Question

For the following exercises, evaluate the integrals, if possible.\(\int_{1}^{\infty} \frac{1}{x^{n}} d x\), for what values of \(n\) does this integral converge or diverge?

Short answer

The integral converges for \( n > 1 \).

Step-by-step solution

Understand the Integral

The integral given is \( \int_{1}^{\infty} \frac{1}{x^{n}} \, dx \). We want to determine the values of \( n \) for which this integral converges or diverges. This type of integral is an improper integral because the upper limit is infinite.

Apply the Integral Test for Convergence

We will use the integral test for convergence. The integral form of the test states that the improper integral \( \int_{1}^{\infty} f(x) \, dx \) converges if and only if the function \( f(x) = \frac{1}{x^{n}} \) has a positive continuous decreasing derivative for \( x \geq 1 \) and the limit exists.

Compute the Antiderivative

The next step is to find the antiderivative. For the function \( f(x) = \frac{1}{x^{n}} \), we can rewrite it as \( x^{-n} \). The antiderivative of \( x^{-n} \) is \( \frac{x^{1-n}}{1-n} + C \) for \( n eq 1 \).

Evaluate the Improper Integral

We compute \( \lim_{b \to \infty} \int_{1}^{b} x^{-n} \, dx \). Plugging in the antiderivative, we get:\[ \lim_{b \to \infty} \left[ \frac{x^{1-n}}{1-n} \right]_{1}^{b} = \lim_{b \to \infty} \left( \frac{b^{1-n}}{1-n} - \frac{1^{1-n}}{1-n} \right) \]

Apply Limit for Convergence Test

We analyze the limit \( \lim_{b \to \infty} \frac{b^{1-n}}{1-n} \). If \( 1-n > 0 \), then as \( b \to \infty \), the term \( b^{1-n} \to 0 \). Thus:- The integral converges if \( n > 1 \).- The integral diverges if \( n \leq 1 \), because the limit does not tend to 0 or becomes undefined.

Key concepts

Antiderivative

An antiderivative is a function that reverses the process of differentiation. Basically, if you have a function \( f(x) \), its antiderivative, usually denoted as \( F(x) \), satisfies \( F'(x) = f(x) \). Figuring out an antiderivative is crucial for solving an integral, as it allows us to reverse-engineer how changes in \( x \) relate to the area under a curve.
For example, if we take the function \( \frac{1}{x^n} \), we can reframe it as \( x^{-n} \). From here, we use the power rule to find the antiderivative. The power rule for integration states that the antiderivative of \( x^m \) is \( \frac{x^{m+1}}{m+1} + C \) (given that \( m eq -1 \)).
So in our case, the antiderivative of \( x^{-n} \) is \( \frac{x^{1-n}}{1-n} + C \), where \( C \) is the constant of integration. When \( n = 1 \), the function looks like \( \frac{1}{x} \), whose antiderivative is \( \ln|x| + C \), because \( \ln x \) is the classic counterexample for the power rule when \( m = -1 \).

Convergence

Convergence in the context of integrals refers to the behavior of an integral as its limits extend to infinity. An integral converges when the area under the curve approaches a specific finite value. This is particularly relevant for improper integrals, which have infinite limits or discontinuities.
To determine the convergence of the integral \( \int_{1}^{\infty} \frac{1}{x^{n}} \, dx \), we use the result of its antiderivative to examine its behavior as \( b \rightarrow \infty \). If the expression tends toward a finite number, the integral converges. For our specific integral, \( \lim_{b \to \infty} \left( \frac{b^{1-n}}{1-n} - \frac{1^{1-n}}{1-n} \right) \) determines convergence.
If \( n > 1 \), the integral converges because \( b^{1-n} \rightarrow 0 \) as \( b \rightarrow \infty \). This means that, even as our upper limit stretches infinitely far, the "accumulated area" underneath remains finite.

Divergence

Divergence is essentially the opposite of convergence when it comes to integral calculus. An integral diverges when it does not settle to a specific number as its limits approach infinity. Instead, the area under the curve expands indefinitely.
For the integral \( \int_{1}^{\infty} \frac{1}{x^{n}} \, dx \), we're particularly interested in the values of \( n \) for which it diverges. According to our analysis, if \( n \leq 1 \), the integral diverges. This is because \( b^{1-n} \rightarrow \infty \) or remains constant if \( 1-n = 0 \), causing the area to expand without bound as \( b \rightarrow \infty \).
This understanding is crucial in many applications of calculus because it tells us when a function’s influence is sufficiently diminished across its domain limits.

Integral Test

The integral test is a handy tool for determining the convergence or divergence of an infinite series. It compares a given function's behavior to a series and uses the result of an integral to make conclusions about that series.
The integral test states that if \( f(x) \) is continuous, positive, and decreasing for \( x \geq 1 \), then the series \( \sum_{n=1}^{\infty} f(n) \) and the integral \( \int_{1}^{\infty} f(x) \, dx \) either both converge or both diverge.
In our exercise, we used the integral test to find that the improper integral \( \int_{1}^{\infty} \frac{1}{x^{n}} \, dx \) converges only if \( n > 1 \). This usage matches the criteria of the test because \( \frac{1}{x^n} \) is continuous, positive, and decreasing for \( x \geq 1 \). The test requires computation of the antiderivative to reach conclusions about convergence and divergence based on the behavior of the function at the limits of integration.