Problem 12 - Advanced Engineering Mathematics

Question

In Problems 11-30, evaluate the Cauchy principal value of the given improper integral. $$ \int_{-\infty}^{\infty} \frac{1}{x^{2}-6 x+25} d x $$

Step-by-Step Solution

Answer

The Cauchy principal value of the integral is \( \frac{\pi}{4} \).

Step 1: Identify the type of singularity

The function inside the integral, \( \frac{1}{x^2 - 6x + 25} \), is a rational function. We need to identify the points where the denominator becomes zero, as these may cause singularities. The quadratic expression \( x^2 - 6x + 25 \) can be analyzed using the discriminant \( b^2 - 4ac \). Here \( a=1 \), \( b=-6 \), and \( c=25 \). The discriminant is \((-6)^2 - 4 \times 1 \times 25 = 36 - 100 = -64\). Since the discriminant is negative, there are no real roots, thus no real singularities.

Step 2: Set up the Cauchy Principal Value

Since there are no real singularities, we consider the limit form to evaluate the improper integral. For the Cauchy Principal Value (PV), we express it as:\[\text{PV} \int_{-\infty}^{\infty} \frac{1}{x^2 - 6x + 25} \, dx = \lim_{a \to \infty} \left( \int_{-a}^{0} \frac{1}{x^2 - 6x + 25} \, dx + \int_{0}^{a} \frac{1}{x^2 - 6x + 25} \, dx \right)\]This setup considers approaching the improper integral symmetrically around the singularity.

Step 3: Completing the Square

Rewrite the quadratic in the denominator to a solvable form. Completing the square: \( x^2 - 6x + 25 \) can be rewritten as:\[ (x - 3)^2 + 16 \]This will help in recognizing it as a standard form in trigonometric or other integral tables.

Step 4: Recognize the Standard Integral Form

The integral now appears in the form: \[ \int \frac{1}{(x-3)^2 + 16} \, dx \]This resembles the standard form of the integral \( \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \). Here, \( a^2 = 16 \) so \( a=4 \). Thus:\[ \int \frac{1}{(x-3)^2 + 4^2} \, dx = \frac{1}{4} \tan^{-1} \left( \frac{x-3}{4} \right) + C \]

Step 5: Calculate the Cauchy Principal Value

Apply the boundaries from the principal value setup:\[ \lim_{a \to \infty} \left( \int_{-a}^{0} \frac{1}{(x-3)^2 + 16} \, dx + \int_{0}^{a} \frac{1}{(x-3)^2 + 16} \, dx \right) \]This equals:\[ \lim_{a \to \infty} \left( \frac{1}{4} \left[ \tan^{-1} \left( \frac{x-3}{4} \right) \right]_{-a}^{0} + \frac{1}{4} \left[ \tan^{-1} \left( \frac{x-3}{4} \right) \right]_{0}^{a} \right) \]As \( a \to \infty \), the arctan limits yield:\[ \tan^{-1}(\infty) = \frac{\pi}{2}, \quad \tan^{-1}(-\infty) = -\frac{\pi}{2}, \quad \text{and} \quad \tan^{-1}(0) = 0 \]Leading to:\[ \frac{1}{4} \left( \frac{\pi}{2} - 0 + 0 - \left(-\frac{\pi}{2}\right) \right) = \frac{1}{4} \times \pi = \frac{\pi}{4} \].

Step 6: Conclusion

Therefore, the Cauchy principal value of the integral \( \int_{-\infty}^{\infty} \frac{1}{x^2 - 6x + 25} \, dx \) is \( \frac{\pi}{4} \).

Key Concepts

Cauchy Principal Value

Sometimes, we encounter improper integrals that are not straightforward to evaluate directly due to their infinite range or points where the function becomes unbounded. The Cauchy Principal Value (PV) provides a way to define the value of such improper integrals, specifically over symmetric intervals around a potential singularity.
This method involves taking a limit to ensure that contributions from both ends of the integral cancel out symmetrically. In essence, it offers a balanced perspective, allowing for more meaningful results, especially when direct integration might prove divergent.

• The Cauchy Principal Value is mainly used when the integral includes singular points or infinite endpoints.
• The method generally requires splitting the integral at the point of singularity and carefully considering limits.

Applying this method helps ensure that the integral's behavior is symmetrically controlled, enhancing our understanding of otherwise divergent scenarios.

Rational Functions

Rational functions are quotients of two polynomials, like the function in the given integral: \( \frac{1}{x^2 - 6x + 25} \). Understanding these functions is key when evaluating integrals, particularly when singularities are involved.
In dealing with rational functions:

• Focus first on the denominator, which can potentially turn zero and create singularities.
• Use the discriminant of the quadratic denominator to discover any real roots, indicating where singularities might lie.
• A negative discriminant, as seen here, means there are no real singularities present.

While these functions may initially seem complex, breaking them down step by step simplifies the process of analyzing and integrating them.

Completing the Square

Completing the square is a valuable technique for transforming a quadratic expression into a form that is easier to handle, especially when integrating. It involves rewriting the quadratic, \(x^2 - 6x + 25\), as \((x - 3)^2 + 16\).
This technique:

• Simplifies expressions, making them recognizable as standard forms in integral tables.
• Maintains the integrity of the original expression while altering its appearance to facilitate calculation.
• Enables easier application of subsequent integrations or substitutions.

By completing the square, you effectively prepare the integral for more efficient evaluation, paving the way for the use of standard integral results.

Trigonometric Substitution

Trigonometric substitution is a powerful method for evaluating integrals involving squares and square roots. It exploits the identity transformations of trigonometric functions to simplify complex algebraic expressions.
In this context, once the equation is rewritten using completing the square, trigonometric substitution allows you to recognize and proceed with integrating the resulting expression. When dealing with \( (x-3)^2 + 16 \), substituting this expression to fit the trigonometric integral form \( \frac{1}{(x-a)^2 + a^2} \) becomes crucial.

• Identify the trigonometric identity that matches the integral's form; it's usually similar to trigonometric forms like \( an^{-1} \).
• The result makes use of inverse trigonometric functions to provide an exact solution.

Using trigonometric substitution transforms a seemingly intricate problem into a manageable calculation, bolstered by well-known integral results.