Question

Determine the convergence of each of the following integrals by comparison with the given integral. If the integral converges, find the number to which it converges.\(\int_{1}^{\infty} \frac{d x}{\sqrt{x}+1} ;\) compare with \(\int_{1}^{\infty} \frac{d x}{2 \sqrt{x}}\)

Short answer

The integral diverges by comparison.

Step-by-step solution

Analyze Comparison Integral

Consider the integral \( \int_{1}^{\infty} \frac{dx}{2\sqrt{x}} \). This integral can be rewritten as \( \frac{1}{2} \int_{1}^{\infty} x^{-1/2} \, dx \). We know that \( \int x^p \, dx \) converges if and only if \( p < -1 \). Here, \( p = -1/2 \). Since \(-1/2 > -1\), this integral diverges.

Set Up the Comparison

We will compare \( \int_{1}^{\infty} \frac{dx}{\sqrt{x}+1} \) with \( \int_{1}^{\infty} \frac{dx}{2\sqrt{x}} \). For \( x \geq 1 \), we have \( \sqrt{x} + 1 \geq \sqrt{x} \), thus \( \frac{1}{\sqrt{x}+1} \leq \frac{1}{\sqrt{x}} \). Dividing by 2, \( \frac{1}{\sqrt{x}+1} \leq \frac{1}{2\sqrt{x}} \) holds true for \( x \geq 1 \).

Apply the Comparison Test

The Comparison Test states that if \( 0 \leq f(x) \leq g(x) \) for all \( x \) in \([a, \infty)\), and \( \int_{a}^{\infty} g(x) \, dx \) diverges, then \( \int_{a}^{\infty} f(x) \, dx \) also diverges. Since \( \int_{1}^{\infty} \frac{dx}{2\sqrt{x}} \) diverges and \( \frac{1}{\sqrt{x}+1} \leq \frac{1}{2\sqrt{x}} \), it follows from the test that \( \int_{1}^{\infty} \frac{dx}{\sqrt{x}+1} \) also diverges.

Key concepts

Convergence Tests

Improper integrals often require us to determine if they converge or diverge as they approach an infinite limit. Convergence tests come in handy for this analysis, providing specific criteria to decide whether an integral converges or not. Essentially, these tests help us to verify if an integral results in a finite number or not when its limit goes to infinity. Some popular convergence tests include the Comparison Test, Limit Comparison Test, and the Integral Test, each one serving a specific type of function or scenario. These tests look at what happens to the function in the limit, guiding us in deducing the behavior of more complex improper integrals. Convergence tests are crucial when evaluating integrals like \[\int_{1}^{\infty} \frac{d x}{\sqrt{x}+1} \] where direct evaluation can be challenging. By using these methods, we can better understand the nature of the integral at infinite limits and simplified approximation using comparable simpler integral functions.

Comparison Test

The Comparison Test is an extremely useful tool when evaluating the convergence of improper integrals. The essence of this test is straightforward: compare the integral in question with another integral that is easier to evaluate. The known behavior of the simpler integral acts like a guideline to interpret the more complex one.For example, comparing \[\int_{1}^{\infty} \frac{d x}{\sqrt{x}+1} \] with \[\int_{1}^{\infty} \frac{dx}{2\sqrt{x}} \] can be insightful. If a function \( f(x) \) is less than or equal to another function \( g(x) \), and \( g(x) \) diverges, then \( f(x) \) also diverges. In our case, \[\frac{1}{\sqrt{x}+1} \leq \frac{1}{2\sqrt{x}}\] for \( x \geq 1 \). We've determined that the integral of \( \frac{1}{2\sqrt{x}} \) diverges, meaning the Comparison Test allows us to conclude the divergence of our original integral too.

Divergence of Integrals

Improper integrals may diverge, indicating that as their limits extend toward infinity, they do not settle to a finite value. Recognizing divergence is as crucial as finding convergence because it defines how functions behave over infinite intervals.The divergence of integrals often results from the function not leveling off fast enough to counterbalance the infinite extent of the range. In terms of convergence tests, a diverging integral provides evidence for comparison in tests like the Comparison Test. For instance, the integral \[\int_{1}^{\infty} \frac{dx}{2\sqrt{x}}\] was found to diverge because its exponent \( p \) in \( x^{-1/2} \) does not satisfy the condition \( p < -1 \) for convergence. Identifying this can impact the results of comparing other integrals that show similar or lesser growth rates. Properly assessing divergence ensures that we correctly classify the behavior of more complex integrals.