Problem 38 - Calculus for Biology and Medicine

Question

(a) Show that $$ \frac{1}{\sqrt{x+\ln x}} \geq \frac{1}{\sqrt{2 x}}>0 $$ for \(x \geq 1\). (b) Use your result in (a) to show that $$ \int_{1}^{\infty} \frac{1}{\sqrt{x+\ln x}} d x $$ is divergent.

Step-by-Step Solution

Answer

The inequality and the integral comparison show the integral diverges.

Step 1: Analyze the inequality

We have the inequality \( \frac{1}{\sqrt{x+\ln x}} \geq \frac{1}{\sqrt{2x}} \). Since both denominators are positive for \( x \geq 1 \), we can square both sides without changing the inequality's direction. This gives us \( x + \ln x \leq 2x \).

Step 2: Simplify the inequality

Subtract \( x \) from both sides of \( x + \ln x \leq 2x \), which results in \( \ln x \leq x \). This inequality is true for any \( x \geq 1 \) because \( \ln x \) grows slower than \( x \).

Step 3: Verify positivity

The inequality \( \frac{1}{\sqrt{2x}} > 0 \) is clearly true for \( x \geq 1 \) since the square root of \( 2x \) and the fraction \( \frac{1}{\text{positive number}} \) are both positive.

Step 4: Set up the integral comparison

According to part (a), we have \( \frac{1}{\sqrt{x+\ln x}} \geq \frac{1}{\sqrt{2x}} \). We will use this comparison to examine the integral \( \int_{1}^{\infty} \frac{1}{\sqrt{x+\ln x}} \, dx \).

Step 5: Analyze the comparison integral

Consider the integral \( \int_{1}^{\infty} \frac{1}{\sqrt{2x}} \, dx \). Make the substitution \( u = \sqrt{2x} \), with \( du = \frac{1}{\sqrt{2x}} \, dx \). Thus, the integral becomes \( \int_{\sqrt{2}}^{\infty} du = \infty \). This integral diverges.

Step 6: Conclusion using comparison test

Since the integral \( \int_{1}^{\infty} \frac{1}{\sqrt{2x}} \, dx \) diverges, and \( \frac{1}{\sqrt{x+\ln x}} \geq \frac{1}{\sqrt{2x}} \), by the comparison test, the integral \( \int_{1}^{\infty} \frac{1}{\sqrt{x+\ln x}} \, dx \) also diverges.

Key Concepts

Inequality Analysis

When working with inequalities in calculus, especially in the context of integral divergence, understanding how to manipulate and analyze inequalities is crucial. In the exercise provided, the inequality in question is \( \frac{1}{\sqrt{x+\ln x}} \geq \frac{1}{\sqrt{2x}} \) for \( x \geq 1 \). To justify this inequality, we use the property that when both sides of an inequality have positive denominators, squaring both sides preserves the inequality direction. This is because both denominators are non-negative for \( x \geq 1 \), resulting in the simplified form \( x + \ln x \leq 2x \). Upon further simplification, subtracting \( x \) from both sides results in \( \ln x \leq x \), which holds true for \( x \geq 1 \). This is due to the logarithmic function \( \ln x \), which increases at a slower rate than the linear function \( x \). As logarithmic inequalities are foundational to analyzing convergence and divergence, understanding this gradual growth is fundamental.

Comparison Test

The comparison test is a valuable tool for determining the convergence or divergence of improper integrals. This technique involves comparing a given integral to another for which the behavior (convergent or divergent) is already known. In this case, we aim to determine the behavior of the integral \( \int_{1}^{\infty} \frac{1}{\sqrt{x+\ln x}} \, dx \) by comparing it to \( \int_{1}^{\infty} \frac{1}{\sqrt{2x}} \, dx \). From the inequality analysis, we establish that \( \frac{1}{\sqrt{x+\ln x}} \geq \frac{1}{\sqrt{2x}} \), meaning the integral of \( \frac{1}{\sqrt{x+\ln x}} \) could potentially be larger than the integral of \( \frac{1}{\sqrt{2x}} \). To utilize the comparison test effectively:

• First, confirm that the reference integral, \( \int_{1}^{\infty} \frac{1}{\sqrt{2x}} \, dx \), diverges. Through substitution, let \( u = \sqrt{2x} \), transforming it into \( \int_{\sqrt{2}}^{\infty} du \). This integral grows without bound and is thus divergent.
• Since \( \frac{1}{\sqrt{x+\ln x}} \) is greater than or equal to \( \frac{1}{\sqrt{2x}} \), and because the comparison integral diverges, our target integral \( \int_{1}^{\infty} \frac{1}{\sqrt{x+\ln x}} \, dx \) must also diverge according to the comparison test.

The comparison test is particularly valuable as it provides a straightforward path to conclude the behavior of a more complex function by relating it to a simpler or familiar integral.

Improper Integrals

Improper integrals are a type of integral where the interval is infinite or the integrand has infinite discontinuities within the interval. These integrals require careful analysis, often involving comparison with known diverging or converging integrals. The integral \( \int_{1}^{\infty} \frac{1}{\sqrt{x+\ln x}} \, dx \) is improper because the upper limit of integration is infinity. Evaluating improper integrals involves looking at limits to determine their convergence or divergence. For the given integral:

• We express it in terms of a limit notation: \( \lim_{b \to \infty} \int_{1}^{b} \frac{1}{\sqrt{x+\ln x}} \, dx \).
• By utilizing the comparison test discussed earlier, the behavior of this integral can be linked to another improper integral that is easier to evaluate or already evaluated as divergent or convergent.

Therefore, understanding improper integrals requires familiarity with techniques like limit operations and comparison tests. This helps determine whether the value approaches a finite bound or diverges as the interval extends to infinity. Engaging with improper integrals helps uncover and confirm the nature of these challenging, unbounded scenarios in real-world contexts.