Question

Evaluate each improper integral or show that it diverges. \(\int_{-\infty}^{\infty} \frac{d x}{\left(x^{2}+16\right)^{2}}\)

Short answer

The integral converges to \(\frac{\pi}{8}\).

Step-by-step solution

Recognize the Symmetry

Notice that the function \( \frac{1}{(x^2+16)^2} \) is an even function, meaning it is symmetric around the y-axis. Thus, the integral from \(-\infty\) to \(\infty\) can be expressed as twice the integral from 0 to \(\infty\).

Evaluate the Improper Integral

As an even function, we can rewrite the integral: \[\int_{-\infty}^{\infty} \frac{dx}{(x^2+16)^2} = 2 \int_{0}^{\infty} \frac{dx}{(x^2+16)^2}.\]Consider the substitution \( x = 4 \tan(\theta) \), which implies \( dx = 4 \sec^2(\theta) d\theta \) and \( x^2 = 16\tan^2(\theta) \). Therefore, the integral becomes \[= 2 \int_{0}^{\pi/2} \frac{4 \sec^2(\theta) d\theta}{(16 \tan^2(\theta) + 16)^2}.\]

Simplify the Integral Using Trigonometric Identity

With \( x = 4 \tan(\theta) \), we have \( x^2 + 16 = 16 \sec^2(\theta) \). Thus, \[2 \int_{0}^{\pi/2} \frac{4 \sec^2(\theta) d\theta}{(16 \sec^2(\theta))^2} = 2 \int_{0}^{\pi/2} \frac{4 \sec^2(\theta) d\theta}{256 \sec^4(\theta)}\]which simplifies to \[\int_{0}^{\pi/2} \frac{d\theta}{4 \sec^2(\theta)}.\]Upon further simplification using \( \sec^{2}(\theta) = 1 + \tan^{2}(\theta) \), this becomes \[\int_{0}^{\pi/2} \cos^2(\theta) d\theta.\]

Evaluate the Simplified Integral

Use the identity \( \cos^{2}(\theta) = \frac{1 + \cos(2\theta)}{2} \): \[ \int_{0}^{\pi/2} \left( \frac{1 + \cos(2\theta)}{2} \right) d\theta = \frac{1}{2} \left( \int_{0}^{\pi/2} d\theta + \int_{0}^{\pi/2} \cos(2\theta) d\theta \right).\]The first integral evaluates to \(\frac{\pi}{4}\), and the second, using substitution \( u = 2\theta \), evaluates to zero, giving a result of \(\frac{\pi}{8}\).

Conclude the Value of the Integral

After simplifying and integrating, the integral evaluates to \(\frac{\pi}{8}\). Therefore, the improper integral converges to \(\frac{\pi}{8}\).

Key concepts

Even Function

When dealing with integrals, recognizing the symmetry of the function can greatly simplify the process. An **even function** is a function where the equation remains unchanged if you replace the variable with its negative. In simple terms, an even function is symmetric about the y-axis. The practical benefit of identifying an even function when evaluating integrals is that you can simplify the limits. For example, if you have an integral from \(-\infty\) to \(\infty\) of an even function, it can be rewritten as twice the integral from 0 to \(\infty\). This is because the area under the curve on the negative side is equal to the area on the positive side.

• The function \( \frac{1}{(x^2+16)^2} \) is even, which means \( f(x) = f(-x) \).
• Therefore, the integral \( \int_{-\infty}^{\infty} \) of this function becomes \( 2 \int_{0}^{\infty} \).

Trigonometric Substitution

**Trigonometric substitution** is a technique often used to simplify certain integrals, especially those involving square roots or quadratic expressions. By replacing variables with trigonometric identities, the integral takes on a form that is more manageable to solve.

In this problem, we use the substitution \( x = 4 \tan(\theta) \). This substitution simplifies the quadratic expression \( x^2 + 16 \) because \( \tan^2(\theta) + 1 = \sec^2(\theta) \). This takes advantage of trigonometric identities to transform our integral to a simpler form:

• With \( x = 4 \tan(\theta) \), then \( dx = 4 \sec^2(\theta) d\theta \).
• \( x^2 + 16 \) becomes \( 16 \sec^2(\theta) \), simplifying the integral.
• The new integral bounds change according to \( \theta \)'s range, often going from 0 to \( \frac{\pi}{2} \).

Convergence of Integrals

An **improper integral** involves integration over an infinite range or an unbounded function. Determining the convergence of these integrals helps us to understand whether the area under the curve is finite or not.

To assess convergence, we often simplify the integral, evaluate its parts over their defined ranges, and then analyze the resulting expressions.

• The function \( \frac{1}{(x^2+16)^2} \) can be integrated from 0 to \( \infty \) yielding finite results by recognizing it converges.
• We used trigonometric substitution leading to \( \int_{0}^{\pi/2} \cos^2(\theta) d\theta \).
• The integral converges, resulting in a finite value of \( \frac{\pi}{8} \), which confirms convergence.

Understanding when an improper integral like this converges helps interpret physical quantities and probabilities in applications.