Problem 28 - Essential Calculus

Question

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{0}^{\infty} \sin \frac{x}{2} d x $$

Step-by-Step Solution

Answer

The improper integral diverges.

Step 1: Identify the improper integral

The given integral \(\int_{0}^{\infty} \sin \frac{x}{2} dx\) is an improper integral as it has an infinity in the interval of integration.

Step 2: Express the integral in terms of limits

Since it is an improper integral, express it in terms of limits: \(\int_{0}^{\infty} \sin \frac{x}{2} dx = \lim_{{t \to \infty}} \int_{0}^{t} \sin \frac{x}{2} dx\)

Step 3: Apply u-substitution

For the integral, u-substitution is easiest. Let \(u = \frac{x}{2}\), then \(du = \frac{1}{2} dx\) and \(dx = 2 du\). The limits of the integral, originally 0 to \(t\), change to 0 to \(t/2\). The integral becomes \(\lim_{{t \to \infty}} \int_{0}^{t/2} 2 \sin u du\)

Step 4 : Integrate

Now, it is a straightforward integral to compute: \(\lim_{{t \to \infty}} [-2 \cos(u)]_{0}^{t/2}\) = \(\lim_{{t \to \infty}} [2 \cos(0) - 2 \cos(t/2)]\) = \(\lim_{{t \to \infty}} [2 - 2 \cos(t/2)]\)

Step 5: Check for convergence/divergence

Since \(\cos(t/2)\) oscillates between -1 and 1 as \(t\) approaches \(\infty\), it means the \(\lim_{{t \to \infty}} [2 - 2 \cos(t/2)]\) does not exist. Therefore, the limit does not exist, meaning the improper integral diverges.