrals

Contour integrals extend the concept of integrals to complex functions over paths, or contours, in the complex plane. They are the backbone of many results in complex analysis, such as the Residue Theorem. To compute a contour integral, we traverse a path \( C \) with endpoints in the complex plane, integrating a complex function over this path.

• For a given function \( f(z) \), the contour integral over \( C \) is \( \int_C f(z) \, dz \).

The magic of contour integrals lies in how they interact with singularities, enabling us to evaluate integrals with complex poles. The powerful concepts of analytic continuity, residues, and Jordan's lemma support the theory, but the basics are approachable through simple calculations of the integral over chosen paths. In practical applications, like in the exercise, converting a real integral into a contour integral involves selecting suitable contours and considering the poles within these paths.

Poles in Complex Analysis

Poles in complex analysis are instances where a function's value "goes to infinity". These are special kinds of isolated singularities. Recognizing and understanding poles are central to using the Residue Theorem effectively. For example, poles can be simple or higher order. Simple poles are where the function behaves like \( \frac{1}{z-z_0} \) near the pole \( z_0 \).

• To compute residues at a simple pole \( z_0 \), you can use the formula: \( \operatorname{Res}(f, z_0) = \lim_{z \to z_0} (z-z_0)f(z) \).

In the original problem, the poles are found by solving \( z^4 + 1 = 0 \), which gives poles at points like \( e^{\pi i/4} \). Focusing only on those in the upper half-plane ensures that the integral evaluation leverages these singularities to simplify the process.

Symmetry in Integrals

Symmetry is a valuable tool in simplifying integrals. It helps save time and effort in both real and complex integrals. For instance, if a function is even, evaluations over symmetric limits can be doubled or halved as needed.

• An even function \( f(x) \) implies that \( f(x) = f(-x) \), allowing integrals like \( \int_{-a}^{a} f(x) \, dx \) to be substituted as \( 2 \int_{0}^{a} f(x) \, dx \).

The exercise used this concept effectively. It recognized the symmetry of the integrand \( \frac{x^2+1}{x^4+1} \), which is even, to simplify calculations. By recognizing the symmetry, it allowed the conversion of the integral over the entire real line into a simpler, halved form, facilitating the application of complex techniques on a more manageable section of the range. This systematic use of symmetry demonstrates its practicality in both theory and practice.