Problem 35 - Calculus for Biology and Medicine

Question

In this problem, we investigate the integral \(\int_{1}^{\infty} \frac{1}{x^{p}} d x\) for \(01\), set \(A(z)=\int_{1}^{z} \frac{1}{x^{p}} d x\) and show that $$A(z)=\frac{1}{1-p}\left(z^{-p+1}-1\right) \quad \text { for } p \neq 1$$ and $$A(z)=\ln z \quad \text { for } p=1$$ (b) Use your results in (a) to show that, for \(0

1\), $$ \lim _{z \rightarrow \infty} A(z)=\frac{1}{p-1} $$

Step-by-Step Solution

Answer

(a) Showed integrals; (b) \(\lim A(z) = \infty\) for \(01\).

Step 1: Define A(z) for General p

Given the function \( A(z) = \int_{1}^{z} \frac{1}{x^{p}} \, dx \). We need to evaluate this integral for different values of \( p \). For \( p eq 1 \), perform integration of \( \frac{1}{x^{p}} \) with respect to \( x \) using the power rule.

Step 2: Integrate A(z) for p Not Equal to 1

The integration of \( \frac{1}{x^p} \) yields:\[ \int \frac{1}{x^p} \, dx = \frac{x^{-p+1}}{-p+1} = \frac{x^{-p+1}}{1-p} \]Evaluating from 1 to \( z \),\[ A(z) = \frac{z^{-p+1} - 1}{1-p} \]

Step 3: Integrate A(z) for p Equals to 1

When \( p = 1 \), the integrand becomes \( \frac{1}{x} \), giving us:\[ \int \frac{1}{x} \, dx = \ln |x| \]Evaluating from 1 to \( z \),\[ A(z) = \ln z \]

Step 4: Evaluate Limit for 0 < p <= 1

For \( 0 < p < 1 \), use the result from Step 2:\[ A(z) = \frac{z^{-p+1} - 1}{1-p} \]Since \( -p+1 > 0 \), as \( z \to \infty \), \( z^{-p+1} \to \infty \), making \( A(z) \to \infty \).For \( p = 1 \), \( A(z) = \ln z \), and as \( z \to \infty \), \( \ln z \to \infty \). Hence, \( \lim_{z \to \infty} A(z) = \infty \) for \( 0 < p \leq 1 \).

Step 5: Evaluate Limit for p > 1

For \( p > 1 \), the integration result from Step 2 is used:\[ A(z) = \frac{z^{-p+1} - 1}{1-p} \]Here, \( -p+1 < 0 \), so \( z^{-p+1} \to 0 \) as \( z \to \infty \). Thus, the limit becomes:\[ \lim_{z \to \infty} A(z) = \frac{0 - 1}{1-p} = \frac{1}{p-1} \]

Key Concepts

Convergence

In the world of improper integrals, understanding convergence is crucial. Convergence essentially tells us whether the integral of a given function over an infinite interval results in a finite or infinite value. When we evaluate an integral like \( \int_{1}^{\infty} \frac{1}{x^p} \, dx \), an important question arises: Does this integral converge or diverge as the upper limit approaches infinity?

• For \( 0 < p \leq 1 \), the integral does not converge. As \( z \to \infty \), the integral blossoming towards infinity indicates that it diverges.
• For \( p > 1 \), a different behavior occurs. Here, the integral converges to \( \frac{1}{p-1} \), which is a finite number. This result means the function stabilizes as the interval expands.

Recognizing convergence tells us a lot about the behavior of functions over extended domains, providing insight into their long-term tendencies.

Divergence

Divergence is the opposite of convergence. It describes scenarios where the value of an integral grows indefinitely. In our specific task of evaluating \( A(z) = \int_{1}^{z} \frac{1}{x^p} \, dx \), understanding divergence is essential.

• When \( 0 < p \leq 1 \), the result is \( A(z) \to \infty \) as \( z \to \infty \). This behavior is characteristic of divergence.
• With \( p = 1 \), the logarithmic function \( A(z) = \ln z \) further exemplifies this outcome, as \( \ln z \) also tends to infinity with increasing \( z \).

Grasping divergence is crucial in calculus as it highlights limits of infinities and boundaries beyond which functions cannot be neatly contained within finite values.

Power Rule

The power rule is a fundamental tool in calculus, particularly valuable when integrating functions of the form \( \frac{1}{x^p} \). Applying it allows us to solve otherwise challenging integrals:

• For \( p eq 1 \), integrating \( \frac{1}{x^p} \) results in \( \frac{x^{-p+1}}{1-p} \). This formula emerges due to the power rule, which simplifies the process of integration by modifying the exponent rules.
• If \( p = 1 \), the power rule is not directly applied due to the indefinite result, leading us to use logarithmic integration instead.

The power rule simplifies the process of integration significantly, offering a straightforward approach in handling polynomial powers during calculus operations.

Logarithmic Function

When dealing with integrands like \( \frac{1}{x} \), we enter the fascinating realm of logarithmic functions. Logarithmic integration offers a distinct path when the power rule does not apply.

• For \( p = 1 \), integrating results in \( A(z) = \ln z \), capturing the behavior of the logarithmic function.
• Logarithmic functions rise slowly, heading towards infinity as their variable increases, making them key in understanding functions that diverge at a slow pace.

Logarithmic integration provides insight and solutions for integrands that aren't easily solved by standard power rule methods, highlighting the versatility required in calculus.