7–58. Improper integrals Evaluate the following integrals or state that they diverge.
28. ∫ (from 1 to ∞) tan⁻¹(s)/(s² + 1) ds

Verified step by step guidance

1. Identify the integral as an improper integral because the upper limit of integration is infinity: \(\int_{1}^{\infty} \frac{\tan^{-1}(s)}{s^{2} + 1} \, ds\).
2. Rewrite the integral as a limit to handle the infinite upper bound: \(\lim_{t \to \infty} \int_{1}^{t} \frac{\tan^{-1}(s)}{s^{2} + 1} \, ds\).
3. Consider using substitution to simplify the integral. Notice that the derivative of \(\tan^{-1}(s)\) is \(\frac{1}{s^{2} + 1}\), which appears in the denominator. Let \(u = \tan^{-1}(s)\), then \(du = \frac{1}{s^{2} + 1} ds\).
4. Rewrite the integral in terms of \(u\): \(\int \frac{\tan^{-1}(s)}{s^{2} + 1} ds = \int u \, du\).
5. Integrate \(\int u \, du\) to get \(\frac{u^{2}}{2} + C\), then substitute back \(u = \tan^{-1}(s)\), and finally evaluate the limit \(\lim_{t \to \infty} \left[ \frac{(\tan^{-1}(t))^{2}}{2} - \frac{(\tan^{-1}(1))^{2}}{2} \right]\) to determine convergence or divergence.

Key Concepts

Improper Integrals

Improper integrals involve integration over an infinite interval or integrands with infinite discontinuities. To evaluate them, we use limits to define the integral as a limit of definite integrals over finite intervals. Determining convergence or divergence is essential before finding the value.

Inverse Trigonometric Functions

Inverse trigonometric functions, like arctangent (tan⁻¹), are the inverses of trigonometric functions and often appear in integrals. Understanding their properties and derivatives helps in simplifying integrals and applying substitution or integration techniques.

Comparison Test for Convergence

The comparison test helps determine if an improper integral converges by comparing it to a known convergent or divergent integral. If the integrand is smaller than a convergent integral's integrand for large values, the integral converges; if larger than a divergent one, it diverges.