Problem 21 - Calculus and its applications

Question

Determine whether each improper integral is convergent or divergent, and find its value if it is convergent. $$ \int_{\pi}^{\infty} \frac{d t}{t^{1.001}} $$

Step-by-Step Solution

Answer

The integral is convergent and evaluates to \( 1000\pi^{-0.001} \).

Step 1: Analyze the Integral Limit

The integral \( \int_{\pi}^{\infty} \frac{1}{t^{1.001}} dt \) has an improper upper limit, as it goes from \( t = \pi \) to \( t = \infty \). To evaluate convergence, we need to analyze the behavior as \( t \to \infty \).

Step 2: Determine Convergence Using p-Test

We apply the p-test for improper integrals of the form \( \int_{a}^{\infty} \frac{1}{t^p} dt \). The integral converges if \( p > 1 \) and diverges if \( p \leq 1 \). Here, \( p = 1.001 \), which is greater than 1, so the integral converges.

Step 3: Set Up the Limit Process

Express the improper integral as a limit: \[\int_{\pi}^{\infty} \frac{1}{t^{1.001}} dt = \lim_{b \to \infty} \int_{\pi}^{b} \frac{1}{t^{1.001}} dt.\]

Step 4: Integrate the Function

Find the antiderivative of \( \frac{1}{t^{1.001}} \). The antiderivative is given by \(-\frac{1}{0.001} t^{-0.001} = -1000t^{-0.001}\). Thus, \[\int \frac{1}{t^{1.001}} dt = -1000t^{-0.001} + C.\]

Step 5: Evaluate the Definite Integral

Evaluate the antiderivative from \( \pi \) to \( b \):\[\int_{\pi}^{b} \frac{1}{t^{1.001}} dt = \left[ -1000t^{-0.001} \right]_{\pi}^{b} = -1000b^{-0.001} + 1000\pi^{-0.001}.\]

Step 6: Take the Limit as b Approaches Infinity

Evaluate the limit as \( b \) approaches infinity:\[\lim_{b \to \infty} (-1000b^{-0.001} + 1000\pi^{-0.001}) = 0 + 1000\pi^{-0.001}=1000\pi^{-0.001}.\]Since the term involving \( b \) vanishes as \( b \to \infty \), the integral converges to \( 1000\pi^{-0.001} \).

Key Concepts

p-Test for Improper Integrals

The p-test is a crucial tool for determining the convergence of improper integrals of specific forms. When you encounter an integral like \( \int_{a}^{\infty} \frac{1}{t^p} \, dt \), the p-test provides a straightforward rule:

• If the exponent \( p > 1 \), the integral converges.
• If \( p \leq 1 \), the integral diverges.

The beauty of this test lies in its simplicity and effectiveness for integrals involving power functions of \( t \). This concept helps eliminate the need for complex calculations just to determine if we can find a finite area under an infinite curve.
In our given problem, the function \( \frac{1}{t^{1.001}} \) has \( p = 1.001 \). Here, since \( 1.001 > 1 \), the integral converges. This quick verification step can save a lot of time and guides you on the path to correctly evaluating the integral's value.

Limit Process in Integrals

When handling improper integrals, setting up a proper limit process is essential. This process involves converting the upper bound of the integral, which can be infinite, into a more manageable form.
The way to do this is by replacing the infinity in the integral with a variable, often denoted by \( b \), and then later taking the limit as \( b \) approaches infinity. For our problem, this setup looks like:
\[\int_{\pi}^{\infty} \frac{1}{t^{1.001}} \, dt = \lim_{b \to \infty} \int_{\pi}^{b} \frac{1}{t^{1.001}} \, dt.\]
This strategic maneuver allows us to treat the integral like it's defined over a finite interval, analyze it smoothly, and then explore the behavior as the interval extends to infinity. By evaluating this limit, we ensure that we remain within the bounds of conventional calculus operations.

Antiderivative of Power Functions

When calculating an improper integral, finding the right antiderivative plays a pivotal role. Power functions like \( \frac{1}{t^{1.001}} \) follow a specific guideline for integration. To find the antiderivative of a power function \( \frac{1}{t^p} \), use the formula \[\int \frac{1}{t^p} dt = \frac{t^{1-p}}{1-p} + C,\]provided that \( p eq 1 \). In our problem, this means: \[-\frac{1}{0.001} t^{-0.001} = -1000t^{-0.001}.\]This computed antiderivative helps us evaluate the definite integral from \( \pi \) to a variable \( b \).
Subsequently, we apply the limits and simplify the expression:
\[\int_{\pi}^{b} \frac{1}{t^{1.001}} \, dt = \left[ -1000 t^{-0.001} \right]_{\pi}^{b} = -1000b^{-0.001} + 1000\pi^{-0.001}.\]
This illustrates how understanding power functions deeply aligns with simplifying and solving integrals efficiently.