Problem 4 - Calculus for Biology and Medicine

Question

All the integrals are improper and converge. Explain in each case why the integral is improper, andevaluate each integral. $$ \int_{e}^{\infty} \frac{d x}{x(\ln x)^{2}} $$

Step-by-Step Solution

Answer

The integral evaluates to 1, as it converges due to the expression \(-\frac{1}{\ln x} + 1\) approaching 1 as \(x\) approaches infinity.

Step 1: Identify Why the Integral is Improper

The integral \( \int_{e}^{\infty} \frac{d x}{x(\ln x)^{2}} \) is improper because it has an infinite upper limit. Evaluating such an integral directly from \( e \) to \( \infty \) is not possible, and thus we need to convert it into a limit.

Step 2: Express the Integral as a Limit

To handle the improper nature, express the integral as a limit: \[ \lim_{b \to \infty} \int_{e}^{b} \frac{d x}{x(\ln x)^{2}} \] This converts the problem to evaluating a definite integral with finite bounds and then taking the limit as \( b \) approaches infinity.

Step 3: Perform a Substitution

Let \( u = \ln x \). Then \( du = \frac{1}{x} \, dx \). This changes our integral to:\[ \int \frac{1}{x (\ln x)^2} \, dx = \int \frac{1}{u^2} \, du \] The limits of integration also change to \( u = \ln e = 1 \) (lower limit) and \( u = \ln b \) (upper limit).

Step 4: Integrate with Respect to u

Compute the indefinite integral:\[ \int \frac{1}{u^2} \, du = -\frac{1}{u} + C \]Therefore, the definite integral becomes:\[ \left[ -\frac{1}{u} \right]_{1}^{\ln b} = -\frac{1}{\ln b} + 1 \]

Step 5: Evaluate the Limit

Substitute back to evaluate the original limit:\[ \lim_{b \to \infty} \left(-\frac{1}{\ln b} + 1 \right) = \lim_{b \to \infty} -\frac{1}{\ln b} + 1 \]As \( b \to \infty \), \( \ln b \to \infty \), causing \( \frac{1}{\ln b} \to 0 \). Thus, the limit simplifies to 1.

Key Concepts

Integration Techniques

In calculus, integration techniques are essential tools that help us find the area under a curve. They allow us to solve integrals that don't immediately yield simple antiderivatives. These techniques can include:

• Basic Integration
• Substitution Method (a.k.a. change of variables)
• Integration by Parts
• Partial Fractions

For improper integrals, the usual direct integration isn't feasible if the interval is infinite or if the integrand is undefined at some point in the interval. In such cases, conversion to a limit, as seen in our exercise, can be very useful. In our example, we converted the improper integral into a limit, taking advantage of the fact that limits can manage infinite behavior by approaching finite values.

Limit Evaluation

The evaluation of limits is a critical aspect of solving improper integrals as it allows us to handle the infinite elements that might otherwise prevent evaluation. In the case of the integral from the exercise:- We translated the problem from evaluating an integral with an infinite domain to instead evaluating a limit.- The integral, which is from \( e \) to \( \infty \), is expressed as a limit: \( \lim_{b \to \infty} \int_{e}^{b} \frac{d x}{x(\ln x)^{2}} \).This step is vital because a limit can simplify the analysis of how the function behaves as the variable approaches a particular value or infinity. In the solution's final step, this understanding allows us to see that as \( b \to \infty \), the value of \( \frac{1}{\ln b} \) approaches zero, leading simply to the result 1.

Substitution Method

The substitution method involves changing the variable of integration, which makes some integrals significantly simpler to evaluate.For the given problem, the substitution method involves:

• Setting \( u = \ln x \) to simplify the expression \( \frac{1}{x(\ln x)^2} \).
• Recognizing that \( du = \frac{1}{x} \, dx \), which allows us to rewrite the integral in terms of \( u \).

Thus, the integral is transformed to:\( \int \frac{1}{u^2} \, du \), which is much easier to integrate.After computing the new simple integral, we remember to change back to the original variable (if necessary) and apply the limits like \( u = \ln e \) and \( u = \ln b \) for full solution evaluation. Such transformations often lead to a more tractable problem, helping us to proceed with the limit evaluation accurately.