Question

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

Short answer

The integral converges to \( 2\pi \).

Step-by-step solution

Define the Integral

The integral given is \( \int_{-\infty}^{\infty} \operatorname{sech} x \, dx \). This is an improper integral as it spans from \(-\infty\) to \(\infty\). We need to evaluate if it converges or diverges.

Split the Integral into Two Limits

We split the integral into two parts: \( \int_{-\infty}^{ ext{a}} \operatorname{sech} x \, dx \) and \( \int_{ ext{a}}^{ ext{b}} \operatorname{sech} x \, dx \), where \( a \to -\infty \) and \( b \to \infty \). This helps in evaluating each part separately.

Evaluate Each Part Using Limits

First, solve \( \lim_{a \to -\infty} \int_{a}^{0} \operatorname{sech} x \, dx \). Similarly, solve \( \lim_{b \to \infty} \int_{0}^{b} \operatorname{sech} x \, dx \). To do this, note that the integral of \( \operatorname{sech} x \) is \( 2 \arctan(e^x) + C \).

Combine and Evaluate Limits

For \( \int_{a}^{0} \operatorname{sech} x \, dx \), the antiderivative evaluated from \( a \) to 0 gives \( \lim_{a \to -\infty} [2\arctan(e^0) - 2\arctan(e^a)] = \pi - 0 = \pi \). Similar evaluation for \( \int_{0}^{b} \operatorname{sech} x \, dx \) gives \( \pi \).

Add the Results

Add the results from both parts: \( \pi + \pi = 2\pi \). Thus, the total integral from \(-\infty\) to \(\infty\) is \( 2\pi \), which indicates that the integral converges.

Key concepts

Improper Convergence

Improper integrals involve integrals where one or both limits of integration are infinite, or the integrand becomes infinite within the limits of integration. They are called "improper" because they don't fit the standard definition of a definite integral due to these infinite bounds or singularities.

To understand whether an improper integral converges or diverges, we divide the integral into sections with defined limits. For example, consider the given integral \( \int_{-\infty}^{\infty} \operatorname{sech} x \, dx \). Since it spans from \(-\infty\) to \(\infty\), we split it into \( \int_{-\infty}^{0} \operatorname{sech} x \, dx \) and \( \int_{0}^{\infty} \operatorname{sech} x \, dx \).

This separation allows us to handle each section individually using limit processes. For the integral to converge, both parts should result in finite numbers when limits are evaluated. In our problem, each part equals \( \pi \) leading to a total, convergent value of \( 2\pi \).

Hyperbolic Functions

Hyperbolic functions, much like trigonometric functions, are based on hyperbolas rather than circles. They include functions like \( \sinh(x) \), \( \cosh(x) \), \( \tanh(x) \), and the hyperbolic secant \( \operatorname{sech}(x) \), which is used in this exercise. The \( \operatorname{sech}(x) \) function is defined as \( \operatorname{sech}(x) = \frac{2}{e^x + e^{-x}} \).

Hyperbolic functions possess properties that are somewhat analogous to trigonometric functions, making them useful in various calculus applications, including solving differential equations and evaluating integrals. In the case of \( \int_{-\infty}^{\infty} \operatorname{sech} x \, dx \), recognizing the hyperbolic nature of the \( \operatorname{sech}(x) \) helps in finding its antiderivative, which involves \( \arctan(e^x) \).

Understanding their properties can make it easier to solve integrals related to hyperbolic functions. They behave in smooth ways that often simplify the process of integration.

Antiderivative Calculation

To integrate a function, we often need to find its antiderivative. This is the reverse process of differentiation. For the function \( \operatorname{sech}(x) \), the antiderivative helps in evaluating our improper integral.

The antiderivative of \( \operatorname{sech}(x) \) is \( 2 \arctan(e^x) + C \), where \( C \) is the constant of integration. This means that if we differentiate \( 2 \arctan(e^x) \), we would arrive back at \( \operatorname{sech}(x) \).

Once the antiderivative is known, it can be applied to the improper integral by evaluating it over given limits, in this case from \(-a\) to \(0\) and \(0\) to \(b\), and then taking the limits \( a \to -\infty \) and \( b \to \infty \). For our integral, both limits return \( \pi \), which we add together to conclude that \( \int_{-\infty}^{\infty} \operatorname{sech} x \, dx = 2\pi \). This efficient use of antiderivatives makes evaluating improper integrals faster and more straightforward, especially with infinite limits.