Problem 59 - Calculus Single Variable

Question

Pach of the improper integrals is improper for two reasons. Determine whether the integral converges or diverges. If it converges, then evaluate it. $$ \int_{1}^{\infty} \frac{2}{x \sqrt{x^{2}-1}} d x $$

Step-by-Step Solution

Answer

The integral converges to \( \pi \).

Step 1: Recognize Improper Integral

The given integral \( \int_{1}^{\infty} \frac{2}{x \sqrt{x^{2}-1}} \, dx \) is improper because the upper limit of integration is infinite. Additionally, the integrand becomes undefined at its lower limit where \( x^2 - 1 = 0 \), i.e., \( x = 1 \).

Step 2: Address Infinite Boundary

Rewrite the integral to address the infinite boundary by replacing \( \infty \) with a limit: \[ \lim_{{b \to \infty}} \int_{1}^{b} \frac{2}{x \sqrt{x^{2}-1}} \, dx. \] This limits the infinity issue to the upper bound.

Step 3: Use Substitution Method

Set \( x = \sec(\theta) \). Then, \( dx = \sec(\theta)\tan(\theta) \, d\theta \) and \( \sqrt{x^2 - 1} = \tan(\theta) \). The new bounds become \( \theta = 0 \) when \( x = 1 \) and \( \theta \to \frac{\pi}{2} \) as \( x \to \infty \). The integral becomes \[ \lim_{{b \to \pi/2^-}} \int_{0}^{b} \frac{2}{\sec(\theta) \tan(\theta)} \cdot \sec(\theta)\tan(\theta) \, d\theta = \lim_{{b \to \pi/2^-}} \int_{0}^{b} 2 \, d\theta. \]

Step 4: Integrate the Simplified Expression

The integral simplifies to \( 2\theta \). Thus, evaluate \[ \lim_{{b \to \pi/2^-}} 2[\theta]_{0}^{b} = \lim_{{b \to \pi/2^-}} 2(b - 0) = 2b. \]

Step 5: Determine Convergence or Divergence

Substitute the limit: \[ \lim_{{b \to \pi/2^-}} 2b = 2 \cdot \frac{\pi}{2} = \pi. \] Thus, the integral converges and evaluates to \( \pi \).

Key Concepts

Understanding Convergence and Divergence

Improper integrals are a special class of definite integrals where one or more conditions make them 'improper.' These can arise due to integration over an infinite range or integrands that are undefined at some points within the interval. In our exercise, both issues are present: the upper limit of integration is infinite, and the integrand is undefined at the lower limit where \( x^2 - 1 = 0 \), meaning \( x = 1 \).
Understanding whether an improper integral converges or diverges is crucial:

• Converges: If as we approach the limits of integration, the value of the integral approaches a finite value, we say the integral converges.
• Diverges: Conversely, if the value of the integral grows without bound, the integral diverges.

To find out if our integral converges or diverges, we can use limits. By substituting the infinity symbol with a limit, we can evaluate the tendency of the integral as it approaches its bounds. In our case, this method revealed that the integral converges to the value \( \pi \).

Mastering Integration Techniques

Integration techniques are methods used to calculate integrals, and choosing the right technique can simplify complex problems considerably. For improper integrals, often a combination of techniques is necessary. For starters, addressing the boundaries of integration is vital when they are infinite or the integrand is undefined at certain points.
To make the integral easier to solve, convert the infinite boundary into a limit: \[ \lim_{{b \to \infty}} \int_{1}^{b} \frac{2}{x \sqrt{x^{2}-1}} \, dx. \] This move eliminates the infinity sign and instead uses the function values as \( b \) approaches infinity.One of the most helpful techniques for solving integrals similar to our exercise is substitution. This method allows us to rewrite the integral in terms that are often simpler to evaluate.

Employing the Substitution Method

The substitution method is a powerful technique, particularly useful when dealing with complex integrands. By changing variables, we can transform the integrand into a more straightforward form. In our exercise, we chose the substitution \( x = \sec(\theta) \), stemming from the trigonometric identity \( \sec^2(\theta) - 1 = \tan^2(\theta) \).
This substitution transforms

• \( dx = \sec(\theta)\tan(\theta) \, d\theta \)
• \( \sqrt{x^2 - 1} = \tan(\theta) \)

The limits of integration also change accordingly: \( x = 1 \) translates to \( \theta = 0 \) and \( x \to \infty \) turns into \( \theta \to \frac{\pi}{2} \). The integral then simplifies considerably, allowing easier computation:\[ \lim_{{b \to \pi/2^-}} \int_{0}^{b} 2 \, d\theta \] Evaluating this, we find the improper integral has a finite value of \( \pi \), confirming the convergence. This substitution transforms a challenging problem into an easier calculation, showcasing the method's effectiveness.