Problem 27 - Essential Calculus

Question

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

Step-by-Step Solution

Answer

The improper integral diverges because the limit as M approaches infinity does not exist.

Step 1: Recognize the Integral

Recognize the given integral as an improper integral because it has an infinite limit. The integral is \(\int_{0}^{\infty} \cos \pi x d x\). Now, the trick here is to convert this integral into a form that is easier to work with.

Step 2: Express Integral with Limits

Consider the limit as M approaches infinity for the integral from 0 to M. This rewrites the integral to \(\lim_{M \rightarrow \infty}\int_{0}^{M} \cos \pi x d x\). This is because the integration cannot be directly performed with an infinite limit.

Step 3: Compute the Antiderivative & Apply Limits

Compute the antiderivative of cos(\(\pi x\)). The antiderivative is \(1/ \pi\) * sin(\(\pi x\)). Apply this to the integral and then separately apply the limits 0 and M: \[ \lim_{M \rightarrow \infty} \left[\frac{1}{\pi}\sin(\pi M) - \frac{1}{\pi}\sin(0)\right] \]

Step 4: Evaluate the Limit

Now, we evaluate the limit as M goes to infinity. The sin function oscillates between -1 and 1. So, \(\sin(\pi M)\) will also oscillate as M approaches infinity. Therefore, this limit does not exist.