Problem 12 - Complex Analysis

Question

Show: (a) $$ \int_{-\infty}^{\infty} \frac{x^{2}}{\left(x^{2}+a^{2}\right)^{2}} d x=\frac{\pi}{2 a}, \quad(a>0) $$ (b) $$ \int_{-\infty}^{\infty} \frac{d x}{\left(x^{2}+4 x+5\right)^{2}}=\frac{\pi}{2} $$ (c) \(\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}=\frac{\pi}{2 a b(a+b)}, \quad(a, b>0)\).

Step-by-Step Solution

Answer

(a) \( \frac{\pi}{2a} \); (b) \( \frac{\pi}{2} \); (c) \( \frac{\pi}{2ab(a+b)} \)."

Step 1: Consider the symmetry of the function

The integrands in both parts (a) and (b) have symmetry properties around the y-axis. Observing these properties helps to recognize that the integral bounds from \(-\infty\) to \(+\infty\) can often be computed by exploring residues within \((-\infty, \infty)\) or by symmetry arguments.

Step 2: Use Residue Theorem for part (a)

Change the variable \( x = a\tan\theta \) for the integrand \( \int_{-\infty}^{fty} \frac{x^{2}}{(x^{2}+a^{2})^{2}} dx \). Then, use contour integration and the residue theorem by considering the residue at the pole \( ia \) to compute the integral.

Step 3: Transform the function for part (a)

The integrand is \( f(x) = \frac{x^{2}}{(x^{2}+a^{2})^{2}} \). By substituting \( z = re^{i\theta} \) and letting \( r \to \infty \), apply Cauchy's residue theorem at \( z = ia \). Then compute: \( \oint_{C} f(z) \, dz = 2\pi i \sum \text{Residues of } f(z) \), focusing on the simple poles.

Step 4: Calculate residues for part (a)

Find the residue by differentiating and solving \( \text{Residue} = \lim_{z \to ia} (z - ia)^2 f(z) \). This simplifies to \( \frac{d}{dz} \left( \frac{z^2}{(z^2+a^2)^2} \right) \bigg|_{z=ia} \). Evaluate the derivative and determine the contribution of the pole at \( ia \) to be \( \frac{\pi}{2a} \).

Step 5: Simplify integrand for part (b)

Complete the square in \( \int_{-\infty}^{fty} \frac{dx}{(x^{2}+4x+5)^{2}} \) by rewriting \( x^2 + 4x + 5 \) as \( (x+2)^2 + 1 \). The new form is \( \int_{-fty}^{fty} \frac{dx}{((x+2)^2+1)^2} \), resembling a squared secant function after a translation.

Step 6: Utilize a substitution for part (b)

Use a substitution \( u = x + 2 \), resulting in \( \int_{-fty}^{fty} \frac{du}{(u^2 + 1)^2} \). This is equivalent to \( rac{1}{2} imes \int_{-fty}^{fty} \frac{du}{u^2 + 1} \), and evaluate using known trigonometric integral results.

Step 7: Complete integration process for part (b)

Compute \( \int_{-\infty}^{fty} \frac{1}{u^2 + 1} \, du = \pi \). Therefore, the integral \( \frac{1}{2} \pi = \frac{\pi}{2} \) confirms the solution: \( \int_{-\infty}^{fty} \frac{du}{((u)^2+1)^2} = \frac{\pi}{2} \).

Step 8: Evaluate integral for part (c) by partial fractions

For \( \int_{0}^{fty} \frac{dx}{(x^2+a^2)(x^2+b^2)} \), decompose into partial fractions: \( \frac{1}{(x^2+a^2)(x^2+b^2)} = \frac{A}{x^2+a^2} + \frac{B}{x^2+b^2} \). Solve for A and B in terms of a and b.

Step 9: Solve partial fraction coefficients for part (c)

Equating coefficients from the identity, solve \( Ax^2 + Ab^2 + Bx^2 + Ba^2 = 1 \) leading to \( A = \frac{1}{b^2-a^2}, \ B = \frac{-1}{b^2-a^2} \). Then integrate separately \( \int_{0}^{\infty} \frac{A \, dx}{x^2+a^2} + \int_{0}^{\infty} \frac{B \, dx}{x^2+b^2} \).

Step 10: Integrate using known results for part (c)

Use results \( \int_{0}^{\infty} \frac{dx}{x^2 + a^2} = \frac{\pi}{2a} \) and \( \int_{0}^{\infty} \frac{dx}{x^2 + b^2} = \frac{\pi}{2b} \). Substitute these into the partial fraction decomposition and simplify to find \( \frac{\pi}{2ab(a+b)} \).

Key Concepts

Contour Integration

Contour Integration takes the evaluation of integrals into the complex plane. This method is particularly helpful for integrals that extend to infinity, as seen in problems like part (a) of the original exercise. The core idea is to transform the real integral into a contour integral by enclosing the poles of the integrand within a contour path.

In contour integration, the Residue Theorem is crucial. It states that the integral of a meromorphic function around a closed contour in the complex plane is equal to \( 2 \pi i \) times the sum of the residues of the poles inside the contour. This transforms what can be a complex real integral into a simpler calculation.

• Identify poles: First, recognize the singularities—or poles—of the function within the defined contour.
• Calculate residues: Determine the contribution of each pole by finding the residue, often achieved by taking appropriate derivatives and limits.
• Apply the Residue Theorem: Sum the residues and multiply by \( 2\pi i \) to evaluate the integral over the contour.

By applying these steps to our integral \( \int_{-\infty}^{\infty} \frac{x^2}{(x^2+a^2)^2} \, dx \), we focus on the pole at \( z = ia \), compute its residue, and utilize contour integration to find the solution.

Partial Fractions

Partial Fractions is a technique utilized to break down complex rational expressions into simpler, more manageable parts. This method is particularly useful when faced with integrands that appear as quotients of polynomials.

In the exercise, partial fractions are applied to part (c), where the integrand is \( \frac{1}{(x^2+a^2)(x^2+b^2)} \). Let's explore the partial fractions technique:

• Decompose the function: Express the integrand as a sum of fractions, each with a simpler denominator. Here, it is split into \( \frac{A}{x^2+a^2} + \frac{B}{x^2+b^2} \).
• Find coefficients: Solve for unknowns (A and B) by equating the original expression to the partial fractions form and matching coefficients of like terms.
• Integrate components: Integrate each fraction separately using familiar integral results, such as \( \int_{0}^{\infty} \frac{dx}{x^2+a^2} = \frac{\pi}{2a} \).

This technique allows for the individual integration of simpler components and recombination to arrive at the final result, demonstrating its utility in solving intricate integral problems.

Definite Integral

A Definite Integral provides the accumulated value of a function across a specified interval. In contrast to indefinite integrals, a definite integral yields a numerical result representing the area under a curve.

In exercises such as parts (a) and (b), we encounter definite integrals on intervals extending to infinity. Here, accessing solutions may require advanced techniques:

• Utilizing symmetry: The symmetry of functions with respect to the y-axis simplifies calculations, especially for integrands in the form of even functions.
• Transformation techniques: Employ substitutions or transformations to convert tricky integrals into standard forms. This is seen in parts (a) and (b), where trigonometric or polynomial substitutions are performed.
• Convergence analysis: Investigate the convergence of integrals as they approach infinity, ensuring the integral remains finite.

In practice, once combined with the techniques of contour integration and partial fractions, definite integrals become far less daunting. Ultimately, by providing a concrete area measurement or sum total, definite integrals play a fundamental role in real analysis and complex function evaluation.