Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each improper integral or show that it diverges. \(\int_{1}^{\infty} \operatorname{csch} x d x\)

Step-by-Step Solution

Answer

The integral converges.

Step 1: Understand the Integral

We are asked to determine the convergence or divergence of the integral \(\int_{1}^{\infty} \csch x \, dx\). The function \(\operatorname{csch} x = \frac{1}{\sinh x}\) is the hyperbolic cosecant function.

Step 2: Set Up for Evaluation

Start the evaluation by considering the definition of the improper integral: \[ \int_{1}^{\infty} \csch x \, dx = \lim_{b \to \infty} \int_{1}^{b} \csch x \, dx \]We can more easily address the convergence by rewriting \(\sinh x\): \(\sinh x = \frac{e^x - e^{-x}}{2}\), making \(\csch x = \frac{2}{e^x - e^{-x}}\).

Step 3: Simplify the Integral

We approximate \(\csch x\) for large \(x\). Notably, for large \(x\), \(\sinh x \approx \frac{1}{2} e^x\). Thus, \(\csch x \approx \frac{2}{e^x}\). This means our integral approximately behaves like \(\int_{1}^{\infty} \frac{2}{e^x} \, dx\).

Step 4: Determine Convergence/Divergence

Consider the integral \(\int_{1}^{\infty} \frac{2}{e^x} \, dx\), which is a known convergent integral since the exponential function grows faster than any polynomial. Hence, we expect \(\int_{1}^{\infty} \csch x \, dx\) to also converge. We can justify this by bounding \(\csch x\) with a function \(\frac{2}{e^x}\) since it behaves similarly for large \(x\).

Step 5: Conclude with Comparison Test

Apply the comparison test by comparing \(\csch x\) to \(\frac{2}{e^x}\). Since for large values of \(x\), \(0 < \csch x \leq \frac{2}{e^x}\), and since \(\int_{1}^{\infty} \frac{2}{e^x} \, dx\) converges, by the comparison test, \(\int_{1}^{\infty} \csch x \, dx\) also converges.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are similar to trigonometric functions, but they are based on hyperbolas rather than circles. These functions are important in calculus, especially when dealing with exponential growth and decay scenarios. One key hyperbolic function is \(\operatorname{csch} x\), the hyperbolic cosecant function.
It is defined as \(\operatorname{csch} x = \frac{1}{\sinh x}\), where \(\sinh x\) is the hyperbolic sine function. The hyperbolic sine function can be expressed by the formula:

• \(\sinh x = \frac{e^x - e^{-x}}{2}\)

The hyperbolic cosecant, therefore, becomes:

• \(\operatorname{csch} x = \frac{2}{e^x - e^{-x}}\)

Understanding these expressions helps when evaluating integrals involving hyperbolic functions, as their behavior especially for large or small values of \(x\) can simplify complex integral evaluation.
In calculus problems such as improper integrals, recognizing the hyperbolic function forms can simplify calculations and help establish convergence or divergence.

Integral Convergence

Integral convergence is a crucial concept in calculus, especially when evaluating improper integrals. An integral converges if it approaches a finite value as its upper limit extends towards infinity. To determine if \(\int_{1}^{\infty} \csch x \, dx\) converges, we analyze the behavior of the function involved.
For the function \(\csch x = \frac{2}{e^x - e^{-x}}\), as \(x\) becomes large, the expression can be approximated by ignoring the \(e^{-x}\) in the denominator because \(e^{-x}\) becomes negligible compared to \(e^x\).
This approximation simplifies our integral to one involving \(\frac{2}{e^x}\). An integral like \(\int_{1}^{\infty} \frac{2}{e^x} \, dx\) is known to converge because the exponential growth in the denominator dominates, ensuring the area under the curve between the function and the x-axis settles to a finite number.
This approach lays the groundwork for comparing our original integral with a known convergent one, helping to establish its convergence through direct observation.

Comparison Test

The comparison test is a method for determining the convergence of an integral by comparing it to another integral whose convergence is already known. In evaluating \(\int_{1}^{\infty} \operatorname{csch} x \ dx\), we use the comparison test by aligning it with \(\int_{1}^{\infty} \frac{2}{e^x} \, dx\).
The test follows a few steps:

• If \(0 \leq f(x) \leq g(x)\) for all \(x\) in the interval of interest, and \(\int_{a}^{\infty} g(x) \, dx\) converges, then \(\int_{a}^{\infty} f(x) \, dx\) also converges.
• Conversely, if \(\int_{a}^{\infty} f(x) \, dx\) diverges, \(\int_{a}^{\infty} g(x) \, dx\) likewise diverges if \(g(x) \leq f(x)\).

Applying this to our problem, we see that for large values of \(x\), \(0 < \operatorname{csch} x \leq \frac{2}{e^x}\). Knowing that \(\int_{1}^{\infty} \frac{2}{e^x} \, dx\) converges, we can comfortably conclude that \(\int_{1}^{\infty} \operatorname{csch} x \, dx\) also converges.
This method proves incredibly useful in simplifying complex integrals by leveraging the properties of other known functions.