Problem 18 - Calculus for Biology and Medicine

Question

Determine whether each integral is convergent. If the integral is convergent, compute its value. $$ \int_{-1}^{0} \frac{1}{x+1} d x $$

Step-by-Step Solution

Answer

The integral is divergent; it does not converge.

Step 1: Identify the Potential Issue in the Integral

The function \( f(x) = \frac{1}{x+1} \) has a potential issue at \( x = -1 \). When approaching \( x = -1 \) from the right, the function heads towards negative infinity. Our task is to find out if the improper integral converges, despite this discontinuity.

Step 2: Use the Cauchy Principal Value

The integral can be rewritten as a limit due to the improper nature at \( x = -1 \). We evaluate the integral as a limit: \[ \lim_{b \to -1^+} \int_{b}^{0} \frac{1}{x+1} \, dx \] This translates our problem to finding the limit as \( b \) approaches \( -1^+ \).

Step 3: Calculate the Antiderivative

Find the antiderivative of \( \frac{1}{x+1} \), which is \( \ln |x+1| + C \). We can now use this antiderivative to evaluate the definite integral: \[ \lim_{b \to -1^+} [ \ln |x+1| ]_b^0\]

Step 4: Evaluate the Definite Integral as a Limit

Substitute the limits into the antiderivative, then evaluate: \[\lim_{b \to -1^+} [ \ln |0+1| - \ln |b+1| ] = \lim_{b \to -1^+} [ \ln 1 - \ln(b+1)] = \lim_{b \to -1^+} - \ln(b+1)\].As \( b \to -1^+ \), \( b+1 \to 0^+ \) and \( \ln(b+1) \to -\infty \). This means the expression \(- \ln(b+1) \to \infty \), showing divergence.

Key Concepts

Cauchy Principal Value

When dealing with improper integrals that have a point of discontinuity, such as \( \int_{-1}^{0} \frac{1}{x+1} \ dx \), the Cauchy Principal Value (CPV) is a technique that helps us to rigorously evaluate the integral. In the given integral, the discontinuity arises at \( x = -1 \), causing challenges because the function \( f(x) = \frac{1}{x+1} \) tends to negative infinity as it approaches \( x = -1 \) from the right.
To determine the CPV, we rewrite the problematic integral as a limit, focusing on the behavior of the function as it approaches the point of discontinuity. In this case, the CPV evaluates:

• \( \lim_{b \to -1^+} \int_{b}^{0} \frac{1}{x+1} \ dx \)

This approach allows us to frame the integral as a limit of increasingly small sections near the discontinuity, improving our understanding of whether the entire integral converges. However, it's important to recognize that even if the CPV exists, it does not guarantee actual convergence of the integral unless the limit resolves to a finite value.

Antiderivative

An antiderivative is a function whose derivative is the original function we are integrating. It is fundamental in solving definite and indefinite integrals. For the function \( f(x) = \frac{1}{x+1} \), the antiderivative is \( \ln |x+1| + C \), where \( C \) is the constant of integration.
In the context of evaluating improper integrals, the antiderivative is used to compute the exact area under the curve over the specified interval. Once the antiderivative \( \ln|x+1| \) is determined, we can utilize the Fundamental Theorem of Calculus to evaluate the definite integral. For example, the expression:

• \( [ \ln |x+1| ]_b^0 = \ln|0+1| - \ln|b+1| \)

Using this, the calculus becomes straightforward, except for the continuation to limits when improper behavior is present. This process underscores the critical role of antiderivatives in assessing the convergence properties of integrals.

Divergence of Integrals

Integrals can be classified based on their convergence properties. In essence, an integral is said to be divergent if it does not resolve to a finite number. Improper integrals, like \( \int_{-1}^{0} \frac{1}{x+1} \ dx \), are particularly prone to divergence due to their discontinuities or infinite limits.
In the example we are discussing, the divergence is evident from the evaluation of the integral as a limit:

• \( \lim_{b \to -1^+} - \ln(b+1) \)

As \( b \) approaches \( -1 \) from the right, \( b+1 \) approaches zero, which leads \( \ln(b+1) \) to approach negative infinity. Consequently, \( -\ln(b+1) \) heads towards positive infinity, confirming that the integral diverges. Even though techniques like the Cauchy Principal Value can sometimes provide insight, they don't change the fundamental divergence behavior, which means this integral does not have a finite value.