Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each improper integral or show that it diverges. \(\int_{-\infty}^{\infty} \frac{1}{x^{2}+2 x+10} d x\)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{3} \).

Step 1: Analyze the Integrand

First, observe the integrand \( \frac{1}{x^2 + 2x + 10} \). Notice that the denominator is a quadratic expression which never equals zero since its discriminant \( \Delta = 2^2 - 4 \times 1 \times 10 = -36 \) is negative, meaning the quadratic has no real roots.

Step 2: Complete the Square

Rewrite the denominator by completing the square. The expression \( x^2 + 2x + 10 \) is transformed by completing the square as follows: \( x^2 + 2x + 1 + 9 = (x+1)^2 + 3^2 \). So the integrand becomes \( \frac{1}{(x+1)^2 + 3^2} \).

Step 3: Recognize the Standard Form

Notice that the rewritten integrand \( \frac{1}{(x+1)^2 + 3^2} \) resembles the standard form of the integral \( \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \), which is a known formula for integrals resembling the arc tangent function.

Step 4: Change of Variable

Perform a change of variable to fit the standard form. Let \( u = x + 1 \), then \( du = dx \), and the limits of integration remain from \(-\infty\) to \(\infty\). The integral then is \( \int_{-\infty}^{\infty} \frac{1}{u^2 + 3^2} \, du \).

Step 5: Integrate Using Arc Tangent

Apply the standard integral formula: \( \int \frac{1}{u^2 + 3^2} \, du = \frac{1}{3} \tan^{-1}\left(\frac{u}{3}\right) + C \). Evaluate the definite integral: \( \lim_{a \to \infty} \left[ \frac{1}{3} \tan^{-1}\left(\frac{u}{3}\right) \right]_{-a}^{a} \).

Step 6: Evaluate Limits

Compute the limits: \( \lim_{a \to \infty} \left[ \frac{1}{3} \tan^{-1}\left(\frac{a}{3}\right) - \frac{1}{3} \tan^{-1}\left(\frac{-a}{3}\right) \right] = \frac{1}{3}(\frac{\pi}{2} - (-\frac{\pi}{2})) = \frac{1}{3} \times \pi = \frac{\pi}{3} \).

Key Concepts

Integral Calculus

Integral calculus is a fundamental concept in mathematics that focuses on determining the total accumulation of quantities. It's often described as the reverse process of differentiation.

Knowing how to evaluate integrals—especially improper ones—is crucial. Improper integrals differ from regular integrals because their limits extend to infinity or they include discontinuities in the interval of integration. For example, \[ \int_{-\infty}^{\infty} \frac{1}{x^{2}+2 x+10} \, dx \]illustrates an improper integral, with both limits stretching to infinity.

• The goal is to discover whether this infinite accumulation converges to a finite number or diverges.
• We handle this by analyzing the integrand's behavior at extreme limits and applying specific techniques like substitution or recognizing standard forms.

Integrals like this often require creative problem-solving, such as transforming them into a common form or examining their limits carefully.

Arc Tangent Function

The arc tangent function, often denoted as \( \tan^{-1}(x) \), is the inverse of the tangent function. Understanding this function is key to solving some integrals.

For instance, if we have a transformed integral that resembles the form \[ \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C, \]we can immediately recognize it as involving the arc tangent.

• This standard form is handy when dealing with quadratic expressions in denominators.
• Completing the square often reveals this form, as seen with \((x+1)^2 + 3^2\) prompting us to think of the arc tangent.

The arc tangent function thus provides a straightforward means to solve otherwise complex integrals. It concludes the improper integral evaluation by translating the algebraic expression into a known inverse trigonometric function.

Definite Integrals

Definite integrals calculate the area under a curve within specific limits. They are denoted as \( \int_{a}^{b} f(x) \, dx \) and allow us to find the total accumulation over an interval, which may include infinite limits.

In the case of an improper integral:\[ \int_{-\infty}^{\infty} \frac{1}{(x+1)^2 + 3^2} \, dx \]the transformation using limits \(-\infty\) to \(\infty\) requires careful consideration of convergence.

• The integration results in an expression involving \( \tan^{-1} \), which we evaluate at the limits.
• The limits as \( x \to \infty \) and \( x \to -\infty \) boil down to computing the difference \( \frac{\pi}{2} - (-\frac{\pi}{2}) \).

Upon calculation, this results in \( \frac{\pi}{3} \). This process captures the essence of definite integrals in resolving both finite and infinite limits through systematic evaluation.