Chapter 8: Problem 19
Evaluate each improper integral or show that it diverges. \(\int_{-\infty}^{\infty} \frac{1}{x^{2}+2 x+10} d x\)
Short Answer
Expert verified
The integral evaluates to \( \frac{\pi}{3} \).
Step by step solution
01
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.
02
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 \( \frac{1}{(x+1)^2 + 3^2} \).
03
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.
04
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 \( \int_{-\infty}^{\infty} \frac{1}{u^2 + 3^2} \, du \).
05
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: \( \lim_{a \to \infty} \left[ \frac{1}{3} \tan^{-1}\left(\frac{u}{3}\right) \right]_{-a}^{a} \).
06
Evaluate Limits
Compute the limits: \( \lim_{a \to \infty} \left[ \frac{1}{3} \tan^{-1}\left(\frac{a}{3}\right) - \frac{1}{3} \tan^{-1}\left(\frac{-a}{3}\right) \right] = \frac{1}{3}(\frac{\pi}{2} - (-\frac{\pi}{2})) = \frac{1}{3} \times \pi = \frac{\pi}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.
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.
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.
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 \( \frac{\pi}{2} - (-\frac{\pi}{2}) \).