Problem 42 - Thomas Calculus

Question

In Exercises \(35-64\) , use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$ \int_{0}^{1} \frac{d t}{t-\sin t}(\text {Hint} : t \geq \sin t \text { for } t \geq 0) $$

Step-by-Step Solution

Answer

The integral diverges by the Direct Comparison Test with \( \frac{1}{t} \).

Step 1: Identify the Problem Type

The problem asks for testing the convergence of the integral \( \int_{0}^{1} \frac{d t}{t-\sin t} \). We’ll use the Direct Comparison Test or the Limit Comparison Test to determine its convergence.

Step 2: Choose a Comparison Function

The hint suggests that \( t \geq \sin t \) for \( t \geq 0 \). Hence, \( t - \sin t \geq 0 \) and therefore, \( \frac{1}{t - \sin t} \leq \frac{1}{t} \) when \( t > 0 \). We can choose \( \frac{1}{t} \) as the comparison function.

Step 3: Analyze the Comparison Function

We know the integral \( \int_{0}^{1} \frac{dt}{t} \) is improper and diverges as it approaches \( t = 0 \) with a vertical asymptote. Since this integral diverges, \( \int_{0}^{1} \frac{d t}{t - \sin t} \) cannot converge by this comparison.

Step 4: Convergence Using Direct Comparison Test

Since \( \frac{1}{t - \sin t} \leq \frac{1}{t} \) and \( \int_{0}^{1} \frac{dt}{t} \) diverges, the Direct Comparison Test implies that \( \int_{0}^{1} \frac{d t}{t - \sin t} \) also diverges.

Key Concepts

Direct Comparison Test

The Direct Comparison Test is a handy tool for determining the convergence or divergence of improper integrals. To use this test, we compare our questionable function to a known easier function. We often select a simpler function that forms a valid comparison, meaning it should be larger or smaller than the function we're examining over the interval.

In the exercise, the integral \[\int_{0}^{1} \frac{d t}{t-\sin t}\] is compared to \[\int_{0}^{1} \frac{1}{t} \ dt.\]
This simpler integral is known to diverge as it approaches zero. The comparison is based on the relationship \(t \geq \sin t\) for \(t \geq 0\). This allows us to establish \(\frac{1}{t-\sin t} \leq \frac{1}{t}\).

Since \(\int_{0}^{1} \frac{1}{t} \ dt\) diverges, the Direct Comparison Test tells us \(\int_{0}^{1} \frac{dt}{t-\sin t}\) diverges as well.

Limit Comparison Test

The Limit Comparison Test is another technique for evaluating the convergence of improper integrals. This test is especially useful when the Direct Comparison Test is challenging to apply directly. It involves taking a limit of the ratio of the function we are investigating and a known comparable function.

For the Limit Comparison Test:

• Suppose we have two functions, \(f(t)\) and \(g(t)\), that are non-negative over some interval \([a, b)\).
• We compute the limit \(\lim_{{t \to a^+}} \frac{f(t)}{g(t)}\).
• If this limit is a positive finite number, \(L > 0\), then both integrals \(\int_{a}^{b} f(t)\ dt\) and \(\int_{a}^{b} g(t)\ dt \) either both converge or both diverge.

In our exercise, the Limit Comparison Test could be applied between \(\frac{1}{t-\sin t}\) and \(\frac{1}{t}\). The test will corroborate the conclusion obtained via the Direct Comparison Test that the given integral diverges.

Improper Integrals

Improper integrals arise when dealing with integrands that have infinite discontinuities or when evaluating over an infinite interval. These integrals require careful analysis because they do not fit the mold of traditional definite integrals.

Consider the integral \(\int_{0}^{1} \frac{d t}{t-\sin t}\): this is an improper integral due to a division by zero issue when \(t = 0\). To evaluate this, we examine its behavior near the problematic point. If the function can be expressed in a comparable form, convergence tests become useful.

For example, knowing that \(\frac{1}{t}\) creates a vertical asymptote at \(t = 0\) helps guide us. By comparing \(\frac{1}{t-\sin t}\) to \(\frac{1}{t}\), we can confidently argue whether it converges or diverges. Understanding such characteristics is essential when evaluating these types of integrals.