Problem 42 - Calculus for Biology and Medicine

Question

Find a comparison function for each integrand and determine whether the integral is convergent. $$ \int_{1}^{\infty} \frac{1}{\left(1+x^{4}\right)^{1 / 3}} d x $$

Step-by-Step Solution

Answer

The integral converges by comparison with \( \frac{1}{x^{4/3}} \).

Step 1: Understand the Integral

We are given the improper integral \( \int_{1}^{\infty} \frac{1}{(1+x^4)^{1/3}} \, dx \). We need to determine if this integral converges or diverges.

Step 2: Identify the Problematic Area

The integral is defined over the interval \([1, \infty)\). The problematic area for convergence is as \(x\) approaches infinity because this is what could cause divergence.

Step 3: Simplify the Integrand for Large x

As \(x \to \infty\), the term \(1 + x^4\) behaves approximately like \(x^4\). Thus, \((1+x^4)^{1/3} \approx (x^4)^{1/3} = x^{4/3}\). Therefore, the integrand simplifies to approximately \(\frac{1}{x^{4/3}}\) for large \(x\).

Step 4: Choose a Comparison Function

We choose the comparison function \(g(x) = \frac{1}{x^{4/3}}\) because it behaves similarly to the integrand for large \(x\). We know that \(\int_{1}^{\infty} \frac{1}{x^{4/3}} \, dx\) determines convergence properties.

Step 5: Test the Comparison Function for Convergence

The integral \(\int_{1}^{\infty} \frac{1}{x^{4/3}} \, dx\) is a p-type integral with \(p = \frac{4}{3} > 1\). Such integrals are known to converge.

Step 6: Apply the Comparison Test

Since \( (1 + x^4)^{1/3} > x^{4/3} \) for \(x\) large, we have \( \frac{1}{(1 + x^4)^{1/3}} < \frac{1}{x^{4/3}} \). By the comparison test, since \( \int_{1}^{\infty} \frac{1}{x^{4/3}} \, dx \) converges, \( \int_{1}^{\infty} \frac{1}{(1+x^4)^{1/3}} \, dx \) also converges.

Key Concepts

Convergence

In the realm of calculus, determining whether an improper integral converges or diverges is a fundamental question. Convergence, in simple terms, means that the integral settles to a finite value despite encompassing an infinite range. When we deal with improper integrals like \( \int_{1}^{\infty} \frac{1}{(1+x^4)^{1/3}} \, dx \), we must investigate how the function behaves as \( x \) approaches infinity.

In this exercise, we are primarily concerned with the behavior of the integrand as the independent variable \( x \) grows very large. If the function within the integral becomes smaller rapidly enough as \( x \) increases, the total area beneath the curve may settle to a finite number, indicating convergence.

Convergence requires careful analysis, often involving comparison with known functions. These comparisons are facilitated using tests such as the Comparison Test, which allows us to determine the convergence status of complex integrals using simpler, well-understood functions.

Comparison Test

The Comparison Test is a valuable tool in calculus for determining the convergence of improper integrals. It is based on comparing the integral of interest to another integral whose convergence properties are already understood.

In this problem, once we simplified \((1+x^4)^{1/3}\) to resemble \(x^{4/3}\) for large \(x\), we selected a comparison function \(g(x) = \frac{1}{x^{4/3}}\). The comparison function should mirror the troublesome integral's behavior at infinity.

Applying the Comparison Test involves the following steps:

• Identifying a comparison function that bounds the integrand from above or below.
• Knowing whether the comparison integral \( \int \frac{1}{x^{4/3}} \, dx \) converges or diverges.
• Using the known behavior of the comparison integral to infer the behavior of the given integral.

For the given integral, since \( (1 + x^4)^{1/3} \) is slightly greater than \( x^{4/3} \) for large \( x \), the inverse relationship \( \frac{1}{(1 + x^4)^{1/3}} \) is less than \( \frac{1}{x^{4/3}} \). Since the latter converges, so does the integral of interest by the Comparison Test.

p-type integral

A p-type integral is characterized by the form \( \int_{a}^{b} \frac{1}{x^p} \, dx \), where the behavior of the integral is largely dictated by the value of \( p \). This simplicity makes them excellent candidates for comparison when using the Comparison Test.

In this solution, after simplifying our integrand to resemble \( \frac{1}{x^{4/3}} \), we recognized it as a classic p-type integral with \( p = \frac{4}{3}\). The integral \( \int_{1}^{\infty} \frac{1}{x^{4/3}} \, dx \) converges because \( p > 1 \). This is a well-known rule:

• If \( p > 1 \), the integral \( \int_{1}^{\infty} \frac{1}{x^p} \, dx \) converges.
• If \( p \leq 1 \), the integral diverges.

Understanding this pattern is crucial for identifying the convergence of integrals at a glance. In our case, having \( p = \frac{4}{3} \) made it clear that both the comparison integral and the original integral converge, providing a straightforward conclusion to the problem.

Remember, p-type integrals are not only foundational in mathematical theory but also serve as effective benchmarks for understanding more complex improper integrals.