Problem 15 - Calculus for Biology and Medicine

Question

All the integrals in problem are improper and converge. Explain in each case why the integral is improper, and evaluate each integral. $$ \int_{-1}^{1} \ln |x| d x $$

Step-by-Step Solution

Answer

The integral is improper at \(x=0\) and evaluates to \(-2\).

Step 1: Identify Improper Points

The function \(\ln |x|\) involves taking the natural log of the absolute value of \(x\). The integral \(\int_{-1}^{1} \ln |x| \, dx\) is improper because \(\ln |x|\) is undefined at \(x = 0\). This point is within the interval of integration, so we need to handle it carefully.

Step 2: Break the Integral at the Improper Point

To handle the improper integral, we split it into two parts around the point where it becomes undefined, i.e., at \(x = 0\). Thus, we express the integral as: \(\int_{-1}^{1} \ln |x| \, dx = \int_{-1}^{0} \ln(-x) \, dx + \int_{0}^{1} \ln x \, dx.\)

Step 3: Evaluate Each Integral Separately

Evaluate \(\int_{-1}^{0} \ln(-x) \, dx\) by substituting \(u = -x\), which makes \(du = -dx\). The limits change from \(-1\) to \(0\) to \(1\) to \(0\), respectively. This gives \(-\int_{1}^{0} \ln u \, du = \int_{0}^{1} \ln u \, du.\)For the second integral \(\int_{0}^{1} \ln x \, dx\), both parts are identical as they evaluate over the same function and interval.

Step 4: Calculate the Definite Integral

We need to evaluate \(\int \ln x \, dx\). We can integrate by parts:\(\int \ln x \, dx = x \ln x - \int x \frac{1}{x} \, dx = x \ln x - x + C.\)Apply limits from \(0\) to \(1\) and recall that the logarithmic term diverges as \(x\) approaches 0 making the contribution to the integral infinite. Thus:\(\left[x \ln x - x\right]_{0}^{1} = \lim_{t \to 0^+} \left[\left(t \ln t - t\right) - (0 \ln 0 - 0) \right] = \lim_{t \to 0^+} (t \ln t - t).\)This evaluates to \(-1\) as \(t \ln t\) approaches 0.

Step 5: Combine the Results

Now combine both results: \(\int_{-1}^{0} \ln(-x) \, dx = -1 \quad \text{and} \quad \int_{0}^{1} \ln x \, dx = -1,\)Add these results to find the solution to the original problem:\(\int_{-1}^{1} \ln |x| \, dx = -1 + (-1) = -2.\)

Key Concepts

Natural Logarithm

The natural logarithm, represented as \( \ln(x) \), is a special logarithm where the base is the constant \( e \approx 2.71828 \). It is widely used in calculus, especially in the context of integrals. Understanding how \( \ln(x) \) behaves is crucial, as it can take both positive and negative values depending on the input. To understand its behavior, consider the following characteristics:

• \( \ln(x) \) is defined for \( x > 0 \).
• For \( x = 1 \), \( \ln(1) = 0 \).
• As \( x \) increases, \( \ln(x) \) also increases.
• As \( x \) approaches 0 from the positive side, \( \ln(x) \) approaches negative infinity.

For our problem, \( \ln |x| \) is of interest since it considers the absolute value of \( x \). This means \( \ln |x| \) is defined for all \( x eq 0 \), thus becoming undefined exactly at \( x = 0 \), which creates the need to split the integral for evaluation.

Integration by Parts

Integration by parts is a useful technique in calculus that extends the process of integrating products of functions. It's based on the product rule for differentiation, and it helps in breaking down more complicated integrals into simpler parts. The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]Here is how it works step-by-step:

• Identify parts of the integrand as \( u \) and \( dv \).
• Differentiate \( u \) to find \( du \) and integrate \( dv \) to find \( v \).
• Substitute into the formula to simplify the integral.

In the context of our problem, integration by parts is applied to \( \int \ln(x) \, dx \). By letting \( u = \ln(x) \) and \( dv = dx \), we find \( du = \frac{1}{x} \, dx \) and \( v = x \), leading to:\[\int \ln(x) \, dx = x \ln(x) - x + C\]After applying limits, the result contributes significantly to the overall solution of the problem.

Convergence of Integrals

Convergence of an integral is a concept that refers to the behavior of the integral over the given interval. An integral is said to converge if it results in a finite value. Otherwise, it diverges, leading to an infinite result. For improper integrals, convergence is particularly vital, as they involve boundaries or points where the function becomes undefined.Here are key ideas around convergence:

• **Improper integrals** occur when an integrand exhibits undefined values or infinite bounds.
• Splitting the integral at improper points can assist in efficient evaluation, as in our example with \( x = 0 \).
• When integrating functions like \( \ln |x| \), special attention is needed where the function tends to infinity, such as near zero.

In our example, the limit process is used to determine the convergence around the improper point \( x = 0 \). By evaluating \( \lim_{t \to 0^+} (t \ln t - t) \), we find that this specific integral contributes \(-1\) to the total integral. Therefore, breaking the integral into parts and assessing each for convergence leads to the conclusion that the integral \( \int_{-1}^{1} \ln |x| \, dx \) converges to \(-2\).