Problem 41 - Calculus for Biology and Medicine

Question

Find a comparison function for each integrand and determine whether the integral is convergent. $$ \int_{1}^{\infty} \frac{1}{\sqrt{1+x}} d x $$

Step-by-Step Solution

Answer

The integral \( \int_{1}^{\infty} \frac{1}{\sqrt{1+x}} \, dx \) diverges.

Step 1: Choose an Appropriate Comparison Function

The first step is to choose a function that is similar to the integrand \( \frac{1}{\sqrt{1+x}} \). As \( x \) becomes very large, \( \sqrt{1+x} \approx \sqrt{x} \). This suggests using the function \( \frac{1}{\sqrt{x}} \) for our comparison. This is because for large \( x \), the behavior of both functions is similar and it simplifies the evaluation.

Step 2: Determine Functional Inequalities

For the comparison test, we need to establish an inequality between \( \frac{1}{\sqrt{1+x}} \) and \( \frac{1}{\sqrt{x}} \). Note that \( 1+x > x \) implies \( \sqrt{1+x} > \sqrt{x} \). Thus, \( \frac{1}{\sqrt{1+x}} < \frac{1}{\sqrt{x}} \) for all \( x \geq 1 \). This comparison will help us determine convergence.

Step 3: Evaluate the Comparison Integral

We need to determine if the integral \( \int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx \) converges. This integral is a well-known p-integral of the form \( \int \frac{1}{x^{1/2}} \, dx \) where \( p = 1/2 \). A p-integral \( \int_{1}^{\infty} x^{-p} \, dx \) converges if \( p > 1 \), which is not the case here since \( p=1/2 \). Thus, \( \int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx \) diverges.

Step 4: Conclude by the Comparison Test

Using the comparison test, since \( \frac{1}{\sqrt{1+x}} < \frac{1}{\sqrt{x}} \) and \( \int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx \) diverges, by the comparison test, \( \int_{1}^{\infty} \frac{1}{\sqrt{1+x}} \, dx \) also diverges.

Key Concepts

Comparison Test

In the realm of improper integrals, particularly those stretching to infinity, the comparison test is a potent tool to determine whether an integral converges (has a finite value) or diverges (grows without bound). To effectively utilize this test, we compare the given integral to a more familiar integral whose behavior we already understand.
During the comparison, our goal is to choose a comparison function with a known convergence or divergence behavior. For example, consider the integral \[\int_{1}^{\infty} \frac{1}{\sqrt{1+x}} \, dx\]
To utilize the comparison test:

• Select a function that is very similar to the integrand. For large values of \(x\), \(\sqrt{1+x}\) behaves like \(\sqrt{x}\), so a logical choice for comparison is \(\frac{1}{\sqrt{x}}\).


• Establish a clear inequality between the two. Here, \(\frac{1}{\sqrt{1+x}} < \frac{1}{\sqrt{x}}\) for \(x \geq 1\), because \(1+x > x\) ensures \(\sqrt{1+x} > \sqrt{x}\).

After establishing this relationship, if the larger function’s integral diverges, so does the original one. Conversely, if it converges, the smaller original one must too.

Convergence

In calculus, convergence of an integral refers to whether the total accumulation of area under the curve is finite as its bounds stretch to infinity. Not all functions yield finite integrals when extended indefinitely; identifying convergence or divergence is crucial.
For example, the integral involving \(\frac{1}{\sqrt{1+x}}\) seeks to find whether there's a finite area beneath as \(x\) approaches infinity.
We establish convergence using:

• Comparing to a simpler integral: By analogy with \(\frac{1}{\sqrt{x}}\), noticing if \(x\) large makes the behavior similar.


• Determining if simple integrals diverge or converge, providing insight for comparison integrals.

An integral converges if it settles at a finite value beyond some point. Conversely, if it stretches endlessly, this indicates divergence.

P-Integrals

P-integrals are a class of integrals in the form \(\int x^{-p} \, dx\), where \(p\) is a constant. They provide valuable insight into convergence behaviors based on their exponents.
In cases where the integral bounds tend towards infinity:

• Specifically, the integral \(\int_{1}^{\infty} \frac{1}{x^{p}} \, dx\) converges if \(p > 1\). This is because the area diminishes fast enough as \(x\) increases.


• If \(0 < p \leq 1\), like the case \(p = 1/2\) in \(\int \frac{1}{\sqrt{x}} \, dx\), the integral diverges since the area under the curve doesn’t decrease rapidly enough to yield a finite result.

Understanding p-integrals aids in predicting the convergence of more complex improper integrals by providing a direct, simplified benchmark for comparison.