§6.5 Improper integrals

Introduction

Improper Integrals Involving Unbounded Intervals

Convergence and Divergence

Using the First FTC, $\int_1^b\frac{1}{x}dx = \ln(b)$. As $b \to \infty$, this shows that $\int_1^\infty\frac{1}{x}dx$

For $\int_1^\infty\frac{1}{x^p}dx$ with $p \gt 0$, the integral converges exactly when

Comparing $y = \frac{1}{x}$ and $y = \frac{1}{x^{3/2}}$ for $x \ge 1$, both approach the $x$-axis, yet

Evaluate $\int_1^\infty\frac{1}{x^2}\,dx$ by computing $\lim_{b\to\infty}\left(-\frac{1}{b}+1\right)$.

Using the antiderivative $-4xe^{-x/4}-16e^{-x/4}$, evaluate $\int_0^\infty xe^{-x/4}\,dx$.

An integrand that approaches $0$ as $x \to \infty$ can still have a divergent improper integral, as $\frac{1}{x}$ shows.